亚洲人成网7777777国产,亚洲欧洲精品无码AV,亚洲AV无码1区2区久久

Java认证之精屁�ȝ��

Fri, 16 Oct 2009 03:50:00 GMT

3.switch里面只能是short,int,char,byte�?
4.if (a="a") �~�译�? if(a=false)�~�译ok,原因是a=false的a代表了个boolean�?br /> 6.�q�种写法Outer.Inner i = new Outer().new Inner(); OK!
7.文�g�? 8 默认是int, 8.0默认是double
8.八进制是 0开�?不是O(英文o)
9.byte -128~127 ,-128是二�q�制的多��?br /> 11. -1>>32�q�是-1, -1>>>32��Z��么还�?1
12. char c='c'; String s ="s"; s+=c;�l�果�?sc!!!
13. boolean b1=true; boolean b2=true; System.out.println(b1|b2); �l�果是true.�~�译没错!
java的默认import包有哪些
AWT
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1. 所有组件flowLayout会被压羃为它们的最��尺�?br /> 2. BorderLayout的North,South,West,East,Center��是�q�个样子,四个单词大小写区�?必须�q�么�?否则�?br /> 2.1如果只是add,而不写东南西北中,默认是中
2.2如果不加component��C��,那中��q��着
2.3如果某个方位有个component,如果你还往上面加component,后来的会覆盖原来�?br /> 3. applet,panel默认布局是flowlayout
4. frame,dialog默认布局是borderlayout
5. 各种awe�cȝ��l�承关系,frame,window,component,...
6. window,frame,dialog不能被嵌入到容器�?注意:window!
7. action event作用于button和textfeild的回车时�?br /> 8. item event作用于list,choice,checkbox的选择改变时刻
9. 如果容器不可�?新layout��理器后前的component��跟随这个新的layout变化,如果容器可见,�q�些component��不受后来的layout影响
10.gridLayout里的component��寸一�?br /> 11.gridBagLayout�?component可以占用多个grid
12.System.exit();在Applet里面不允许调�?
AWT事�g
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.事�g�c�L��
(symantic)语义事�g:ActionEvent,AdjustEvent,ItemEvent,TextEvent
低��事�g:ComponentEvent,ContainerEvent,FocusEvent,InputEvent,KeyEvent,MouseEvent,PaintEvent,WindowEvent
2.监听�?
ActionListener,AdjustListener,CompentListener,ContainerListener,FocusListener,ItemListener,
KeyListener,MouseListener,MouseMotionListener,TextListener,WindwosListener, �?1个Listener,
七个adpter,��的4个是ActionLisenter,AdjustListener,ItemListener,TextListener,它们只有一个方�?
3,鼠标MouseListener�?个方�?clicked,pressed,released,entered,exited
4.鼠标MouseMotionListener�?个方�?mouseDragged,mouseMoved
�c�d��对象(Class and Object)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
0.最外层的类可以声明成final: "$file a":< final class a{}> ok!,但是不能是private和static�?
1.overload是同�c�里同样的方法名,override是父子的�l�承
2.override的返回结果必��M��?否则�~�译错哦
The return type of an overriding method is identical to the return type of the method it overrides.
2.1 override的modifier可以扩大,但是不能�~�小.比如父类private void test(){} 子类:public void test(){} ,没问�?如果反了,��死��翘�?
3.super.super(),�?错的,没这么写�?br /> 4.static和非static之间永远无法override!
5. 看程�?br /> public class A{
void test1() throws BaseEx{hi();}
void hi(){System.out.println("say hi,a");}
}
class AA extends A{
void hi(){System.out.println("say hi,aa");}
}
class test{
static void main(String b[]) throws Exception{
A a = new AA();
a.test1();
}
}
�l�果�?"say hi,aa",�q�说明什么说�?�Ҏ��永远跟着�cȝ��原来面目�?�?变量恰恰相反!
6.一个非抽象�Ҏ��L��也别想override成一个抽象方�?br /> 7.override的子�cȝ��Ҏ��抛出的的异常只能是父�c�L��法抛出异常的子异常类,或者无!
8.构造器不能是native,final,static,synchronized�?可以是public,private,什么都没有�?呵呵
9.构造器函数里还可以写return�?但后面什么都不许�?甚至null(�q�不是废话吗,哈哈)
10.构造器不能�q�回�?�q�大安��知道,但如果有�?构造器"反��g��,别紧�?它就不是构造器�?只是个普通函�?br /> 11.super();this();�q�两个函数只能在构造函数里调用.
12,成员变量声明时候赋�?比构造函数还�?int i=1; ealier than Test(){}
13.�Ҏ��的参数变量可以是final.
14. hashCode�q�回一个int
15. void wait() throws InterruptException wait扔InterruptException异常
16. java.lang.Void 是void的包装类
17. Byte,Interger,Double...所有的跟数有关的包装类都是�l�承于Number

接口Interface)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.接口的所有方法默认都是public,abstract和non-static�?br /> 2.接口的所有变量默认都是public,static,final�?所�?接口的变量不能改�?在它的实现类�?
3.接口的实现类实现的方法必��d��接口的方法抛出同��L��异常,不许是子�c?�q�和override不一�?同样,如果接口�Ҏ��没有�?实现�Ҏ��也不能抛.
4.实现�c�d��现的�Ҏ��必须昑ּ�的声明成public,什么都不写都不�?�?!!
5.接口不能声明成final,要不它怎么被实现呢(�q�好像又是废话啊
6.一个类实现两个接口,如果两个接口有相同的�Ҏ��,实现�c�d��实现�q�个�Ҏ��,没问题的.

内嵌�c�Inner Class)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.内嵌�c�d��以访问outer�cȝ��M��变量,包括�U�有�?
2.静态inner�c?只能讉K��outer�cȝ��M��static变量
2.1内嵌�c�d��以是final,abstract�?br /> 3.我靠,�Ҏ��内的内嵌�c�M��能�ؓstatic: void test(){ static class A{}} XXXXX!!!!
4.我靠,�Ҏ��内的内嵌�c�M��不能带�Q何modifier,void test(){ public class A{}} XXXXX!!!!
5.我靠,�Ҏ��内的内嵌�c�d��能访问方法内的final变量,但是,可以讉K��outer�cȝ��M��变量.
6.匿名�c�M��能有构造器,但声明时候带参数,相当于构造器的参��C��?
class ABC{}
class ABCD{private ABCD(int i){}}
ABC test3(){return new ABC(){};}
ABCD test4(){return new ABCD(3){};}
interface iii{}
iii test5(){return new iii(){};}
//class BCD extends ABCD{} compile error,因�ؓ,
看上面就知道,new iii(){};实际上匿名类实现了iii接口;new ABC(){};实际上是匿名�cȝ��承了ABC.
8.
class A {private A(){System.out.println("a!");}}
class B extends A{}
我靠,没错!B实例的时候会��d��调用父类A的构�?即��是private�?看来也没问题!!!
9.内部�c�d��以有synchronized�Ҏ��,那么锁是�q�个内部�c?跟外部类没一点关�p?内外分别�?在锁的问题上.
10.外部�c�M��能通过this被访�?this�q�时候应该指的是内部�c?享用外部�cȝ��成员��q��接用,不用加�Q何限定词
11.如何用this呢请�?
class Outer{ int i;
class Inner{
class InnerInner{
void Test(){
Outer.this.i=1;
}
}
}
}
看见了吧,�c�d��.this.变量�?可以引用到i,�W�一�ơ看到吧,嘿嘿,孤陋寡闻.
12.注意�q�两�U�写法都可以
Class Outer.Inner i = new Outer().new Inner();
或�? Class o= new Outer(); Class Outer.Inner i=o.new Inner();

�U�程Thread)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
�ȝ��Thread的API!!!!
1.�U�程启动必须通过start函数.
2.run函数不许也只能是public�?
3.�U�程有优先��?�?0,通过Thread.setPriority(int);来改�?不能��过10,否则会出现运行异�?br /> 4.�U�程优先�U�默认是5,即NORM_PRIORITY.NORM_PRIORITY是Thread的静态变量吗
5.Thread.yeild();是静态方�?所�?使用格式是Thread.yield();她强�q�当前的�q�程攑ּ�CUP.
6.sleep(1000),是说�U�程睡觉1�U?然后,�q�入Ready状�?注意,不是�q�行状�?它还要等OS来调度来获得CUP.

java.lang.*;
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.数组的科�? int ia[][]={{1,2},null}; int ib[][]=(int[][])ia.clone();
2.什么是NaN然后看ceil(NaN),floor(NaN),...
3.Math.floor(-1.1f);//-2.0
Math.ceil(-1.1f);//-1.0
Math.round(-1.6d)//-2
4.0=<1
5.Math,Interger,Boolean...�{�类型包装类都是final�?不可�l�承
6.int round(float); long round(double);�?round永远�q�回不了��数�?br /> 7.static double ceil(double)
8.static double floor(double)注意�Q�ceil,floor的只有这个double版本,什么都转成double!
9.static double sin(double 弧度); �q�有cos,tan
10. new String; 可以是byte[];char[];String;StringBuffer
11. String的一些函�? int length(); char charAt(int); String toUpperCase(); String toLowerCase();
12. String("Abc").equals(String("abc"))不相�{�的,不然��׃��会有boolean equalsIgnoreCase(String)函数
13."012345678"是一个串的顺序号�?indexOf('1'),indexOf("1")都返�?,subString(1,5)�?345,嘿嘿:�?[)"的感�?br /> 14, trim()�q�tab都退�?"\t\n java ",trim()一下就只剩�?java"�?br /> 15. 关于对象的科�?��d��喳API
16. "abcd".trim(),"abcd" + new String("ef")都是合理的写�?br /> 17. StringBuffer�?个构造器: ()初始化容量�ؓ16,(int 初始化容�?,(String),初始化容量�ؓ串长加上16
18. StringBuffer的一些函�? String toString(); append();reverse();insert();delete(int start,int end);deleteCharAt(int);setLength(int newLength);
19. String s=""; StringBuffer sb=new StringBuffer(); if (s==sb){}�~�译�?因�ؓ,s,sb�c�d��不一�?不能比较
集合:
1.各接口和�cȝ��关系,只有最后一个是�c?br /> Collection:List:vector,ArrayList,LinkedList
Map:SortedMap:TreeMap
Collection:Set:SortedSet:TreeSet
Map:HashTable
Collection:Set:HashSet

基础Base)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.java application的main可以不是public�?但必��L��static�?br /> 2.一个文件只能有一个public�c?而且�q�得跟文件名一�?包括大小�?br /> 3.变量只能是字�?$,_开�?接下来的�W�二个可以是,数字
4.ch\u0061r='a'; char \u0063='b'; char c='\u0063';都是合法�?br /> 5.1e-5d,合法.e-5d不合�?必须有前面的�p�L��
6.int[] i[]={null{1,2}}正确! int i[]={1,2,3,} 正确!","和没有的效果一�?br /> 7.局部array,跟变量一�?使用前要初始�?br /> 8.main�Ҏ��可以为final

操作�W�和分配�W?Operator and Assignment)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.System.out.printl(1+2+"3");//33 System.out.printl("1"+2+3);//123
2.int i=0; i=i++;i=i++;i=i++; �l�果,i=0!
3.int i[]; index=0; i[index]=index=5; �l�果�?i[0]=5;!!!
4.byte b=10;可以,因�ؓ,10可以被自动从int转成byte
5.接下�?byte b=b+10;不可�?!因�ؓ,10+b后的int无法被自动从int转成byte,问我,我也不知道�ؓ什�?�?
6.byte b1 = 4; byte b2 = 6; b1 = b1 + b2;�?�~�译�?b1 + b2�?必须强制转成byte,b1x1+b2);
7.XOR 一��L��?�Q�不一样�ؓ1 1,1=0;0,0=0;1,0/0,1=1
8. x == Float.NaN�~�译错，应该是Float.IsNaN
9. x == Double.POSITIVE_INFINITY�~�译可以
10.-1�?111.... 1111,<<永远双��Ӟ��>>正补�Ӟ��负补一�Q?gt;>>恒补�?br /> 10.1 -1>>多少位都�?1 ; 1<<31变成了最��负�?1000....0000
11.最大正数是01111....1111
12.最��负数是1000....0000(-2147483648)
13. a instanceof b,b必须是类/接口�Q�不能是实例
--------补充------------------------------
1. byte,short,char 都存�?var = -var;�~�译错误,因�ؓ,�q�时候的-var已经自动转成个int�c�d��?br /> 2. int/0会扔出ArithmeticException
double,float/0得INF或�?INF
0/0得NaN
3. int a-b-c;是不�W�合命名规则的变量名�~�译会出�?
4. char a='\u0001';�? char b=\u0001;�~�译�?
5. boolean b1,b2,b3,b4,b5;
b1 = b2==b3;
b1 = b2<=b3 && b4==b5;
b1 = b2==b3==true
都是对的!�?变�?
b1 = b2==b3==b4 XXXXXXX�~�译�?
6. 1>>1 �?
7. %= <<= =>> =>>>都是合法�W�号
8. --1-10*4 �q�种写法没错,��是 (--1)-10*4
9. k=1;++k + k++ + +k ;�l�果�?,相当�?(++2)+(2++)+(+3)
10.标号不能标示声明.
hi:
if {
break hi;
//break hi1;不行,不能向后调�{
}
//hi1:不行,不能攑֜�声明�?br /> int i;
hi1:
i=1;
11.public static void main(String s[]) throws Exception{}可以�?main可以扔出异常
12. hi:
if(b==true){break hi;}
break 标号�Q�可以用在if里面.别的��M��都不�?包括break,continue 标号.
13.int x = i*-j; 我靠,没问题的!!!�~�译没错! int x = i*j++ + +i++; �q�个也没问题,

变量修饰�W?Modifier)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.最外面一层的�c�M��能声明成protect和pravite�?br /> 2.同一个类的两个实例之间可以访问彼此的�U�有�Ҏ��和私有变�?�?br /> 3.protect和无修饰�W�的具体区别关于外包可以讉K��被子�c�访问的是哪�?br /> 4.成员变量被修饰成final�?必须声明时候就赋初值或者在构造器里面赋初�?别指望她可以得到default�?
5.抽象�Ҏ��不能是static�?!!
6.静态方法将随着�cȝ��变化而变�?看例�?
class Parent{
static void test(){System.out.println("hi,parent")};
}
class Child extends Parent{
static void test(){System.out.println("hi,child")};
}
Parent p = new Child();
p.test(); //打出来的是hi,parent!
7.静态方法可以通过�cȝ��实例调用.
new Child().test(); �?Child.test(); 都OK!
8.transient只能用在�cȝ��成员变量�?不能用在�Ҏ��?
9.transient变量不能是final和static�?br /> 10.native�Ҏ��可以是private,abstractd�?/p>

��程控制
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1。不可到辄��语句声明为错�Q�while(false){} ;for(;false{};if(false){}都无法编�?br /> 2。for(�W�一部分;的第一部分可以用来声明或者赋��|��但不能两者都
3。byte b; switch { case 200: // 200 not in range of byte�Q�因�?00��过b的范��_��编译错�?br /> 4。带标签的continue回达到标�{��位置�Q�从新进入紧接在标签后面的��@�?br /> 5。带标签的break会中断当前��@环，�q��{�U�d��标签标示的的循环的末��?/p>

转型和上�?Converting and Casting)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Binary operators convert the smaller(less precise) operand to the type of the larger(more precise) operand.
All operators will convert things smaller than ints to ints or larger. This includes char 's!
1.byte,char,short默认转成int
2.byte->short->int->long->float->double
char^
�q�是默认可以转型的方�?反了必须要显式cast! 特别注意:long->float是默认的,别看long64,float32,嘿嘿
�q�有��是看得出来,char�?byte,short之间无法互相默认转换
3.float f=1/3; OK!float f=1.0/3.0;�~�译出错,因�ؓ1.0/3.0�l�果是double�?噢噢~,错喽!!
4.int i=1; byte b=i;�?需要显式cast.
final i=1; byte b=i;��ok! 我也不知道�ؓ什�?final��可�?而且,据我实验只有int和byte的关�p�这�?其他不行.
5.int i[]; Object[] obj=i;�? Object obj=i;�? 数组只能转成Object,而不能是Object[]
6.int i[]; Object[] obj;i=(int[])obj; �? 对象可以通过昑ּ�来�{成一个数�l?

I/O
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.File�cȝ��一些重要方�?isFile();isDirectory();String[] list();exists();getAbsolutePath();getParent();
2.通过delete();mkdir();rename(File newname);可以操纵文�g,但是却改变不了文件的内容
2.1 File�c�L��法改变当前目�?除非重新创徏一个File对象.
3.InputStreamReader(InputStream in,String encodingName);
OutputStreamReader(OutputStream in,String encodingName);
Encoding: 8859_1是Latin-1,包含ASCII
4.关闭close一个流,��p��动调用了flush.
5.System.in,System.out,System.err,由JVM自动创徏
6.RandomAccessFile(File file,String mode);mode�?r,rw

李云�?/a> 2009-10-16 11:50 发表评论

�W�九部分�Q�集合类框架

Fri, 16 Oct 2009 03:44:00 GMT

�W�九部分�Q�集合类框架

知道如何在特定的条�g下选择适合的集合类/接口�?
能正��地实现hashcode�Ҏ��?

�W�一�?nbsp; 所有接口和class定义

1:Set

public interface Set extends Collection

A collection that contains no duplicate elements. More formally, sets contain no pair of elements e1 and e2 such that e1.equals(e2), and at most one null element. As implied by its name, this interface models the mathematical set abstraction.

2:SortedSet

public interface SortedSet extends Set.

A set that further guarantees that its iterator will traverse the set in ascending element order, sorted according to the natural ordering of its elements (see Comparable), or by a Comparator provided at sorted set creation time.

3:HashSet

public class HashSet extends AbstractSet implements Set, Cloneable, Serializable

This class implements the Set interface, backed by a hash table (actually a HashMap instance). It makes no guarantees as to the iteration order of the set; in particular, it does not guarantee that the order will remain constant over time. This class permits the null element.

4:TreeSet

public class TreeSet extends AbstractSet implements SortedSet, Cloneable, Serializable

This class implements the Set interface, backed by a TreeMap instance. This class guarantees that the sorted set will be in ascending element order, sorted according to the natural order of the elements (see Comparable), or by the comparator provided at set creation time, depending on which constructor is used.

5:LinkedHashSet:

public class LinkedHashSet extends HashSet implements Set, Cloneable, Serializable

Hash table and linked list implementation of the Set interface, with predictable iteration order. This implementation differs from HashSet in that it maintains a doubly-linked list running through all of its entries. This linked list defines the iteration ordering, which is the order in which elements were inserted into the set (insertion-order). Note that insertion order is not affected if an element is re-inserted into the set. (An element e is reinserted into a set s if s.add(e) is invoked when s.contains(e) would return true immediately prior to the invocation.)

5:Map

public interface Map

An object that maps keys to values. A map cannot contain duplicate keys; each key can map to at most one value.

6:SortedMap

public interface SortedMap extends Map

A map that further guarantees that it will be in ascending key order, sorted according to the natural ordering of its keys (see the Comparable interface), or by a comparator provided at sorted map creation time. (This interface is the map analogue of the SortedSet interface.)

7:HashMap(允许key和value为NULL)

public class HashMap extends AbstractMap implements Map, Cloneable, Serializable

Hash table based implementation of the Map interface. This implementation provides all of the optional map operations, and permits null values and the null key. (The HashMap class is roughly equivalent to Hashtable, except that it is unsynchronized and permits nulls.) This class makes no guarantees as to the order of the map; in particular, it does not guarantee that the order will remain constant over time

8:TreeMap

public class TreeMap extends AbstractMap implements SortedMap, Cloneable, Serializable

Red-Black tree based implementation of the SortedMap interface. This class guarantees that the map will be in ascending key order, sorted according to the natural order for the key's class (see Comparable), or by the comparator provided at creation time, depending on which constructor is used.

9:HashTable(不允许key和value为NULL)

public class Hashtable extends Dictionary implements Map, Cloneable, Serializable.

This class implements a hashtable, which maps keys to values. Any non-null object can be used as a key or as a value.

9:LinkedHashMap

public class LinkedHashMap extends HashMap

Hash table and linked list implementation of the Map interface, with predictable iteration order. This implementation differs from HashMap in that it maintains a doubly-linked list running through all of its entries. This linked list defines the iteration ordering, which is normally the order in which keys were inserted into the map (insertion-order). Note that insertion order is not affected if a key is re-inserted into the map. (A key k is reinserted into a map m if m.put(k, v) is invoked when m.containsKey(k) would return true immediately prior to the invocation.)

9:IdentityHashMap

public class IdentityHashMap extends AbstractMap implements Map, Serializable, Cloneable

This class implements the Map interface with a hash table, using reference-equality in place of object-equality when comparing keys (and values). In other words, in an IdentityHashMap, two keys k1 and k2 are considered equal if and only if (k1==k2). (In normal Map implementations (like HashMap) two keys k1 and k2 are considered equal if and only if (k1==null ? k2==null : k1.equals(k2)).)

This class is not a general-purpose Map implementation! While this class implements the Map interface, it intentionally violates Map's general contract, which mandates the use of the equals method when comparing objects. This class is designed for use only in the rare cases wherein reference-equality semantics are required.

10:List

public interface List extends Collection

An ordered collection (also known as a sequence). The user of this interface has precise control over where in the list each element is inserted. The user can access elements by their integer index (position in the list), and search for elements in the list.

Unlike sets, lists typically allow duplicate elements. More formally, lists typically allow pairs of elements e1 and e2 such that e1.equals(e2), and they typically allow multiple null elements if they allow null elements at all.

11:ArrayList

public class ArrayList extends AbstractList implements List, Cloneable, Serializable

Resizable-array implementation of the List interface. Implements all optional list operations, and permits all elements, including null. In addition to implementing the List interface, this class provides methods to manipulate the size of the array that is used internally to store the list. (This class is roughly equivalent to Vector, except that it is unsynchronized.)

12:LinkedList

public class LinkedList extends AbstractSequentialList implements List, Cloneable, Serializable

Linked list implementation of the List interface. Implements all optional list operations, and permits all elements (including null). In addition to implementing the List interface, the LinkedList class provides uniformly named methods to get, remove and insert an element at the beginning and end of the list. These operations allow linked lists to be used as a stack, queue, or double-ended queue (deque).

13:Vector

public class Vector extends AbstractList implements List, Cloneable, Serializable

The Vector class implements a growable array of objects. Like an array, it contains components that can be accessed using an integer index. However, the size of a Vector can grow or shrink as needed to accommodate adding and removing items after the Vector has been created.

14:Collection

public interface Collection

The root interface in the collection hierarchy. A collection represents a group of objects, known as its elements. Some collections allow duplicate elements and others do not. Some are ordered and others unordered. The SDK does not provide any direct implementations of this interface: it provides implementations of more specific subinterfaces like Set and List. This interface is typically used to pass collections around and manipulate them where maximum generality is desired.

15:Collections

public class Collections extends Object

This class consists exclusively of static methods that operate on or return collections. It contains polymorphic algorithms that operate on collections, "wrappers", which return a new collection backed by a specified collection, and a few other odds and ends.

李云�?/a> 2009-10-16 11:44 发表评论

Fri, 16 Oct 2009 03:43:00 GMT

�W�七部分�Q�线�E?/p>

能用java.lang,Thread或java.lang.Runnable两种�Ҏ��定义�Q�实例化和开始一个新的线�E��?
知道哪些情况下可能阻止一个线�E�的执行�?
能��用synchronized,wait,notify和notifyAll去避免�ƈ发访问的问题�Q�以及线�E�间�怺�通讯的问题�?
当执行synchronized,wait,notify和notifyAll�Ӟ��知道�U�程和对象锁之间的交互作用�?

§1.1.1 一

What will happen when you attempt to compile and run the following code?

class MyThread extends Thread

{

public void run()

{

System.out.println("MyThread: run()");

}

public void start()

{

System.out.println("MyThread: start()");

}

class MyRunnable implements Runnable

{

public void run()

{

System.out.println("MyRunnable: run()");

}

public void start()

{

System.out.println("MyRunnable: start()");

}

public class MyTest

{

public static void main(String args[])

{

MyThread myThread = new MyThread();

MyRunnable myRunnable = new MyRunnable();

Thread thread = new Thread(myRunnable);

myThread.start();

thread.start();

}

Choices:

a. Prints : MyThread: start() followed by MyRunnable:run()

b. Prints : MyThread: run() followed by MyRunnable:start()

c. Prints : MyThread: start() followed by MyRunnable:start()

d. Prints : MyThread: run() followed by MyRunnable:run()

e. Compile time error

f. None of the above

――――――――――――――�?/p>

A is the correct choice. In the above code there is not any compilation error. Thus choice E is incorrect. Inside main() method, objects of MyThread and MyRunnable class are created followed by creation of Thread with object of MyRunnable class. Note that MyThread class extends Thread class and overrides the start() method of the Thread class. Thus on execution of "myThread.start()" statement, the start() method of the MyThread class will be executed and as a result "MyThread:start()" will be printed. Had the start() method not there in MyThread class, the start() method of the Thread class would be called which in turn would call the run() method of the MyThread class. On execution of "thread.start();", the start() method of the Thread class would be called which in turn will call the run() method of the class which is passed to Thread constructor (i.e. MyRunnable class). Thus "MyRunnable:run()" will be printed out. Thus choice A is correct.

§1.1.2 �?/a>

What will be the output on compiling/running the following code?

public class MyThread implements Runnable

{

String myString = "Yes ";

public void run()

{

this.myString = "No ";

}

public static void main(String[] args)

{

MyThread t = new MyThread();

new Thread(t).start();

for (int i=0; i < 10; i++)

System.out.print(t.myString);

}

Choices:

a. Compilation Error

b. Prints : Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes and so on.

c. Prints : No No No No No No No No No No and so on.

d. Prints : Yes No Yes No Yes No Yes No Yes No and so on.

e. The Output cannot be determined.

E is correct. Please note that there will not be any compilation error when the above code is compiled. Also note that calling start() method on a Thread doesn't start the Thread. It only makes a Thread ready to be called. Depending on the operation system and other running threads, the thread on which start is called will get executed. In the above case it is not guaranteed that the thread will be executed(i.e. run() method will be called), always before "for" loop is executed. Thus the output cannot be determined.

§1.1.3 �?/a>

Multiple objects of MyClass (given below) are used in a program that uses

multiple Threadsto create new integer count. What will happen when other threads

use the following code?

class MyClass

{

static private int myCount = 0;

int yourNumber;

private static synchronized int nextCount()

{

return ++myCount;

}

public void getYourNumber()

{

yourNumber = nextCount();

}

Choices:

a. The code will give compilation error.

b. The code will give runtime error.

c. Each thread will get a unique number.

d. The uniqueness of the number among different Threads can't be guaranteed.

――――――――――――�?/p>

C is correct. The use of synchronized ensures that the number generated will not be duplicated, no matter how many Threads are trying to create the number. Thus D is incorrect. A and B are incorrect as the above code will not give any compiletime or runtime error.

§1.1.4 �?/a>

What will happen when you attempt to compile and run the following code?

public class MyThread extends Thread

{

String myName;

MyThread(String name)

{

myName = name;

}

public void run()

{

for(int i=0; i<100;i++)

{

System.out.println(myName);

}

public static void main(String args[])

{

try

{

MyThread mt1 = new MyThread("mt1");

MyThread mt2 = new MyThread("mt2");

mt1.start();

// XXX

mt2.start();

}

catch(InterruptedException ex)

{

}

Choices:

a.The above code in its current condition will not compile.

b. In order to make the MyThread class prints "mt1" (100 times) followed by

"mt2" (100 times), mt1.join(); can be placed at //XXX position.

c. In order to make the MyThread class prints "mt1" (100 times) followed by

"mt2" (100 times), mt1.sleep(100); can be placed at //XXX position.

d. In order to make the MyThread class prints "mt1" (100 times) followed by

"mt2" (100 times), mt1.run(); can be placed at //XXX position.

e. In order to make the MyThread class prints "mt1" (100 times) followed by

"mt2" (100 times), there is no need to write any code.

―――――――――――――�?/p>

A and B are correct. In its current condition, the above code will not compile as "InterruptedException" is never thrown in the try block. The compiler will give following exception: "Exception java.lang.InterruptedException is never thrown in the body of the corresponding try statement." Note that calling start() method on a Thread doesn't start the Thread. It only makes a Thread ready to be called. Depending on the operating system and other running threads, the thread on which start is called will get executed. After making the above code to compile (by changing the InterruptedException to some other type like Exception), the output can't be predicted (the order in which mt1 and mt2 will be printed can't be guaranteed). In order to make the MyThread class prints "mt1" (100 times) followed by "mt2" (100 times), mt1.join() can be placed at //XXX position. The join() method waits for the Thread on which it is called to die. Thus on calling join() on mt1, it is assured that mt2 will not be executed before mt1 is completed. Also note that the join() method throws InterruptedException, which will cause the above program to compile successfully. Thus choice A and B are correct.

§1.1.5 �?/a>

What will happen when you attempt to compile and run the following code?

public class MyThread extends Thread{

String myName;

MyThread(String name){

myName = name;

}

public void run(){

for(int i=0; i<100;i++){

System.out.println(myName);

}

public static void main(String args[]){

try{

MyThread mt1 = new MyThread("mt1");

MyThread mt2 = new MyThread("mt2");

mt1.start();

// XXX

mt2.start();

}catch(InterruptedException ex){}

}

A. compile error

B. mt1.join();

C. mt1.sleep(100);

D. mt1.run()

E. nothing need

Choice A and B are correct. In its current condition, the above code will not compile as "InterruptedException" is never thrown in the try block. The compiler will give following exception: "Exception java.lang.InterruptedException is never thrown in the body of the corresponding try statement."

Note that calling start() method on a Thread doesn't start the Thread. It only makes a Thread ready to be called. Depending on the operating system and other running threads, the thread on which start is called will get executed. After making the above code to compile (by changing the InterruptedException to some other type like Exception), the output can't be predicted (the order in which mt1 and mt2 will be printed can't be guaranteed). In order to make the MyThread class prints "mt1" (100 times) followed by "mt2" (100 times), mt1.join() can be placed at //XXX position. The join() method waits for the Thread on which it is called to die. Thus on calling join() on mt1, it is assured that mt2 will not be executed before mt1 is completed. Also note that the join() method throws InterruptedException, which will cause the above program to compile successfully. Thus choice A and B are correct.

§1.1.6 �?/a>

Multiple objects of MyClass (given below) are used in a program that uses multiple Threads to create new integer count. What will happen when other threads use the following code?

class MyClass{

static private int myCount = 0;

int yourNumber;

private static synchronized int nextCount(){

return ++myCount; //myCount为static

}

public void getYourNumber(){

yourNumber = nextCount();

}

A. the code ill give ompilation error

B. the code ill give runtime error

C. each thread will get a unique number

D. the uniqueness of the number different Threads can’t be guaranteed.

§1.1.7 �?/a>

What will be the output on compiling/running the following code?

public class MyThread implements Runnable {

String myString = "Yes ";

public void run() {

this.myString = "No ";

}

public static void main(String[] args) {

MyThread t = new MyThread();

new Thread(t).start();

for (int i=0; i < 10; i++)

System.out.print(t.myString);

}

A. compile error

B. prints: yes yes yes yes yes yes and so on

C. prints: no no no no no no no no and so on

D. prints: yes no yes no ye no ye no and so on

E. the output cannot be determinated

§1.1.8 �?/a>

Which statements about thread are true?

　　A. Once a thread is created, it can star running immediately.

　　B. To use the start() method makes a thread runnable, but it does not

necessarily start immediately.

　　C. When a thread stops running because of pre-emptive, it is placed at

the front end of the runnable queue.

　　D. A thread may cease to be ready for a variety of reasons.

　　(bd)

　　题目�Q�有关线�E�的哪些叙述是对的�?/p>

　　A. 一旦一个线�E�被创徏�Q�它��q��卛_��始运行�?/p>

　　B. 使用start()�Ҏ��可以使一个线�E�成为可�q�行的，但是它不一定立卛_��始运行�?/p>

　　C. 当一个线�E�因为抢先机制而停止运行，它被攑֜�可运行队列的前面�?/p>

　　D. 一个线�E�可能因��Z��同的原因停止�Q�cease�Q��ƈ�q�入��q�A状态�?/p>

　　一个新创徏的线�E��ƈ不是自动的开始运行的�Q�必��调用它的start()�Ҏ��使之��线�E�放�?/p>

可运行态（runnable

state�Q�，�q�只是意味着该线�E�可为JVM的线�E�调度程序调度而不是意味着它可以立卌��行�?/p>

�U�程的调度是抢先式的�Q�而不是分旉��片式的。具有比当前�q�行�U�程高优先��的线�E�可以��?/p>

前线�E�停止运行而进入就�l�状态，不同优先�U�的�U�程间是抢先式的�Q�而同�U�线�E�间是轮转式�?/p>

。一个线�E�停止运行可以是因�ؓ不同原因�Q�可能是因�ؓ更高优先�U�线�E�的抢占�Q�也可能是因�?/p>

调用sleep()�Ҏ��Q�而即使是因�ؓ抢先而停止也不一定就�q�入可运行队列的前面�Q�因为同�U�线

�E�是轮换式的�Q�它的运行可能就是因��换，而它因抢占而停止后只能在轮换队列中排队而不

能排在前面�?/p>

§1.1.9 �?/a>

Which two CANNOT directly cause a thread to stop executing? (Choose Two)

A.Existing from a synchronized block

B.Calling the wait method on an object

C.Calling notify method on an object

D.Calling read method on an InputStream object

E.Calling the SetPriority method on a Thread object

Answer�Q�AC。同55�?/p>

§1.1.10 �?/a>

public class SyncTest{

public static void main(String[] args) {

final StringBuffer s1= new StringBuffer();

final StringBuffer s2= new StringBuffer();

new Thread () {

public void run() {

synchronized(s1) {

s2.append(“A”);

synchronized(s2) {

s2.append(“B”);

System.out.print(s1);

System.out.print(s2);

}

}.start();

new Thread() {

public void run() {

synchronized(s2) {

s2.append(“C”);

synchronized(s1) {

s1.append(“D”);

System.out.print(s2);

System.out.print(s1);

}

}.start();

}

Which two statements are true? (Choose Two)

A. The program prints “ABBCAD”

B. The program prints “CDDACB”

C. The program prints “ADCBADBC”

D. The output is a non-deterministic point because of a possible deadlock condition

E. The output is dependent on the threading model of the system the program is running on.

Answer�Q�DE

§1.1.11 十一

What will happen when you attempt to compile and run the following code?

public class Test{

int i = 0;

public static void main(String argv[]) {

Test t = new Test();

t.myMethod();

}

public void myMethod(){

while(true) {

try {

wait();

}catch (InterruptedException e) {}

i++;

}

A. Compile time error, no matching notify within the method.

B. Compile and run but an infinite looping of the while method.

C. Compilation and run without any output.

E. Runtime Exception "IllegalMonitorStatException".

Answer: E

Note: The wait/notify protocol can only be used within code that is synchronized. In this case calling code does not have a lock on the object(not synchronized) and will thus cause an Exception at runtime.

§1.1.12 十二

1.10 What is the result of compiling and executing the following code?

public class ThreadTest extends Thread {

public void run() {

System.out.println("In run");

yield();

System.out.println("Leaving run");

}

public static void main(String args []) {

(new ThreadTest()).start();

}

A. The code fails to compile in the main() method.

B. The code fails to compile in the run() method.

C. Only the text "In run" will be displayed.

D. The text "In run" followed by "Leaving run" will be displayed.

E. The code compiles correctly, but nothing is displayed.

Answer: D

§1.1.13 十三

Which of the following will definitely stop a thread from executing?A. wait()B. notify()C. yield()D. suspend()E. sleep()Answer: ACDE

§1.1.14 十四

Which of the following will definitely stop a thread from executing?

A. wait()

B. notify()

C. yield()

D. suspend()

E. sleep()

Answer: ACDE

§1.1.15 十五

Which of the following statements about threading are true?

A. You can only obtain a mutually exclusive lock on methods in a class that extends Thread or implements runnable.

B. You can obtain a mutually exclusive lock on any object.

C. You can't obtain a mutually exclusive lock on any object.

D. Thread scheduling algorithms are platform dependent.

Answer: BD

Consider the following statement:

Thread myThread = new Thread();

Which of the following statements are true regarding myThread?

A. The thread myThread is now in a runnable state.

B. The thread myThread has a priority of 5.

C. On calling the start() method on myThread, the run method in the Thread class will be executed.

D. On calling the start() method on myThread, the run method in the calling class will be executed.

Answer: C

Note: the priority of myThread will be inherited from the Thread that called the constructor.

§1.1.16 十六

What is the effect of issuing a wait() method on an object�Q�（Mutiple�Q?/p>

1) If a notify() method has already been sent to that object then it has no effect

2) The object issuing the call to wait() will halt until another object sends a notify() or notifyAll() method

3) An exception will be raised

4) The object issuing the call to wait() will be automatically synchronized with any other objects using the receiving object.

ANSWER 1)

10:

Pick all the true statements below.

1) If a thread wants to call wait() on an object, the thread must own that object's lock.

2) There is a method that you can call on an instance of the Thread class that puts the instance to sleep for a specified number of milliseconds.

3) At the moment when a thread is notified, it automatically gets the lock of the object for which it was waiting.

ANSWER 1

李云�?/a> 2009-10-16 11:43 发表评论

Fri, 16 Oct 2009 03:43:00 GMT

�W�八部分�Q�在java.lang包中的基��c?/p>

能够应用Math�c�M��的abs,ceil,floor,max,min,random,round,sin,cos,tan,sqrt�Ҏ��?
正确理解String�c�M��可改变的意义�?
当执行一�D늨�序，中间包含有wrapper�cȝ��一个实例，知道它运行的前提条�g�q�行�l�果。能用下面wrapper�c�（例如Integer,Double,�{�等�Q�的�Ҏ��来写�E�序�Q?

doublevalue

floatvalue

intvalue

longvalue

parseXxx

getXxx

toString

toHexString

李云�?/a> 2009-10-16 11:43 发表评论

Fri, 16 Oct 2009 03:42:00 GMT

�W�六部分�Q�覆盖，重蝲�Q�运行时期类型及光��向对�?/p>

知道面向对象设计中封装的好处以及如何实现�Q�能知道is a和has a的意义�?
能正��用覆盖和重蝲的方法，能正��调用父�c�L��重蝲的构造方�?constructor)�Q�知道调用这些方法后的结果�?
能正��实例化�c�L��内部�c?

§1.1.1 一

3. What will happen when you attempt to compile and run the following code?

class MyParent

{

int x, y;

MyParent(int x, int y)

{

this.x = x;

this.y = y;

}

public int addMe(int x, int y)

{

return this.x + x + y + this.y;

}

public int addMe(MyParent myPar)

{

return addMe(myPar.x, myPar.y);

}

class MyChild extends MyParent

{

int z;

MyChild (int x, int y, int z)

{

super(x,y);

this.z = z;

}

public int addMe(int x, int y, int z)

{

return this.x + x + this.y + y + this.z + z;

}

public int addMe(MyChild myChi)

{

return addMe(myChi.x, myChi.y, myChi.z);

}

public int addMe(int x, int y)

{

return this.x + x + this.y + y;

}

public class MySomeOne

{

public static void main(String args[])

{

MyChild myChi = new MyChild(10, 20, 30);

MyParent myPar = new MyParent(10, 20);

int x = myChi.addMe(10, 20, 30);

int y = myChi.addMe(myChi);

int z = myPar.addMe(myPar);

System.out.println(x + y + z);

}

Choices:

a. 300

b. 240

c. 120

d. 180

e. Compilation error

f. None of the above

―――――――――――�?/p>

A is the correct choice. In the code, MyChild class overrides the addMe(int x, int y) method of the MyParent class. And in both the MyChild and MyParent class, addMe() method is overloaded. There is no compilation error anywhere in the above code. On execution, first, the object of MyChild class will be constructed. Please note that there is a super() call from the constructor of MyChild class, which will call the constructor of MyParent class. This will cause the value of z variable of MyChild class to be 30 and x, y variables of MyParent class will become 10 and 20 respectively. The next statement will again call the constructor of MyParent class with same x and y values. This is followed by execution of addMe() method of MyChild class with x as 10, y as 20 and z as 30. Also x and y are inherited by MyChild class from the MyParent class. Thus in the addMe() method of the MyChild class, the value of this.x will be 10, this.y will be 20 and this.z will be 30. The return value of this method will be "10 + 10 + 20 + 20 + 30 + 30", which is equal to 120. Thus x will become 120. This is followed by the invocation of the other addMe() method which takes object reference of the MyChild class. From this method, the method which was called earlier is invoked. This call is exactly the same as the earlier one. Thus the value of y will also be 120 like x. Now the addMe() method of MyParent class is invoked. This method invokes another addMe() method of the same class. Its equivalent to the invocation of addMe(int x, int y) method with x as 10 and y as 20. Also the value of instance variables x and y of My Parent class is 10 and 20 respectively. The value of z will be evaluated to "10 + 10 + 20 + 20", which is equal to 60. Thus the value of x, y and z after all the invocations will be 120, 120 and 60 respectively. As a result of this finally, "120 + 120 + 60" which is equal to 300 will be printed. Thus A is the correct choice.

§1.1.2 �?/a>

Given the code below, and making no other changes, which access modifiers

(public, protected or private) can legally be placed before myMethod() on line 3?

If line 3 is left as it is, which keywords can legally be placed before myMethod

on line 8?

1.class HumptyDumpty

2.{

3.void myMethod() {}

4.}

6.class HankyPanky extends HumptyDumpty

7.{

8.void myMethod() {}

9.}

Choices:

a. private or nothing(i.e. leaving it as it is) on line 3.

Nothing(i.e. leaving it as it is) or protected or public

on line 8.

b. public or protected on line 3. private or nothing(i.e. leaving it

as it is) on line 8.

c. nothing(i.e. leaving it as it is) or protected or public on

line 3. private or nothing(i.e. leaving it as it is) on line 8.

d. None of the above.

A is correct. The basic principle is that a method cannot be overridden to be more private. Since the method is being overridden to be friendly(default modifier) it can only be private or friendly in the superclass. Secondly if the method in superclass is left as it is(i.e. friendly access) the method in subclass can be friendly, protected or public.

§1.1.3 �?/a>

What results from the following code?

1.class MyClass

2.{

3.void myMethod(int i) {System.out.println("int version");}

4.void myMethod(String s) {System.out.println("String version");}

5.public static void main(String args[])

6.{

7.MyClass obj = new MyClass();

8.char ch = 'c';

9.obj.myMethod(ch);

10.}

11.}

Choices:

a. Line 4 will not compile as void methods can't be overridden.

b. An exception at line 9.

c. Line 9 will not compile as there is no version of myMethod which takes a char as argument.

d. The code compiles and produces output: int version.

e. The code compiles and produces output: String version.

―――――――――�?/p>

D is correct. A is incorrect as void methods can be overridden without any problem. B is incorrect as char ch declaration is valid. C is incorrect as char type in java is internally stored as integer and there is a method which takes int as an input. D is correct, on line 9 char ch is widened to an int and passed to int version of the myMethod(). E is incorrect as int version of myMethod() is called.

§1.1.4 �?/a>

What is displayed when the following is executed?

class Parent

{

private void method1()

{

System.out.println("Parent's method1()");

}

public void method2()

{

System.out.println("Parent's method2()");

method1();

}

class Child extends Parent

{

public void method1()

{

System.out.println("Child's method1()");

}

public static void main(String args[])

{

Parent p = new Child();

p.method2();

}

Choices:

a. Compile time error

b. Run time error

c. prints : Parent's method2()

Parent's method1()

d. prints : Parent's method2()

Child's method1()

――――――�?/p>

C is correct. The code will compile without any error and also will not give any run time error. The variable p refers to the Child class object. The statement p.method2() on execution will first look for method2() in Child class. Since there is no method2() in child class, the method2() of Parent class will be invoked and thus "Parent's method2()" will be printed. Now from the method2() , there is a call to method1(). Please note that method1() of Parent class is private, because of which the same method (method1() of Parent class) will be invoked. Had this method(method1() of Parent class) been public/protected/friendly (default), Child's class method1() would be called. Thus C is correct answer.

§1.1.5 �?/a>

What is displayed when the following is executed?

class Parent{

private void method1(){

System.out.println("Parent's method1()");

}

public void method2(){

System.out.println("Parent's method2()");

method1();

}

class Child extends Parent{

public void method1(){

System.out.println("Child's method1()");

}

public static void main(String args[]){

Parent p = new Child();

p.method2();

}

A. compile time error

B. run time error

C. prints: parent’s method2() parent’s method1()

D. prints: parent’s method2() child’s method1()

§1.1.6 �?/a>

　　1) class Person {

　　2) public void printValue(int i, int j) { }

　　3) public void printValue(int i){ }

　　4) }

　　5) public class Teacher extends Person {

　　6) public void printValue() { }

　　7) public void printValue(int i) {}

　　8) public static void main(String args[]){

　　9) Person t = new Teacher();

　　10) t.printValue(10);

　　11) }

　　12) }

　　Which method will the statement on line 10 call?

　　A. on line 2

　　B. on line 3

　　C. on line 6

　　D. on line 7

　　��译

　　�W�十行的声明��调用哪些方法�?/p>

　　�{�案　D

　　解析　变量t是一个Person对象�Q�但是它是用Teacher实例化的�Q�这个问题涉及到java�?/p>

�~�译时多态和�q�行时多态的问题�Q�就�~�译时多态来��_��t实际上是一个Person�c�，�q�涉及到�c?/p>

型的自动转换�Q�将一个子�cȝ��实例赋值给一个父�cȝ��变量是不用进行强制类型�{换，反之则需

要进行强制类型�{换，而且被赋值的变量实际上应该是一个子�cȝ��对象�Q�，如果对t调用了子

�c�M��新增的方法则造成�~�译旉��误编译将不能通过�Q�而在�q�行�Ӟ��q�行时系�l�将�Ҏ��t实际�?/p>

向的�c�d��调用对应的方法，对于本例来说�Q�t.print(10)��调用t实际指向的Teacher�cȝ��对应

�Ҏ��。在java中，可以用一个子�cȝ��实例实例化父�cȝ��一个变量，而变量在�~�译时是一个父�c?/p>

实例�Q�在�q�行时可能是一个子�c�d��例�?/p>

§1.1.7 �?/a>

35、public class Parent {

　　public int addValue( int a, int b) {

　　int s;

　　s = a+b;

　　return s;

　　}

　　class Child extends Parent {

　　}

　　Which methods can be added into class Child?

　　A. int addValue( int a, int b ){// do something...}

　　B. public void addValue (){// do something...}

　　C. public int addValue( int a ){// do something...}

　　D. public int addValue( int a, int b )throws MyException {//do

something...}

　　(bc)

　　题目�Q�哪些方法可以加入类Child中�?/p>

　　此题涉及�Ҏ��重蝲�Q�overload�Q�，�Ҏ��重写�Q�override�Q�以及类�z��时方法重写的规则

。方法重载的规则是：一、参数列表必��M��同，个数的不同完全可以，如果个数相同则参数类

型的不同不能引�v歧意�Q�例如int

和long,float和double��׃��能作为唯一的类型不同；二、返回值可以不同，但是不能是重�?/p>

时唯一的不同点�Q�这点和c++中不同，c++中返回类型必��M��_��。方法重写发生在�cȝ��承时�Q?/p>

子类可以重写一个父�c�M��已有的方法，必须在返回类型和参数列表一��h��才能说是重写�Q�否�?/p>

��是重蝲�Q�java中方法重写的一个重要而且�Ҏ��被忽略的规则是重写的�Ҏ��的访问权限不能比

被重写的�Ҏ��的访问权限低�Q�重写的另一个规则是重写的方法不能比被重写的�Ҏ��抛弃(thro

ws)更多�U�类的异常，其抛弃的异常只能��，或者是其子�c�，不能以抛弃异常的个数来判断种

�c�，而应该是异常�c�d��ơ结果上的种�c�R��此题中�{�案a的错误就是重写的讉K��权限比被重写�?/p>

�Ҏ��的低�Q�而b,c都属于重载，d的错误在于比被重写的�Ҏ��抛弃了更多种�cȝ��异常�?/p>

李云�?/a> 2009-10-16 11:42 发表评论

Fri, 16 Oct 2009 03:41:00 GMT

�W�五部分�Q�操作与赋�?/p>

能知道各�U�操作符�Q�包括赋值操作符和intanceof操作�W�）应用于不同类型的�l�果�?
知道String�Q�Boolean和类使用equals()�Ҏ��后的�l�果�?
知道当对已经知道值的变量�q�行&,|,&&,||操作�Ӟ��哪些操作数被�q�算了，表达式最�l�的�l�果是怎样的�?
知道对象和原始类型数据传入方法后�Q�在�Ҏ��内部�q�行赋值或其它修改操作的结果�?

§1.1.1 一

Which of the following lines will print false?

1.public class MyClass

2.{

3.static String s1 = "I am unique!";

4.public static void main(String args[])

5.{

6.String s2 = "I am unique!";

7.String s3 = new String(s1);

8.System.out.println(s1 == s2);

9.System.out.println(s1.equals(s2));

10.System.out.println(s3 == s1);

11.System.out.println(s3.equals(s1));

12.System.out.println(TestClass.s4 == s1);

13.}

14.}

15.

16.class TestClass

17.{

18.static String s4 = "I am unique!";

19.}

Choices:

a. Lines 10 and 12

b. Line 12 only

c. Line 8 and 10

d. None of these

―――――――――�?/p>

D is correct. Only line 10 will print false. Strings are immutable objects. That is, a string is read only once the string has been created and initialized, and Java optimizes handling of string literals; only one anonymous string object is shared by all string literals with the same contents. Hence in the above code the strings s1, s2 and s4 refer to the same anonymous string object, initialized with the character string: "I am unique!". Thus s1 == s2 and TestClass.s4 will both return true and obviously s1.equals(s2) will return true. But creating string objects using the constructor String(String s) creates a new string, hence s3 == s1 will return false even though s3.equals(s1) will return true because s1 and s3 are referring to two different string objects whose contents are same.

§1.1.2 �?/a>

What is displayed when the following code is compiled and executed?

String s1 = new String("Test");

String s2 = new String("Test");

if (s1==s2)

System.out.println("Same");

if (s1.equals(s2))

System.out.println("Equals");

Choices:

a. Same

Equals

b. Equals

c. Same

d. The code compiles, but nothing is displayed upon execution.

e. The code fails to compile.

B is correct. Here s1 and s2 are two different object references, referring to different objects in memory. Please note that operator == checks for the memory address of two object references being compared and not their value. The "equals()" method of String class compares the values of two Strings. Thus s1==s2 will return "false" while s1.equals(s2) will return "true". Thus only "Equals" will be printed.

§1.1.3 �?/h3>
Given the following code, what will be the output?

class Value

{

public int i = 15;

}

public class Test

{

public static void main(String argv[])

{

Test t = new Test();

t.first();

}

public void first()

{

int i = 5;

Value v = new Value();

v.i = 25;

second(v, i);

System.out.println(v.i);

}

public void second(Value v, int i)

{

i = 0;

v.i = 20;

Value val = new Value();

v = val;

System.out.println(v.i + " " + i);

}

}

Choices:

a. 15 0

20

b. 15 0

15

c. 20 0

20

d. 0 15

20

――――――――――――――�?/p>
A is correct. When we pass references in Java what actually gets passed is the value of that reference (i.e. memory address of the object being referenced and not the actual object referenced by that reference) and it gets passed as value (i.e a copy of the reference is made). Now when we make changes to the object referenced by that reference it reflects on that object even outside of the method being called but any changes made to the reference itself is not reflected on that reference outside of the method which is called. In the example above when the reference v is passed from method first() to second() the value of v is passed. When we assign the value val to v it is valid only inside the method second() and thus inside the method second() what gets printed is 15 (initial value of i in the object referenced by val), then a blank space and then 0 (value of local variable i). After this when we return to the method first() v actually refers to the same object to which it was referring before the method second() was called, but one thing should be noted here that the value of i in that object (referred by v inside the method first()) was changed to 20 in the method second() and this change does reflect even outside the method second(), hence 20 gets printed in the method first(). Thus overall output of the code in consideration is 15 0 20

§1.1.4 �?/h3>
What will happen when you attempt to compile and run the following code?

interface MyInterface

{

}

public class MyInstanceTest implements MyInterface

{

static String s;

public static void main(String args[])

{

MyInstanceTest t = new MyInstanceTest();

if(t instanceof MyInterface)

{

System.out.println("I am true interface");

}

else

{

System.out.println("I am false interface");

}

if(s instanceof String)

{

System.out.println("I am true String");

}

else

{

System.out.println("I am false String");

}

}

}

Choices:

a. Compiletime error

b. Runtime error

c. Prints : "I am true interface" followed by " I am true String"

d. Prints : "I am false interface" followed by " I am false String"

e. Prints : "I am true interface" followed by " I am false String"

f. Prints : "I am false interface" followed by " I am true String"

―――――――――――�?/p>
E is the correct choice. The "instanceof" operator tests the class of an object at runtime. It returns true if the class of the left-hand argument is the same as, or is some subclass of, the class specified by the right-hand operand. The right-hand operand may equally well be an interface. In such a case, the test determines if the object at left-hand argument implements the specified interface. In the above case there will not be any compiletime or runtime error. The result of "t instance of MyInterface" will be true as "t" is the object of MyInstanceTest class which implements the MyInstance interface. But the result of "s instanceof String" will be false as "s" refers to null. Thus the output of the above program will be : "I am true interface" followed by " I am false String". Thus choice E is correct and others are incorrect.

§1.1.5 �?/h3>
What will happen when you attempt to compile and run the following code snippet?

String str = "Java";

StringBuffer buffer = new StringBuffer(str);

if(str.equals(buffer)){

System.out.println("Both are equal");

}else{

System.out.println("Both are not equal");

}

A. it will print – both are not equal

B. it will print – both are equal

C. compile time error

D. Runtime error

A is the correct choice. The equals method overridden in String class returns true if and only if the argument is not null and is a String object that represents the same sequence of characters as this String object. Hence, though the contents of both str and buffer contain "Java", the str.equals(buffer) call results in false.

The equals method of Object class is of form -public boolean equals(Object anObject). Hence, comparing objects of different classes will never result in compile time or runtime error.

§1.1.6 �?/a>

10. Which of the following statements are true?

　　A. The equals() method determines if reference values refer to the same

object.

　　B. The == operator determines if the contents and type of two separate

objects match.

　　C. The equals() method returns true only when the contents of two

objects match.

　　D. The class File overrides equals() to return true if the contents and

type of two separate objects match.

　　��译

　　下面的哪些叙�q�Cؓ真�?/p>

　　A. equals()�Ҏ��判定引用值是否指向同一对象�?/p>

　　B. == 操作�W�判定两个分立的对象的内容和�c�d��是否一致�?/p>

　　C. equals()�Ҏ��只有在两个对象的内容一致时�q�回true�?/p>

　　D. �c�File重写�Ҏ��equals()在两个分立的对象的内容和�c�d��一致时�q�回true�?/p>

　　�{�案　A,D

　　解析　严格来说�q�个问题的答案是不确定的�Q�因为equals()�Ҏ��是可以被重蝲的，但是

按照java语言的本意来��_��如果没有重写�Q�override�Q�新�cȝ��equals()�Q�则该方法和

操作�W�一样在两个变量指向同一对象时返回真�Q�但是java推荐的是使用equals()�Ҏ��来判�?/p>

两个对象的内�Ҏ��否一��P��像String�cȝ��equals()�Ҏ��所做的那样�Q�判定两个String对象�?/p>

内容是否相同�Q��?=操作�W�返回true的唯一条�g是两个变量指向同一对象。从�q�个意义上来�?/p>

选择�l�定的答案。从更严格的意义来说正确�{�案应该只有d�?/p>

§1.1.7 �?/a>

Use the operators "<<", ">>", which statements are true?

　　A. 0000 0100 0000 0000 0000 0000 0000 0000<<5 gives

　　1000 0000 0000 0000 0000 0000 0000 0000

　　B. 0000 0100 0000 0000 0000 0000 0000 0000<<5 gives

　　1111 1100 0000 0000 0000 0000 0000 0000

　　C. 1100 0000 0000 0000 0000 0000 0000 0000>>5 gives

　1111 1110 0000 0000 0000 0000 0000 0000

　　D. 1100 0000 0000 0000 0000 0000 0000 0000>>5 gives

　　0000 0110 0000 0000 0000 0000 0000 0000

　　��译

　　使用"<<"�?">>"操作�W�的哪些陈述是对的�?/p>

　　�{�案　A,C

　　解析　Java的移位操作符一共有三种�Q�分别是”>>”,”>>>”,”<<”,执行的操作分�?/p>

是有�W�号右移�Q�无�W�号右移�Q�左�U�，有符号右�Uȝ��意思是说移入的最高位和原最高符号位相同

�Q�无�W�号右移是移入位始终补零�Q�左�U�L��最低位始终补零�Q�最高位被舍弃。移位操作符另一�?/p>

非常值得注意的特�Ҏ��其右操作数是取模�q�算的，意思是说对于一个int型数据而言�Q�对它移

�?2位的�l�果是保持不变而非变成�Ӟ��卻I��a>>32的结果是a而不�?�Q�同理，对long型数是对

��x��作数�?4的模�Q�a>>64==a�Q�还有一炚w��要注意的是移位操作符”>>>”只对int型和long�?/p>

有效�Q�对byte或者short的操作将��D��自动�c�d��转换�Q�而且是带�W�号的�?/p>

§1.1.8 �?/a>

　　String s= "hello";

　　String t = "hello";

　　char c[] = {'h','e','l','l','o'} ;

　　Which return true?

　　A. s.equals(t);

　　B. t.equals(c);

　　C. s==t;

　　D. t.equals(new String("hello"));

　　E. t==c.

　　(acd)

　　题目�Q�哪些返回true�?/p>

　　�q�个在前面第10题的equals()�Ҏ��?=操作�W�的讨论中论�q�过�?=操作�W�比较的是操�?/p>

�W�两端的操作数是否是同一个对象，而String的equals()�Ҏ��比较的是两个String对象的内�?/p>

是否一��P��其参数是一个String对象时才有可能返回true�Q�其它对象都�q�回假。需要指出的�?/p>

�׃��s和t�q��使用new创徏的，他们指向内存池中的同一个字�W�串帔R��Q�因此其地址实际上是

相同的（�q�个可以从反�~�译一个简单的��试�E�序的结果得刎ͼ�限于��幅不列出测试代码和反编

译的分析�Q�，因此�{�案c也是正确的�?/p>

§1.1.9 �?/a>

Class Teacher and Student are subclass of class Person.

　　Person p;

　　Teacher t;

　　Student s;

　　p, t and s are all non-null.

　　if(t instanceof Person) { s = (Student)t; }

　　What is the result of this sentence?

　　A. It will construct a Student object.

　　B. The expression_r is legal.

　　C. It is illegal at compilation.

　　D. It is legal at compilation but possible illegal at runtime.

　　(c)

　　题目�Q�类Teacher和Student都是�c�Person的子�c?/p>

　　…

　　p,t和s都是非空�?/p>

　　…

　　�q�个语句��D��的结果是什�?/p>

　　A. ��构造一个Student对象�?/p>

　　B. 表达式合法�?/p>

　　C. �~�译旉��法�?/p>

　　D. �~�译时合法而在�q�行时可能非法�?/p>

　　instanceof操作�W�的作用是判断一个变量是否是��x��作数指出的类的一个对象，�׃��ja

va语言的多态性��得可以用一个子�cȝ��实例赋值给一个父�cȝ��变量�Q�而在一些情况下需要判�?/p>

变量到底是一个什么类型的对象�Q�这时就可以使用instanceof了。当左操作数是右操作数指�?/p>

的类的实例或者是子类的实例时都返回真�Q�如果是��一个子�cȝ��实例赋值给一个父�cȝ��变量�Q?/p>

用instanceof判断该变量是否是子类的一个实例时也将�q�回真。此题中的if语句的判断没有问

题，而且��返回真�Q�但是后面的�c�d��转换是非法的�Q�因为t是一个Teacher对象�Q�它不能被强�?/p>

转换��Z��个Student对象�Q�即使这两个�c�L��共同的父�c�R��如果是��t转换��Z��个Person对象则可

以，而且不需要强制�{换。这个错误在�~�译时就可以发现�Q�因此编译不能通过�?/p>

李云�?/a> 2009-10-16 11:41 发表评论

�W�四部分�Q�语�a�基础

Fri, 16 Oct 2009 03:40:00 GMT

�W�四部分�Q�语�a�基础

能正��声明包、import、类�Q�包括内部类�Q�、接口、方法（包括用于开始一个类的执行的main�Ҏ��Q�、变量，�q�正��用修饰符�?
能够正确使用实现了接口的�c�，包括java.lang.Runnable接口或题目中指定的接口�?
知道命��o行参数是如何传递给�cȝ��main�Ҏ��的�?
知道哪些是JAVA的合法keyword。注意：考试中不会出现要你解释keyword和各�U�常数这�U�深奥的问题�?
明白如果没有昑ּ�地赋��|��各种变量或者数�l�的默认值是什么�?
掌握所有原始数据类型的取��D��_��如何声明各种数据�c�d��?

§1.1.1 一

1. What will happen when you attempt to compile and run the following code?

public class Static

{

static

{

int x = 5;

}

static int x,y;

public static void main(String args[])

{

x--;

myMethod();

System.out.println(x + y + ++x);

}

public static void myMethod()

{

y = x++ + ++x;

}

Choices:

a. Compiletime error

b. prints : 1

c. prints : 2

d. prints : 3

e. prints : 7

f. prints : 8

―――――――――――――――――――�?/p>

1) D is the correct choice. The above code will not give any compilation error. Note that "Static" is a valid class name. Thus choice A is incorrect. In the above code, on execution, first the static variables (x and y) will be initialized to 0. Then static block will be called and finally main() method will be called. The execution of static block will have no effect on the output as it declares a new variable (int x). The first statement inside main (x--) will result in x to be -1. After that myMethod() will be executed. The statement "y = x++ + ++x;" will be evaluated to y = -1 + 1 and x will become 1. In case the statement be "y =++x + ++x", it would be evaluated to y = 0 + 1 and x would become 1. Finally when System.out is executed "x + y + ++x" will be evaluated to "1 + 0 + 2" which result in 3 as the output. Thus choice D is correct.

§1.1.2 �?/a>

Considering the following code, Which variables may be referenced correctly at line 12?

1.public class Outer

2.{

3.public int a = 1;

4.private int b = 2;

5.public void method(final int c)

6.{

7.int d = 3;

8.class Inner

9.{

10.private void iMethod(int e)

11. {

12.

13.}

14.}

15.}

16.}

Choices:

a. a

b. b

c. c

d. d

e. e

A, B, C and E are correct. Since Inner is not a static inner class, it has a reference to an enclosing object, and all the variables of that object are accessible. Therefore A and B are correct, even if b is private. Variables in the enclosing method are only accessible when they are marked as final hence c is accessible but not d. E is obviously correct as it is a parameter to the method containing line 12 itself.

§1.1.3 �?/a>

What will be the result of executing the following code?

// Filename; SuperclassX.java

package packageX;

public class SuperclassX

{

protected void superclassMethodX()

{

}

int superclassVarX;

}

// Filename SubclassY.java

1.package packageX.packageY;

3.public class SubclassY extends SuperclassX

4.{

5.SuperclassX objX = new SubclassY();

6.SubclassY objY = new SubclassY();

7.void subclassMethodY()

8.{

9.objY.superclassMethodX();

10.int i;

11.i = objY.superclassVarX;

12.}

13.}

Choices:

a.Compilation error at line 5

b. Compilation error at line 9

c. Runtime exception at line 11

d. None of these

―――――――�?/p>

D is correct. When no access modifier is specified for a member, it is only accessible by another class in the package where its class is defined. Even if its class is visible in another package, the member is not accessible there. In the question above the variable superclassVarX has no access modifier specified and hence it cannot be accessed in the packageY even though the class SuperclassX is visible and the protected method superclassMethodX() can be accessed. Thus the compiler will raise an error at line 11.

§1.1.4 �?/a>

Consider the class hierarchy shown below:

--------------------------------------------------------------------

class FourWheeler implements DrivingUtilities

class Car extends FourWheeler

class Truck extends FourWheeler

class Bus extends FourWheeler

class Crane extends FourWheeler

----------------------------------------------------------------------

Consider the following code below:

1.DrivingUtilities du;

2.FourWheeler fw;

3.Truck myTruck = new Truck();

4.du = (DrivingUtilities)myTruck;

5.fw = new Crane();

6.fw = du;

Which of the statements below are true?

Choices:

a. Line 4 will not compile because an interface cannot refer to an object.

b. The code will compile and run.

c. The code will not compile without an explicit cast at line 6, because going

down the hierarchy without casting is not allowed.

d.The code at line 4 will compile even without the explicit cast.

e.The code will compile if we put an explicit cast at line 6 but will throw an exception at runtime.

――――――――――�?/p>

C and D are correct. A and B are obviously wrong because there is nothing wrong in an interface referring to an object. C is correct because an explicit cast is needed to go down the hierarchy. D is correct because no explicit cast is needed at line 4, because we are going up the hierarchy. E is incorrect because if we put an explicit cast at line 6, the code will compile and run perfectly fine, no exception will be thrown because the runtime class of du (that is Truck) can be converted to type FourWheeler without any problem.

§1.1.5 �?/a>

What will be printed when you execute the following code?

class X

{

Y b = new Y();

X()

{

System.out.print("X");

}

class Y

{

Y()

{

System.out.print("Y");

}

public class Z extends X

{

Y y = new Y();

Z()

{

System.out.print("Z");

}

public static void main(String[] args)

{

new Z();

}

Choices:

a. Z

b. YZ

c. XYZ

d. YXYZ

――――――――�?/p>

D is correct. A difficult but a fundamental question, please observe carefully. Before any object is constructed the object of the parent class is constructed(as there is a default call to the parent's constructor from the constructor of the child class via the super() statement). Also note that when an object is constructed the variables are initialized first and then the constructor is executed. So when new Z() is executed , the object of class X will be constructed, which means Y b = new Y() will be executed and "Y" will be printed as a result. After that constructor of X will be called which implies "X" will be printed. Now the object of Z will be constructed and thus Y y = new Y() will be executed and Y will be printed and finally the constructor Z() will be called and thus "Z" will be printed. Thus YXYZ will be printed.

§1.1.6 �?/a>

What will happen when you attempt to compile and run the following code?

class Base

{

int i = 99;

public void amethod()

{

System.out.println("Base.amethod()");

}

Base()

{

amethod();

}

public class Derived extends Base

{

int i = -1;

public static void main(String argv[])

{

Base b = new Derived();

System.out.println(b.i);

b.amethod();

}

public void amethod()

{

System.out.println("Derived.amethod()");

}

Choices:

a. Derived.amethod()

-1

Derived.amethod()

b. Derived.amethod()

c.Derived.amethod()

d. Derived.amethod()

e.Compile time error

―――――――�?/p>

B is correct. The reason is that this code creates an instance of the Derived class but assigns it to a reference of a the Base class. In this situation a reference to any of the fields such as i will refer to the value in the Base class, but a call to a method will refer to the method in the class type rather than its reference handle. But note that if the amethod() was not present in the base class then compilation error would be reported as at compile time, when compiler sees the statement like b.amethod(), it checks if the method is present in the base class or not. Only at the run time it decides to call the method from the derived class.

§1.1.7 �?/a>

Given the following code fragment:

　　1) public void create() {

　　2) Vector myVect;

　　3) myVect = new Vector();

　　4) }

　　Which of the following statements are true?

　　A. The declaration on line 2 does not allocate memory space for the

variable myVect.

　　B. The declaration on line 2 allocates memory space for a reference to a

Vector object.

　　C. The statement on line 2 creates an object of class Vector.

　　D. The statement on line 3 creates an object of class Vector.

　　E. The statement on line 3 allocates memory space for an object of class

Vector

　　��译

　　�l�出下面的代码片断。。。下面的哪些陈述为true(�??

　　A. �W�二行的声明不会为变量myVect分配内存�I�间�?/p>

　　B. �W�二行的声明分配一个到Vector对象的引用的内存�I�间�?/p>

　　C. �W�二行语句创��Z��个Vector�c�d��象�?/p>

　　D. �W�三行语句创��Z��个Vector�c�d��象�?/p>

　　E. �W�三行语句�ؓ一个Vector�c�d��象分配内存空间�?/p>

　　�{�案　A,D,E

　　解析

　　SL-275中指出：要�ؓ一个新对象分配�I�间必须执行new

Xxx()调用�Q�new调用执行以下的操作：

　　1�Q��ؓ新对象分配空间�ƈ��其成员初始化�ؓ0或者null�?/p>

　2�Q�执行类体中的初始化。（例如在类中有一个成员声明int a=10;在第一步后a=0

�Q�执行到�W�二步后a=10�Q?/p>

　　3�Q�执行构造函数�?/p>

　　4�Q�变量被分配��Z��个到内存堆中的新对象的引用�?/p>

§1.1.8 �?/a>

Which of the following statements about variables and their scopes

are true?

　　A. Instance variables are member variables of a class.

　　B. Instance variables are declared with the static keyword.

　　C. Local variables defined inside a method are created when the method

is executed.

　　D. Local variables must be initialized before they are used.

　　(acd)

　　题目�Q�下面关于变量及其范围的陈述哪些是对的�?/p>

　　A. 实例变量是类的成员变量�?/p>

　　B. 实例变量用关键字static声明�?/p>

　　C. 在方法中定义的局部变量在该方法被执行时创�?/p>

　　D. 局部变量在使用前必��被初始化�?/p>

　　�c�M��有几�U�变量，分别是：局部变量（英文可以为：local\automatic\temporary\stac

variable�Q�是定义在方法里的变量；实例变量�Q�英文�ؓ�Q�instance

variable�Q�是在方法外而在�c�d��明内定义的变量，有时也叫成员变量�Q�类变量�Q�英文�ؓ�Q�cl

ass

variable�Q�是用关键字static声明的实例变量，他们的生存期分别是：局部变量在定义该变

量的�Ҏ��被调用时被创建，而在该方法退出后被撤销�Q�实例变量在使用new

Xxxx()创徏该类的实例时被创建，而其生存期和该类的实例对象的生存期相同；�c�d��量在�?/p>

�c�被加蝲时被创徏�Q�不一定要用new

Xxxx()创徏�Q�所有该�cȝ��实例对象�׃�n该类变量�Q�其生存期是�cȝ��生存期。�Q何变量在使用

前都必须初始�?但是需要指出的是局部变量必��L��式初始化�Q�而实例变量不必，原始�c�d��?/p>

实例变量在该�cȝ��构造方法被调用时�ؓ它分配的�~�省的��|��整型�?�Q�布��型是false�Q�而��Q�?/p>

型是0.0f�Q�引用类型（�cȝ��型）的实例变量的�~�省值是null�Q�没有进行实际的初始化，对它�?/p>

使用��引起NullPointException�Q�，�c�d��量的规则和实例变量一��P��不同的是�c�d��量的初始�?/p>

是在�c�被加蝲时�?/p>

§1.1.9 �?/a>

　　public class Parent {

　　int change() {…}

　　}

　　class Child extends Parent {

　　}

　　Which methods can be added into class Child?

　　A. public int change(){}

　　B. int chang(int i){}

　　C. private int change(){}

　　D. abstract int chang(){}

　　(ab)

　　题目�Q�哪些方法可被加入类Child�?/p>

　　需要注意的是答案D的内容，子类可以重写父类的方法�ƈ��之声明为抽象方法，但是�q�引发的问题是类必须声明为抽象类�Q�否则编译不能通过�Q�而且抽象�Ҏ��不能有方法体�Q�也��是�Ҏ��声明后面不能带上那两个大括号�Q�{}�Q�，�q�些D都不能满��?/p>

李云�?/a> 2009-10-16 11:40 发表评论

Fri, 16 Oct 2009 03:39:00 GMT

�W�三部分�Q�垃圾收�?/p>

了解垃圾攉��机制能保证执行什么�?
能通过代码操作一个对象，使它能被垃圾攉��器收集�?
知道在程序的哪一行，垃圾攉��器能合法地收集某个对象�?

�W�一�?nbsp; 垃圾攉��解析

垃圾攉��器是Java语言区别于其他程序设计语�a�的一大特艌Ӏ�它把程序员从手工回收内存空间的�J�重工作中解�׃��出来。在SUN公司的Java�E�序员（Java Programmer�Q�认证考试中，垃圾攉��器是必考的内容�Q�一般最多可以占��d��值的6%左右。但是由于SUN公司的Java Programming Language SL-275 评��的标准教材中�Q�对有关垃圾攉��器的内容只做了非常简单的介绍�Q�而另外的一些关于Java技术的书籍�Q�比如《Java 2 核心技术》（Core Java 2�Q�、《Java�~�程思想》（Thinking in Java�Q�、《精通Java 2》等�{�，里面关于垃圾攉��器的内容也几乎没有，或者只是简单地提两句，所以很多参加Java Programmer认证考试的中国考生�Q�在垃圾攉��器这一部分的得分都�?分（�W�者曾认识一位SUN公司授权的中国Java培训班的老师�Q�其考试��d��?9%�Q�但垃圾攉��器的部分竟然也�ؓ0分）。鉴于此�Q�笔者�ȝ��了这个垃圾收集器的专题，希望对广大Java技术的爱好者和准备认证考试的考生们有所帮助�?/p>

我们知道�Q�许多程序设计语�a�都允许在�E�序�q�行期动态地分配内存�I�间。分配内存的方式多种多样�Q�取决于该种语言的语法结构。但不论是哪一�U�语�a�的内存分配方式，最后都要返回所分配的内存块的�v始地址�Q�即�q�回一个指针到内存块的首地址�?/p>

当已�l�分配的内存�I�间不再需要时�Q�换句话说当指向该内存块的句柄超��Z��使用范围的时候，该程序或其运行环境就应该回收该内存空��_��以节省宝�늚�内存资源�?/p>

在C�Q�C++或其他程序设计语�a�中，无论是对象还是动态配�|�的资源或内存，都必��ȝ��E�序员自行声明��生和回收�Q�否则其中的资源��消耗，造成资源的浪费甚��x��机。但手工回收内存往往是一��复杂而艰巨的工作。因��预先��定占用的内存空间是否应该被回收是非常困隄��Q�如果一�D늨�序不能回收内存空��_��而且在程序运行时�pȝ��中又没有了可以分配的内存�I�间�Ӟ��q�段�E�序��只能崩溃。通常�Q�我们把分配出去后，却无法回收的内存�I�间�U�Cؓ"内存渗漏体（Memory Leaks�Q?�?/p>

以上�q�种�E�序设计的潜在危险性在Java�q�样以严谨、安全著�U�的语言中是不允许的。但是Java语言既不能限制程序员�~�写�E�序的自由性，又不能把声明对象的部分去除（否则��׃��是面向对象的�E�序语言了）�Q�那么最好的解决办法��是从Java�E�序语言本��n的特性入手。于是，Java技术提供了一个系�l��的线�E�（Thread�Q�，卛_��圾收集器�U�程�Q�Garbage Collection Thread�Q�，来跟�t�每一块分配出�ȝ��内存�I�间�Q�当Java 虚拟机（Java Virtual Machine�Q�处于空闲��@环时�Q�垃圾收集器�U�程会自动检查每一快分配出�ȝ��内存�I�间�Q�然后自动回收每一快可以回收的无用的内存块�?/p>

垃圾攉��器线�E�是一�U�低优先�U�的�U�程�Q�在一个Java�E�序的生命周期中�Q�它只有在内存空闲的时候才有机会运行。它有效地防止了内存渗漏体的出现�Q��ƈ极大可能地节省了宝贵的内存资源。但是，通过Java虚拟机来执行垃圾攉��器的�Ҏ��可以是多�U�多��L��?/p>

下面介绍垃圾攉��器的特点和它的执行机�Ӟ��

垃圾攉��器系�l�有自己的一套方案来判断哪个内存块是应该被回收的�Q�哪个是不符合要求暂不回收的。垃圾收集器在一个Java�E�序中的执行是自动的�Q�不能强制执行，即�ɽE�序员能明确地判断出有一块内存已�l�无用了�Q�是应该回收的，�E�序员也不能强制垃圾攉��器回收该内存块。程序员唯一能做的就是通过调用System. gc �Ҏ��?��"执行垃圾攉��器，但其是否可以执行�Q�什么时候执行却都是不可知的。这也是垃圾攉��器的最主要的缺炏V��当然相对于它给�E�序员带来的巨大方便性而言�Q�这个缺�Ҏ��瑕不掩瑜的�?/p>

垃圾攉��器的主要特点有：

1�Q�垃圾收集器的工作目标是回收已经无用的对象的内存�I�间�Q�从而避免内存渗漏体的��生，节省内存资源�Q�避免程序代码的崩溃�?/p>

2�Q�垃圾收集器判断一个对象的内存�I�间是否无用的标准是�Q�如果该对象不能再被�E�序中�Q何一�?�z�d��的部�?所引用�Q�此时我们就��_��该对象的内存�I�间已经无用。所�?�z�d��的部�?�Q�是指程序中某部分参与程序的调用�Q�正在执行过�E�中�Q�尚未执行完毕�?/p>

3�Q�垃圾收集器�U�程虽然是作��Z��优先�U�的�U�程�q�行�Q�但在系�l�可用内存量�q�低的时候，它可能会�H�发地执行来挽救内存资源。当然其执行与否也是不可预知的�?/p>

4�Q�垃圾收集器不可以被强制执行�Q�但�E�序员可以通过调用System. gc�Ҏ��来徏议执行垃圾收集器�?/p>

5�Q�不能保证一个无用的对象一定会被垃圾收集器攉��Q�也不能保证垃圾攉��器在一�D�Java语言代码中一定会执行。因此在�E�序执行�q�程中被分配出去的内存空间可能会一直保留到该程序执行完毕，除非该空间被重新分配或被其他�Ҏ��回收。由此可见，完全��d��地根�l�内存渗漏体的��生也是不可能的。但是请不要忘记�Q�Java的垃圾收集器毕竟使程序员从手工回收内存空间的�J�重工作中解�׃��出来。设想一个程序员要用C或C++来编写一�D?0万行语句的代码，那么他一定会充分体会到Java的垃圾收集器的优点！

6�Q�同��h��有办法预知在一�l�均�W�合垃圾攉��器收集标准的对象中，哪一个会被首先收集�?/p>

7�Q��@环引用对象不会媄响其被垃圾收集器攉��?/p>

8�Q�可以通过��对象的引用变量�Q�reference variables�Q�即句柄handles�Q�初始化为null��|��来暗�C�垃圾收集器来收集该对象。但此时�Q�如果该对象�q�接有事件监听器�Q�典型的 AWT�l��g�Q�，那它�q�是不可以被攉��。所以在设一个引用变量�ؓnull��g��前，应注意该引用变量指向的对象是否被监听�Q�若有，要首先除�ȝ��听器�Q�然后才可以赋空倹{�?/p>

9�Q�每一个对象都有一个finalize( )�Ҏ��Q�这个方法是从Object�cȝ��承来的�?/p>

10�Q�finalize( )�Ҏ��用来回收内存以外的系�l�资源，��像是文件处理器和网�l�连接器。该�Ҏ��的调用顺序和用来调用该方法的对象的创建顺序是无关的。换句话��_��书写�E�序时该�Ҏ��的顺序和�Ҏ��的实际调用顺序是不相�q�的。请注意�q�只是finalize( )�Ҏ��的特炏V�?/p>

11�Q�每个对象只能调用finalize( )�Ҏ��一�ơ。如果在finalize( )�Ҏ��执行时��生异常（exception�Q�，则该对象仍可以被垃圾攉��器收集�?/p>

12�Q�垃圾收集器跟踪每一个对象，攉��那些不可到达的对象（卌��对象没有被程序的��M��"�zȝ��部分"所调用�Q�，回收其占有的内存�I�间。但在进行垃圾收集的时候，垃圾攉��器会调用finalize( )�Ҏ��Q�通过让其他对象知道它的存在，而��不可到达的对象再��?复苏"为可到达的对象。既然每个对象只能调用一�ơfinalize( )�Ҏ��Q�所以每个对象也只可�?复苏"一�ơ�?/p>

13�Q�finalize( )�Ҏ��可以明确地被调用�Q�但它却不能�q�行垃圾攉��?/p>

14�Q�finalize( )�Ҏ��可以被重载（overload�Q�，但只有具备初始的finalize( )�Ҏ��特点的方法才可以被垃圾收集器调用�?/p>

15�Q�子�cȝ��finalize( )�Ҏ��可以明确地调用父�cȝ��finalize( )�Ҏ��Q�作��子类对象的最后一�ơ适当的操作。但Java�~�译器却不认��是一�ơ覆盖操作（overriding�Q�，所以也不会对其调用�q�行��查�?/p>

16�Q�当finalize( )�Ҏ��未被调用时�Q�System. runFinalization( )�Ҏ��可以用来调用finalize( )�Ҏ��Q��ƈ实现相同的效果，�Ҏ��用对象进行垃圾收集�?/p>

17�Q�当一个方法执行完毕，其中的局部变量就会超��Z��用范��_��此时可以被当作垃圾收集，但以后每当该�Ҏ��再次被调用时�Q�其中的局部变量便会被重新创徏�?/p>

18�Q�Java语言使用了一�U?标记交换区的垃圾攉��法"。该��法会遍历程序中每一个对象的句柄�Q��ؓ被引用的对象做标讎ͼ�然后回收��未做标记的对象。所谓遍历可以简单地理解�?��查每一�?�?/p>

19�Q�Java语言允许�E�序员�ؓ��M��Ҏ��d��finalize( )�Ҏ��Q�该�Ҏ��会在垃圾攉��器交换回收对象之前被调用。但不要�q�分依赖该方法对�pȝ��资源�q�行回收和再利用�Q�因��Ҏ��调用后的执行�l�果是不可预知的�?/p>

通过以上对垃圾收集器特点的了解，你应该可以明��垃圾收集器的作用，和垃圾收集器判断一块内存空间是否无用的标准。简单地��_��当你��Z��个对象赋��gؓnull�q�且重新定向了该对象的引用者，此时该对象就�W�合垃圾攉��器的攉��标准�?/p>

判断一个对象是否符合垃圾收集器的收集标准，�q�是SUN公司�E�序员认证考试中垃圾收集器部分的重要考点�Q�可以说�Q�这是唯一的考点�Q�。所以，考生在一�D늻�定的代码中，应该能够判断出哪个对象符合垃圾收集器攉��的标准，哪个不符合。下面结合几�U�认证考试中可能出现的题型来具体讲解：

Object obj = new Object ( ) ;

我们知道�Q�obj为Object的一个句柄。当出现new关键字时�Q�就�l�新建的对象分配内存�I�间�Q�而obj的值就是新分配的内存空间的首地址�Q�即该对象的�?��L��别注意，对象的值和对象的内�Ҏ��不同含义的两个概念：对象的值就是指其内存块的首地址�Q�即对象的句柄；而对象的内容则是其具体的内存�?。此时如果有 obj = null�Q�则obj指向的内存块此时��无用了�Q�因��Z��面再没有调用该变量了�?/p>

请再看以下三�U�认证考试时可能出现的题型�Q?/p>

�E�序�D?�Q?/p>

1�Q�fobj = new Object ( ) ;

2�Q�fobj. Method ( ) ;

3�Q�fobj = new Object ( ) ;

4�Q�fobj. Method ( ) ;

问：�q�段代码中，�W�几行的fobj �W�合垃圾攉��器的攉��标准�Q?/p>

�{�：�W?行。因为第3行的fobj被赋了新��|��产生了一个新的对象，��x��了一块新的内存空��_��也相当于为第1行中的fobj赋了null倹{��这�U�类型的题在认证0考试中是最��单的�?/p>

�E�序�D?�Q?/p>

1�Q�Object sobj = new Object ( ) ;

2�Q�Object sobj = null ;

3�Q�Object sobj = new Object ( ) ;

4�Q�sobj = new Object ( ) ;

问：�q�段代码中，�W�几行的内存�I�间�W�合垃圾攉��器的攉��标准�Q?/p>

�{�：�W?行和�W?行。因为第2行�ؓsobj赋��gؓnull�Q�所以在此第1行的sobj�W�合垃圾攉��器的攉��标准。而第4行相当于为sobj赋��gؓnull�Q�所以在此第3行的sobj也符合垃圾收集器的收集标准�?/p>

如果有一个对象的句柄a�Q�且你把a作�ؓ某个构造器的参敎ͼ��?new Constructor ( a )的时候，即��你给a赋��gؓnull�Q�a也不�W�合垃圾攉��器的攉��标准。直到由上面构造器构造的新对象被赋空值时�Q�a才可以被垃圾攉��器收集�?/p>

�E�序�D?�Q?/p>

1�Q�Object aobj = new Object ( ) ;

2�Q�Object bobj = new Object ( ) ;

3�Q�Object cobj = new Object ( ) ;

4�Q�aobj = bobj;

5�Q�aobj = cobj;

6�Q�cobj = null;

7�Q�aobj = null;

问：�q�段代码中，�W�几行的内存�I�间�W�合垃圾攉��器的攉��标准�Q?/p>

�{�：�W?行。注意这�c�题型是认证考试中可能遇到的最��N��型了�?/p>

�?-3分别创徏了Object�cȝ��三个对象�Q�aobj�Q�bobj�Q�cobj

�?�Q�此时对象aobj的句柄指向bobj�Q�所以该行的执行不能使aobj�W�合垃圾攉��器的攉��标准�?/p>

�?�Q�此时对象aobj的句柄指向cobj�Q�所以该行的执行不能使aobj�W�合垃圾攉��器的攉��标准�?/p>

�?�Q�此时仍没有��M��一个对象符合垃圾收集器的收集标准�?/p>

�?�Q�对象cobj�W�合了垃圾收集器的收集标准，因�ؓcobj的句柄指向单一的地址�I�间。在�W?行的时候，cobj已经被赋��gؓnull�Q�但由cobj同时�q�指向了aobj�Q�第5行）�Q�所以此时cobj�q�不�W�合垃圾攉��器的攉��标准。而在�W?行，aobj所指向的地址�I�间也被赋予了空值null�Q�这��p��明了�Q�由cobj所指向的地址�I�间已经被完全地赋予了空倹{��所以此时cobj最�l�符合了垃圾攉��器的攉��标准。但对于aobj和bobj�Q�仍然无法判断其是否�W�合攉��标准�?/p>

��M��Q�在Java语言中，判断一块内存空间是否符合垃圾收集器攉��标准的标准只有两个：

1�Q�给对象赋予了空值null�Q�以下再没有调用�q��?/p>

2�Q�给对象赋予了新��|��既重新分配了内存�I�间�?/p>

最后再�ơ提醒一下，一块内存空间符合了垃圾攉��器的攉��标准�Q��ƈ不意味着�q�块内存�I�间��׃��定会被垃圾收集器攉��?/p>

李云�?/a> 2009-10-16 11:39 发表评论

Fri, 16 Oct 2009 03:38:00 GMT

�W�二部分�Q�流�E�控�Ӟ��断言和异常处�?/p>

能够正确使用if,switch语句�Q��ƈ且能正确使用合法的参数类型�?
能够正确使用所有带标签或不带标�{��循环语句�Q�能正确使用break,continue�Q�能计算在��@环中或��@环后循环计数器的倹{�?
能够正确使用异常和异常处理语句（try,catch,finally�Q�。能正确声明抛出异常的方法，�q�知道怎样覆盖它�?
了解代码�D�中的异常对�E�序跌��{的媄响。注意：异常可能是一个运行时异常�Q�runtime exception�Q?一个已�l�定义的异常�Q�checked exception�Q�，也可能是一个error�?
能正��用断�a��Q��ƈ了解关于断言机制的正��说法�?

能够正确使用if,switch语句�Q��ƈ且能正确使用合法的参数类型�?
能够正确使用所有带标签或不带标�{��循环语句�Q�能正确使用break,continue�Q�能计算在��@环中或��@环后循环计数器的倹{�?
能够正确使用异常和异常处理语句（try,catch,finally�Q�。能正确声明抛出异常的方法，�q�知道怎样覆盖它�?
了解代码�D�中的异常对�E�序跌��{的媄响。注意：异常可能是一个运行时异常�Q�runtime exception�Q?一个已�l�定义的异常�Q�checked exception�Q�，也可能是一个error�?
能正��用断�a��Q��ƈ了解关于断言机制的正��说法�?

�W�一�?nbsp; 断言 assert

§1.1.1 assertion的语法和语义

J2SE 1.4在语�a�上提供了一个新�Ҏ��，��是assertion(断言)功能�Q�它是该版本在Java语言斚w��最大的革新。在软�g开发中�Q�assertion是一�U�经典的调试、测试方式，本文��深入解析assertion功能的��用以及其设计理念�Q��ƈ�l�出相关的例子�?/p>

assertion(断言)在��Y件开发中是一�U�常用的调试方式�Q�很多开发语�a�中都支持�q�种机制�Q�如C�Q�C++和Eiffel�{�，但是支持的�Ş式不��相同，有的是通过语言本��n、有的是通过库函数等。另外，从理��Z��来说�Q�通过assertion方式可以证明�E�序的正��性，但是�q�是一��相当复杂的工作�Q�目前还没有太多的实跉|��义�?/p>

在实��C��Q�assertion��是在程序中的一条语句，它对一个boolean表达式进行检查，一个正��程序必��M��证这个boolean表达式的��gؓtrue�Q�如果该��gؓfalse�Q�说明程序已�l�处于不正确的状态下�Q�系�l�将�l�出警告或退出。一般来��_��assertion用于保证�E�序最基本、关键的正确性。assertion��查通常在开发和��试时开启。�ؓ了提高性能�Q�在软�g发布后，assertion��查通常是关闭的。下面简单介�l�一下Java中assertion的实现�?/p>

1�Q?/strong>1) 语法表示

在语法上�Q��ؓ了支持assertion�Q�Java增加了一个关键字assert。它包括两种表达式，分别如下�Q?/p>
1.       assert expression_r1;

2.       assert expression_r1: expression_r2;

在两�U�表辑ּ�中，expression_r1表示一个boolean表达式，expression_r2表示一个基本类型或者是一个对�?Object) �Q�基本类型包括boolean,char,double,float,int和long。由于所有类都�ؓObject的子�c�，因此�q�个参数可以用于所有对象�?/p>
1�Q?/strong>2) 语义含义

在运行时�Q�如果关闭了assertion功能�Q�这些语句将不�v��M��作用。如果打开了assertion功能�Q�那么expression_r1的值将被计��，如果它的��gؓfalse�Q�该语句强抛��Z��个AssertionError对象。如果assertion语句包括expression_r2参数�Q�程序将计算出expression_r2的结果，然后��这个结果作为AssertionError的构造函数的参数�Q�来创徏AssertionError对象�Q��ƈ抛出该对象；如果expression_r1��gؓtrue�Q�expression_r2��不被计��?/p>
一�U�特�D�情冉|��Q�如果在计算表达式时�Q�表辑ּ�本��n抛出Exception�Q�那么assert��停止运行，而抛��个Exception�?/p>
1�Q?/strong>3) 一�?/strong>assertion例子

下面是一些Assert的例子�?/p>
1.       assert　　0 < value;

2.       assert　　0 < value:"value="+value;

3.       assert　　ref != null:"ref doesn't equal null";

4.       assert　　isBalanced();

1�Q?/strong>4) �~�译

�׃��assert是一个新关键字，使用老版本的JDK是无法编译带有assert的源�E�序。因此，我们必须使用JDK1.4(或者更�?的Java�~�译器，在��用Javac命��o�Ӟ��我们必须加上-source 1.4作�ؓ参数�?source 1.4表示使用JDK 1.4版本的方式来�~�译源代码，否则�~�译��׃��能通过�Q�因为缺省的Javac�~�译器��用JDK1.3的语法规则�?/p>
一个简单的例子如下�Q?/p>
javac      -source   1.4    test.java

1�Q?/strong>5) �q�行

�׃��带有assert语句的程序运行时�Q��用了新的ClassLoader和Class�c�，因此�Q�这�U�程序必��d��JDK1.4(或者更高版�?的JRE下运行，而不能在老版本的JRE下运行�?/p>
1.       �׃��我们可以选择开启assertion功能�Q�或者不开启，另外我们�q�可以开启一部分�c�L��包的assertion功能�Q�所以运行选项变得有些复杂。通过�q�些选项�Q�我们可以过滤所有我们不兛_��的类�Q�只选择我们兛_��的类或包来观察�?/p>
§1.1.2                     assertion的设计问�?/h3>
首先�Q�我们认为assertion是必要的。因为，如果没有�l�一的assertion机制�Q�Java�E�序通常使用if-then-else或者switch-case语句�q�行assertion��查，而且��查的数据�c�d��也不完全相同。assertion机制让Java�E�序员用�l�一的方式处理assertion问题�Q�而不是按自己的方式处理。另外，如果用户使用自己的方式进行检查，那么�q�些代码在发布以后仍然将起作用，�q�可能会影响�E�序的性能。而从语言�a�层次支持assertion功能�Q�这��把assertion�Ҏ��能带来的负面媄响降到最��?/p>
Java是通过增强一个关键字assert实现支持assertion�Q�而不是��用一个库函数支持�Q�这说明Java认�ؓassertion对于语言本��n来说是非帔R��要的。实际上�Q�在Java的早期的规范中，Java是能够支持assert的，但是�׃��一些实现的限制�Q�这些特性从规范中除��M��。因此，assert的再�ơ引入应该是恢复了Java对assert的支持。C语言��是通过Assert.h函数库实现断�a�的支持�?/p>
Java的assertion的开启也和C语言不太一��P��我们都知道在C语言中，assertion的开启是在编译时候决定的。当我们使用debug方式�~�译�E�序时候，assertion被开启，而��用release方式�~�译时候，assertion自动被关闭。而Java的assertion却是在运行的时候进行决定的。其实，�q�两�U�方式是各有优缺炏V��如果采用编译时军_��方式�Q�开发�h员将处理两种�c�d��的目标码�Q�debug版本和release版本�Q�这加大了文��管理的隑ֺ��Q�但是提高了代码的运行效率。Java采用�q�行时决定的方式�Q�这��h��有的assertion信息��置于目标代码中�Q�同一目标代码可以选择不同方式�q�行�Q�增强目标代码的灉|��性，但是它将牺牲因�ؓassertion而引起一部分性能损失。Java专家��组认�ؓ�Q�所牺牲的性能相当��，因此java采用了运行时军_��方式�?/p>
另外�Q�我们注意到AssertionError作�ؓError的一个子�c�，而不是RuntimeException。关于这一点，专家�l�也�q�行了长期的讨论。Error代表一些异常的错误�Q�通常是不可以恢复的，而RuntimeException��该错误在�q�行时才发生的特炏V��AssertionError通常为非常关键的错误�Q�这些错误往往是不�Ҏ��恢复的，而且assertion机制也不鼓励�E�序员对�q�种错误�q�行恢复。因此，��Z��assertion的含义，Java专家��组选择了让AssertError为Error的子�c�R�?/p>
§1.1.3                     assertion与��?/h3>
在本节，我们��考虑assertion与��承的关系�Q�研�I�assert是如何定位的。如果开启一个子�cȝ��assertion�Q�那么它的父�cȝ��assertion是否执行�Q?/p>
下面的例子将昄��如果一个assert语句在父�c�，而当它的子类调用它时�Q�该assert为false。我们看看在不同的情况下�Q�该assertion是否被处理�?/p>
class Base

{

public void baseMethod()

{ // ��L��assertion��p�|

    assert      false : "Assertion failed:This is base ";

    System.out.println("Base Method");

}

}

class Derived

extends Base

{

public void derivedMethod()

{ // ��L��assertion��p�|

    assert false: "Assertion failed:This is derive";

    System.out.println( "Derived Method" );

}

public static void main( String[] args )

{

    try

    {

      Derived derived = new Derived();

      derived.baseMethod( );

     derived.derivedMethod();

    }

    catch( AssertionError ae )

    {

      System.out.println(ae);

    }

}

}

�q�行命��o

含义

�l�果

Java Derived

不启用assertion

Base Method
Derived Method

Java -ea Derived

开启所有assertion

Java.lang.AssertionError:Assertion Failed:This is base

Java -da Derived

关闭所有assertion

Base Method
Derived Method

Java -ea:Base Derived

仅打开Base的assertion

Java.lang.AssertionError:Assertion Failed:This is base

Java -ea:Derived Derived

仅打开Derived的assertion

Base Method
Java.lang.AssertionError:Assertion Failed:This is derived

从这个例子我们可以看出，父类的assert语句��只有在父类的assert开启才起作用，如果仅仅开启子�cȝ��assert�Q�父�cȝ��assert仍然不运行。例如，我们执行java -ea:Derived Derived的时候，Base�cȝ��assert语句�q�不执行。因此，我们可以认�ؓ�Q�assert语句不具有��承功能�?/p>
§1.1.4                     assertion的��?/h3>
assertion的��用是一个复杂的问题�Q�因��涉及到�E�序的风��|��assertion�q�用的目标，�E�序的性质�{�问题。通常来说�Q�assertion用于��查一些关键的��|��q�且�q�些值对整个�E�序�Q�或者局部功能的完成有很大的影响�Q��ƈ且这�U�错误不�Ҏ��恢复的。assertion表达式应该短��、易懂，如果需要评估复杂的表达式，应该使用函数计算。以下是一些��用assertion的情�늚�例子�Q�这些方式可以让java�E�序的可靠性更高�?/p>
1.       ��查控制流�Q�在if-then-else和swith-case语句中，我们可以在不应该发生的控制支��上加上assert false语句。如果这�U�情况发生了�Q�assert能够��查出来�?br /> 例如�Q�x取值只能��1,2,3�Q�我们的�E�序可以如下表示

2.                   switch (x)

3.                   { case 1: …;

4.                     case 2: …;

5.                     case 3: …

6.                     default: assert false:"x value is invalid: "+x;

7.           }

8.       在私有函数计��前�Q�检查输入参数是否有效；对于一�U�有些函敎ͼ�要求输入满��一些特定的条�g�Q�那么我们可以在函数开始处使用assert�q�行参数��查。对于公共函敎ͼ�我们通常不��用assertion��查，因�ؓ一般来��_��公共函数必须�Ҏ��效的参数�q�行��查和处理。而私有函数往往是直接��用的�?br /> 例如�Q�某函数可能要求输入的参数必��M��为null。那么我们可以在函数的一开始加�?assert parameter1!=null : "paramerter is null in test method";

9.       在函数计��后�Q�检查函数结果是否有效；对于一些计��函敎ͼ�函数�q�行完成后，某些值需要保证一定的性质�Q�因此我们可以通过assert��查该倹{�?br /> 例如�Q�我们有一个计��绝对值的函数�Q�那么我们就可以在函数的�l�果处，加上一个语句：

assert value>=0:"Value should be bigger than 0:"+value;

通过�q�种方式�Q�我们可以对函数计算完的�l�果�q�行��查�?/p>
10.   ��查程序不变量�Q�有些程序中�Q�存在一些不变量�Q�在�E�序的运行生命周期，�q�些不变量的值都是不变的。这些不变量可能是一个简单表辑ּ��Q�也可能是一个复杂的表达式。对于一些关键的不变量，我们可以通过assert�q�行��查�?br /> 例如�Q�在一个胦会系�l�中�Q�公司的支出和收入必��M��持一定的�q��关系�Q�因此我们可以编写一个表辑ּ��查这�U��^衡关�p�，如下表示�?/p>
11.                 private boolean isBalance() {

12.                  ……

13.                }



在这个系�l�中�Q�在一些可能媄响这�U��^衡关�pȝ��Ҏ��的前后，我们都可以加上assert验证�Q�assert isBalance():"balance is destoried";

§1.1.5                     �l�论

assertion为开发�h员提供了一�U�灵�z�d��调试和测试机�Ӟ��它的使用也非常简单、方�ѝ��但是，如何规范、系�l�地使用assertion(特别是在Java语言�?仍然是一个亟待研�I�的问题

李云�?/a> 2009-10-16 11:38 发表评论

�W�一部分�Q�声明和讉K��控制(�l�习�?

Fri, 16 Oct 2009 03:37:00 GMT

�W�一部分�Q�声明和讉K��控制

知道怎么样声明、创建和初始化各�U�类型的数组
知道怎么样声明类、内部类、方法、成员变量、静态变量和�Ҏ��内部变量�Q��ƈ会��用合法的修饰�W�（如public,final,static,abstract,�{�等�Q�。了解这些修饰符单独使用或组合��用的含义�Q��ƈ且知道它们对于包作用域的影响�?
了解�cȝ��构造器�Q�constructor�Q?
�l�定一个方法，能判断它的合法返回类�?

§1.1.1     一

In the following pieces of code, A and D will compile without any error. True/False?

A: StringBuffer sb1 = "abcd";

B: Boolean b = new Boolean("abcd");

C: byte b = 255;

D: int x = 0x1234;

E: float fl = 1.2;

Choices:

A. True

B. False

――――――――――――――――――――――――――――――�?/p>
�{�案 B. The code segments B and D will compile without any error. A is not a valid way to construct a StringBuffer, you need to creat a StringBuffer object using "new". B is a valid construction of a Boolean (any string other than "true" or "false" to the Boolean constructor will result in a Boolean with a value of "false"). C will fail to compile because the valid range for a byte is -128 to +127 (ie, 8 bits,signed). D is correct, 0x1234 is the hexadecimal representation in java. E fails to compile because the compiler interprets 1.2 as a double being assigned to a float (down-casting), which is not valid. You either need an explicit cast (as in "(float)1.2") or "1.2f", to indicate a float.

§1.1.2       �?/a>

What will be the result of executing the following code?

Given that Test1 is a class.

1. Test1[] t1 = new Test1[10];

2. Test1[][] t2 = new Test1[5][];

3. if (t1[0] == null)

4. {

5.t2[0] = new Test1[10]

6.t2[1] = new Test1[10]

7.t2[2] = new Test1[10]

8.t2[3] = new Test1[10]

9.t2[4] = new Test1[10]

10. }

11. System.out.println(t1[0]);

12. System.out.println(t2[1][0]);

Choices:

a. The code will not compile because the array t2 is not initialized in an unconditional statement before use.

b. The code will compile but a runtime exception will be thrown at line 12.

c. The code will compile but a runtime exception will be thrown at line 11.

d. None of these.

――――――――――――――――――――�?/p>
D is correct. Though we cannot use local variables without initializing them (compilation error), there is an exception to it. In case of arrays initialization is supposed to be complete when we specify the leftmost dimension of the array. The problem occurs at runtime if we try to access an element of the array which has not been initialized (specification of size). In the question above the array t2 is initialized before use, therefore there will be no problem at runtime also and the lines 11 and 12 will both print null.

§1.1.3       �?/a>

Which declarations of identifiers are legal?

　A. $persons

　　B. TwoUsers

　　C. *point

　　D. this

    E. _endline

　　

　　�{�案　A,B,E

　　

　　解析　Java的标识符可以以一个Unicode字符�Q�下滑线�Q�_�Q�，��元�W�（$�Q�开始，后箋�?/p>
�W�可以是前面的符号和数字�Q�没有长度限�Ӟ��大小写敏感，不能是保留字�?/p>

§1.1.4       �?/a>

Which of the following answer is correct to express the value 8 in octal number?

　　A. 010

　　B. 0x10

　　C. 08

　　D. 0x8

　　��译

　　下面的哪些答案可以用以表�C�八�q�制�?�?/p>
　　�{�案　A

　　解析　　八进制��g��0开��_��?x开头的为十六进制��|��八进制中不能出现数字8�Q�最大只�?�?/p>

§1.1.5       �?/a>

Which are not Java keywords?

　　A. TRUE

　　B. sizeof

　　C. const

　　D. super

　　E. void

　　��译

　　哪些不是Java关键字�?/p>

　　�{�案　A,B

　　解析

　 A�Q�不是，Java中有true,但是�q�也不是关键字而是字面量（literal�Q��?/p>
　B�Q�不是，Java中不需要这个操作符�Q�所有的�c�d��Q�原始类型）的大��都是固定的�?/p>
　C、D、E都是�Q�需要说明的是const是java中未被��用的关键字�?/p>

§1.1.6       �?/a>

Which statements about Java code security are true?

　　A. The bytecode verifier loads all classes needed for the execution of a program.

　　B. Executing code is performed by the runtime interpreter.

　　C. At runtime the bytecodes are loaded, checked and run in an interpreter.

　　D. The class loader adds security by separating the namespaces for the

classes of the local file system from those imported from network sources.

―――――――――――――�?/p>
�{�案 BCD

§1.1.7       �?/h3>
　　题目�Q�下面有关java代码安全性的叙述哪些是对的�?/p>
　　A. 字节码校验器加蝲查询执行需要的所有类�?/p>

　　B. �q�行时解释器执行代码�?/p>

　　C. 在运行时�Q�字节码被加载，验证然后在解释器里面�q�行�?/p>

　　D. �c�d��载器通过分离本机文�g�pȝ��的类和从�|�络导入的类增加安全性�?/p>

　　SL275中描�q�的Java�E�序�q�行的过�E�是�q�样的：�c�d��载器�Q�class loader�Q�加载程序运行所需要的所有类�Q�它通过区分本机文�g�pȝ��的类和网�l�系�l�导入的�c�d��加安全性，�q�可以限制�Q何的�Ҏ��伊木马程序，因�ؓ本机�c�L��L��先被加蝲�Q�一旦所有的�c�被加蝲完，执行文�g的内存划分就固定了，在这个时候特定的内存地址被分配给对应的符号引用，查找表（lookuo table�Q�也被徏立，�׃��内存划分发生在运行时�Q�解释器在受限制的代码区增加保护防止未授权的讉K��Q�然后字节码校验器（byte code verifier�Q�进行校验，主要执行下面的检查：�cȝ��合JVM规范的类文�g格式�Q�没有违反访问限�Ӟ��代码没有造成堆栈的上溢或者下溢，所有操作代码的参数�c�d��都是正确的，没有非法的数据类型�{换（例如��整型数转换成对象类型）发生�Q�校验通过的字节码被解释器�Q�interpreter�Q�执行，解释器在必要旉��过�q�行时系�l�执行对底层��g的合适调用。后三个�{�案是SL275中的原话�?/p>
§1.1.8       �?/h3>
Which fragments are correct in Java source file?

　　A. package testpackage;

　　public class Test{//do something...}

　　B. import java.io.*;

　　package testpackage;

　　public class Test{// do something...}

　　C. import java.io.*;

　　class Person{// do something...}

　　public class Test{// do something...}

　　D. import java.io.*;

　　import java.awt.*;

　　public class Test{// do something...}

------------------------

�{�案 ACD

　

§1.1.9       �?/h3>
Which of the following statements are legal?

　　A. long l = 4990;

　　B. int i = 4L;

　　C. float f = 1.1;

　　D. double d = 34.4;

　　E. double t = 0.9F.

----------------------------

�{�案　ADE

李云�?/a> 2009-10-16 11:37 发表评论

�q�行命��o	含义	�l�果
Java Derived	不启用assertion	Base Method Derived Method
Java -ea Derived	开启所有assertion	Java.lang.AssertionError:Assertion Failed:This is base
Java -da Derived	关闭所有assertion	Base Method Derived Method
Java -ea:Base Derived	仅打开Base的assertion	Java.lang.AssertionError:Assertion Failed:This is base
Java -ea:Derived Derived	仅打开Derived的assertion	Base Method Java.lang.AssertionError:Assertion Failed:This is derived

亚洲人成网7777777国产,亚洲欧洲精品无码AV,亚洲AV无码1区2区久久

Java认证之精屁�ȝ��

�W�九部分�Q�集合类框架

�W�一�?nbsp; 所有接口和class定义

§1.1.1 一

�W�四部分�Q�语�a�基础

�W�一�?nbsp; 垃圾攉���解析

�W�一�?nbsp; 断言 assert

§1.1.1 assertion的语法和语义

�W�一部分�Q�声明和讉K��控制(�l�习�?

§1.1.9 �?/h3> Which of the following statements are legal? A. long l = 4990; B. int i = 4L; C. float f = 1.1; D. double d = 34.4; E. double t = 0.9F. ---------------------------- �{�案 ADE

�W�一�?nbsp; 垃圾攉��解析

§1.1.9 �?/h3>
Which of the following statements are legal?

　　A. long l = 4990;

　　B. int i = 4L;

　　C. float f = 1.1;

　　D. double d = 34.4;

　　E. double t = 0.9F.

----------------------------

�{�案　ADE