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yuxh — Wed, 29 Mar 2017 08:01:00 GMT

最�q�在�|�上看到的相��x��料都比较陈旧�Q�也太简略，参看了一下其他�h的内容，针对Hive2.1.1做点分��n�Q?br />

1�Q�下载apache-hive-2.1.1-bin.tar.gz

2�Q�解压羃�Q�下面的命��o(h��)行如启动报错�Q�请自行查略Hive启动配置

3�Q�启�?span style="line-height: 1.5;">hiveserver2 �Q�非必须�Q��用jdbc讉K��的时候才使用�Q?br />

bin目录�?/div>

hive --service hiveserver2 -p10001来启�?span style="line-height: 1.5;">hiveserver2 服务(默认�?0000端口)

nohup hive --service hiverserver2 -p10001可以在后台跑

4�Q?span style="line-height: 1.5;">hive脚本�q�行��程

bin目录下，使用命��o(h��)�Ҏ(gu��)��为：(x��)

./hive --service serviceName

举例�Q�hive --debug :

查看bin/hive文�g

��程中会(x��)判断$1=‘--debug’�?DEBUG=‘--debug’

if [ "$DEBUG" ]; then

if [ "$HELP" ]; then //如还�?-help�Q�就�?x��)执行debug_help�Ҏ(gu��)��?/span>

debug_help

exit 0

else

get_debug_params "$DEBUG"

export HADOOP_CLIENT_OPTS="$HADOOP_CLIENT_OPTS $HIVE_MAIN_CLIENT_DEBUG_OPTS"http://讄��HIVE_MAIN_CLIENT_DEBUG_OPTS的参��C��加入debug相应参数

if [ "$SERVICE" = "" ] ; then

if [ "$HELP" = "_help" ] ; then

SERVICE="help"

else

SERVICE="cli" //默认赋值cli

�q�个shell脚本很多变量应该是在其他sh文�g中定义，其中$SERVICE_LIST��是其他很多sh文�g的最开始�Ş成的�Q�export SERVICE_LIST="${SERVICE_LIST}${THISSERVICE} "

hive脚本最后的$TORUN "$@" �Q�默认情况下TORUN其实��是cli,��x��?ext/cli.sh脚本�Q�该脚本中主要是调用/ext/util/execHiveCmd.sh 来执行最后的CliDriver�?/div>

�?span style="line-height: 1.5;">shell脚本中的$*�Q?@�?#

举例��_(d��)��(x��)

脚本名称叫test.sh 入参三个: 1 2 3

�q�行test.sh 1 2 3�?/div>

$*�?1 2 3"�Q�一赯��引号包住�Q?/div>

$@�?1" "2" "3"�Q�分别被包住�Q?/div>

$#�?�Q�参数数量）

�?/span>

即exec $HADOOP jar ${HIVE_LIB}/$JAR $CLASS $HIVE_OPTS "$@" //1

其中:

$HADOOP=$HADOOP_HOME/bin/hadoop 【hive脚本中定�?/span>HADOOP=$HADOOP_HOME/bin/hadoop�?/span>

$CLASS=org.apache.hadoop.hive.cli.CliDriver【传入的�W�一个参敎ͼ�在cli.sh中有定义�?/div>

hadoop脚本�Q?.7.3��Z��Q�中最�l�会(x��)执行�Q?/div>

# Always respect HADOOP_OPTS and HADOOP_CLIENT_OPTS

HADOOP_OPTS="$HADOOP_OPTS $HADOOP_CLIENT_OPTS"

#make sure security appender is turned off

HADOOP_OPTS="$HADOOP_OPTS -Dhadoop.security.logger=${HADOOP_SECURITY_LOGGER:-INFO,NullAppender}"

export CLASSPATH=$CLASSPATH

exec "$JAVA" $JAVA_HEAP_MAX $HADOOP_OPTS $CLASS "$@" //2

hive的debug参数��是在启动hive脚本时放到HADOOP_OPTS中的

1�?处结合可得到最�l�的�q�行命��o(h��)�Q�查看一下运行结果：(x��)ps -ef|grep CliDriver

/usr/java/jdk1.8.0_101/bin/java -Xmx256m -Djava.net.preferIPv4Stack=true -Dhadoop.log.dir=.. -Dhadoop.log.file=hadoop.log -Dhadoop.home.dir=.. -Dhadoop.id.str=root -Dhadoop.root.logger=INFO,console -Djava.library.path=.. -Dhadoop.policy.file=hadoop-policy.xml -Djava.net.preferIPv4Stack=true -Xmx512m -Dproc_hivecli -XX:+UseParallelGC -agentlib:jdwp=transport=dt_socket,server=y,address=8000,suspend=y -Dlog4j.configurationFile=hive-log4j2.properties -Djava.util.logging.config.file=..
-Dhadoop.security.logger=INFO,NullAppender org.apache.hadoop.util.RunJar /yuxh/app/apache-hive-2.*/lib/hive-cli-2.*.jar org.apache.hadoop.hive.cli.CliDriver

yuxh 2017-03-29 16:01 发表评论

JBoss Forge 2入门教程

yuxh — Wed, 23 Sep 2015 13:14:00 GMT

摘要: 阅读全文

yuxh 2015-09-23 21:14 发表评论

appfuse3.5打印Hibernate sql参数

yuxh — Wed, 22 Jul 2015 06:11:00 GMT

appfuse3.5使用Hibernate4.3.6, 而Hibernate日志框架使用了jboss-logging�Q�想在后台打出sql的参��C��直无法生效�?/div>

��(g��)查配�|�文�Ӟ��框架里面的两个配�|�文�Ӟ��src/test/resources/log4j2.xml�Q�单元测试时配置�Q�，src/main/resources/log4j2.xml�Q�运行时配置�Q?/div>

搞清log4j2的配�|�后�Q�各�U�修改（主要�?/div>

)
用junit��试�ȝ��无法打印出真实参数。根据这些实践，��定log4j2是��用无误生效的�Q�只是org.hibernate�q�部分的logger一直未��h��

参考国内外�|�站�Q�一直无人回�{�hibernate4的这个问题，有�h指出�q�部分Hibernate官方文档只是提了一句，一直未更新相关内容。最后有人提到应该是 jboss-logging �?/span>LoggerProviders�q�个�cȝ��问题�Q�看实现对log4j2已经做支持。最后发�?nbsp;jboss-logging使用的是3.2.0.beta�Q�对比相关类的源代码�Q�更改�ؓ(f��)3.2.0.Final�Q�生效！

P.S 把这个问题提交给Appfuse官网�Q�issue APF-1478�Q�作者标志�ؓ(f��)4.0版本修复�?/span>

yuxh 2015-07-22 14:11 发表评论

Andriod webview实现文�g上传

yuxh — Mon, 12 Mar 2012 04:54:00 GMT

Andriod �?.2版本为止�Q�webview方式下��?lt;input type="file" />点击后都没有反应。实际上��层是有隐含的接口没实现的，可以自己重写�q�个�Ҏ(gu��)��来实现。以phonegap��Z��Q?br />

public class App extends DroidGap {
    private ValueCallback<Uri> mUploadMessage;
    private final static int FILECHOOSER_RESULTCODE = 1;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        super.init();
        // WebView wv = new WebView(this);
        // wv.setWebViewClient(new WebViewClient());
        this.appView.setWebChromeClient(new CordovaChromeClient(App.this) {
            // For Android 3.0+
            public void openFileChooser(ValueCallback<Uri> uploadMsg, String acceptType) {
                mUploadMessage = uploadMsg;
                Intent i = new Intent(Intent.ACTION_GET_CONTENT);
                i.addCategory(Intent.CATEGORY_OPENABLE);
                i.setType("image/*");
                App.this.startActivityForResult(Intent.createChooser(i, "File Chooser"), FILECHOOSER_RESULTCODE);
            }

            // The undocumented magic method override
            // Eclipse will swear at you if you try to put @Override here
            public void openFileChooser(ValueCallback<Uri> uploadMsg) {
                mUploadMessage = uploadMsg;
                Intent i = new Intent(Intent.ACTION_GET_CONTENT);
                i.addCategory(Intent.CATEGORY_OPENABLE);
                i.setType("image/*");
                App.this.startActivityForResult(Intent.createChooser(i, "File Chooser"), App.FILECHOOSER_RESULTCODE);
            }
        });

        // setContentView(wv);
        super.loadUrl("file:///android_asset/www/login.html");
    }

    @Override
    protected void onActivityResult(int requestCode, int resultCode, Intent intent) {
        if (requestCode == FILECHOOSER_RESULTCODE) {
            if (null == mUploadMessage)
                return;
            Uri result = intent == null || resultCode != RESULT_OK ? null : intent.getData();
            mUploadMessage.onReceiveValue(result);
            mUploadMessage = null;
        }
    }
}

如果直接的webview方式�Q�extends WebChromeClient卛_��? 参考：(x��)http://stackoverflow.com/questions/5907369/file-upload-in-webview

yuxh 2012-03-12 12:54 发表评论

RCP与Document风格

yuxh — Sat, 08 Oct 2011 13:47:00 GMT

有两�U�SOAP message风格�Q?em>document �?em>RPC�Q?/em>他们定义了SOAP message body的格式。��用document风格�Ӟ��包括wrapped和unwrapped�Q�，在wsdl中有一个非�I�的types部分�Q�这个部分用XML Schema language定义了web service要用到的�c�d��?em>wsgen工具从SIB�Q�有SEI��p��够了�Q�中生成与XSD对应的java�c�R��用java代码生成WSDL文�g的时候需要一些java�c�，wsgen工具可以生成�q�些Java�c�，生成的这些java�c�被�U�Cؓ(f��)wsgen artifacts�Q�底层的JWS�c�d��?x��)用到这些类�Q�特别是JAX-B�p�d��的包�Q�会(x��)用来转换(marshal)java�c�d��?that is, Java in-memory objects)为XML�c�d��的XML实例�Q�满��XML Schema document的XML文档实例�Q�，

The inverse operation is used to convert (unmarshal) an XML document instance to an in-memory
object, an object of a Java type or a comparable type in some other language。因�?em>wsgen工具生成的artifacts�Q�支持了Java为基��的web service的互操作性。JAX-B�c�d��提供了Java和XSD�c�d��转换的底层支持�?/p>

For the most part, the wsgen utility can be used without our bothering to inspect the artifacts that it produces. For the most part, JAX-B remains unseen infrastructure.

wsgen artifacts实际上是wsdl message的数据类型，他们和XML Schema type�l�定�Q�每个message的XML Schema types从这些java�c�d��得来的。注�Q�在当前的jdk1.6.24中，已经融入wsgen自动生成的过�E�，不需手动调用�?/p>

wsgen工具可用来生成wsdl文�g�Q�如�Q? wsgen -cp "." -wsdl ch01.ts.TimeServerImpl 。这为TimeServer服务生成了wsdl。用wsgen生成的wsdl和通过讉K��发布的服务生成的wsdl 有个很大的区别：(x��)wsgen生成的没有endpoint,因�ؓ(f��)�q�个URL是在发布服务的时候决定的。其他地方两个wsdl是相同的�?/p>

wsimport(以前叫wsdl2java�?java2wsdl更�Ş�?em>)工具可��用WSDL生成用来帮助写客��L(f��ng)��的artifacts .

1 先契�U�再�~�码方式
一个例子：(x��)得到一个tempConvert.wsdl文�g�Q��用命�?wsimport -keep -p ch02.tc tempConvert.wsdl �Q�命令根据wsdl的portType生成一个SEI�c�，把SEI的interface换�ؓ(f��)class�Q�再把方法改为实现就可变为SIB。把该SIB发布�Q�再使用命��o(h��)wsimport -keep -p clientTC http://localhost:5599/tc?wsdl�Q�来生成客户端辅助类
2 �~�码优先
服务被发布之后，�?x��)自动生成WSDL供客��L(f��ng)��使用。然而，使用annotations可以控制WSDL或WSDL-generated artifacts的生成�?/p>

yuxh 2011-10-08 21:47 发表评论

区分getPath(), getAbsolutePath(), getCanonicalPath()

yuxh — Fri, 24 Jun 2011 05:42:00 GMT

来自

http://stackoverflow.com/questions/1099300/whats-the-difference-between-getpath-getabsolutepath-and-getcanonicalpath

C:\temp\file.txt" - this is a path, an absolute path, a canonical path

.\file.txt This is a path, It's not an absolute path nor canonical path.

C:\temp\myapp\bin\..\\..\file.txt This is a path, and an absolute path, it's not a canonical path

Canonical path is always an absolute path.

Converting from a path to a canonical path makes it absolute (通常�?x��)处理改变当前目录，所以像. ./file.txt 变�ؓ(f��)c:/temp/file.txt). The canonical path of a file just "purifies" the path, 去除和解析类�?#8220; ..\” and resolving symlinks(on unixes)

In short:

getPath() gets the path string that the File object was constructed with, and it may be relative current directory.
getAbsolutePath() gets the path string after resolving it against the current directory if it's relative, resulting in a fully qualified path.
getCanonicalPath() gets the path string after resolving any relative path against current directory, and removes any relative pathing (. and ..), and any file system links to return a path which the file system considers the canonical means to reference the file system object to which it points.

Also, each of this has a File equivalent which returns the corresponding File object.

The best way I have found to get a feel for things like this is to try them out:

import java.io.File;
public class PathTesting {
        public static void main(String [] args) {
                File f = new File("test/.././file.txt");
                System.out.println(f.getPath());
                System.out.println(f.getAbsolutePath());
                try {
                        System.out.println(f.getCanonicalPath());
                }
                catch(Exception e) {}
        }
}

Your output will be something like:

test\..\.\file.txt
C:\projects\sandbox\trunk\test\..\.\file.txt
C:\projects\sandbox\trunk\file.txt

So, getPath() gives you the path based on the File object, which may or may not be relative; getAbsolutePath() gives you an absolute path to the file; and getCanonicalPath() gives you the unique absolute path to the file. Notice that there are a huge number of absolute paths that point to the same file, but only one canonical path.

When to use each? Depends on what you're trying to accomplish, but if you were trying to see if two Files are pointing at the same file on disk, you could compare their canonical paths.

yuxh 2011-06-24 13:42 发表评论

XML介绍

yuxh — Thu, 24 Mar 2011 01:05:00 GMT

DTDs

Introduced as part of the XML 1.0 specification, DTDs are the oldest constraint model around in the XML world. They're simply to use, but this simplicity comes at a price: DTDs are inflexible, and offer you little for data type validation as well.

XML Schema (XSD)

XML Schema is the W3C's anointed successor to DTDs. XML Schemas are literally orders of magnitude more flexible than DTDs, and offer an almost dizzying array of support for various data types. However, just as DTDs were simple and limited, XML Schemas are flexible, complex, and (some would argue) bloated. It takes a lot of work to write a good schema, even for 50- or 100-line XML documents. For this reason, there's been a lot of dissatisfaction with XML Schema, even though they are widely being used.

[prefix]:[element name]

元素�Q?br>root元素必须包含所有文档中的元素，只能有一个root元素。元素名只能以下划线或字母开��_(d��)��不能有空��|��区分大小写。开元素必须有对应闭元素�Q�也有类似html的简写，�?lt;img src="http://www.tkk7.com/images/xml.gif" />�Q�。文档由DTD或schema来限制它是否合格�?/pre>属性：(x��)
什么时候用属性？基本原则�Q�多个值的数据用元素，单值的数据用元素。如果数据有很多值或者比较长�Q�数据最可能属于元素。他主要被当作文本，�?lt;rss:author>Doug Hally Neil Gaiman易搜索，好用。比如一本书的章节描�q�。然而如果数据主要作为单值处理的话，最好作为属性。如果搞不清楚，可以安全的��用元素�?/pre>命名�I�间Namespaces�Q?br>xml的命名空间是一�U�用一个特定的URI来关联XML文档里的一个或多个元素的方法。意味着元素是由名字和命名空间一��h��识别的。许多复杂的XML文�g里，同一个名字会(x��)有多个用途。比如，一个RSS feed有一个作者，�q�个作者同时是每个日记的。虽然这些数据都用author元素来表�C�，但他们不应该被当作同一�?em>�c�d��的数据。命名空间很好的解决了这个问题，命名�I�间说明书要求一个前�~�和唯一的URI联合��h��区分不同命名�I�间里的元素。如http://www.neilgaiman.com/entries作�ؓ(f��)URI�Q�联合前�~�journal用来表示日志相关的元素�?br>�Q�然后就可��用了�Q?br>Doug Hally Neil Gaiman实际上在使用命名�I�间前缀的时候再定义也可以的�Q?br>Doug Hally
如果名字没有命名�I�间�Q�不代表在默认命名空间中�Q�而是xml处理器以在�Q何命名空间以外方式解释他。要声明默认命名�I�间的话��׃��用后面冒号部分，�?lt;Insured xmlns=" 
使用的一些术�?/strong>:



The name of a namespace (such as http://www.ibm.com/software) is the namespace URI.

The element or attribute name can include a prefix and a colon (as in prod:Quantity). A name in that form is called a qualified name, or QName, and the identifier that follows the colon is called a local name. If a prefix is not in use, neither is the colon, and the QName and local name are identical.

An XML identifier (such as a local name) that has no colon is sometimes called an NCName. (The NC comes from the phrase no colon.)
Entity references:
用来处理转义字符�Q�语法是& [entity name] ;XML解析器碰到这�U�entity reference�Q�就�?x��)用对应的值替换掉他。如<�Q?lt;�Q?>(>),&(&),"("),'(')。注意entity reference是用户可定义的。比如多处用到版权提�C�，自己定义后以后更改就方便了：(x��)&OReillyCopyright;。除了用来表�C�数据中的复杂或�Ҏ(gu��)��字符外，entity reference�q�会(x��)有更多用途�?/p>
不解析的数据�Q?br>当传输大量数据给应用且不用xml解析的时候，CDATA���有用了。当大量的字�W�需要用entity reference转义的时候，或空格必���M��留的时候，使用CDATA�?lt;![CDATA[�?]]>


yuxh 2011-03-24 09:05 发表评论

js分割csv文�g

yuxh — Fri, 03 Dec 2010 01:06:00 GMT
转自http://www.bennadel.com/blog/1504-Ask-Ben-Parsing-CSV-Strings-With-Javascript-Exec-Regular-Expression-Command.htm�Q�把csv文�g按分隔符切割后放在数�l�中�?
// This will parse a delimited string into an array of arrays.
   // The default delimiter is the comma, but this
// can be overriden in the second argument.
    CSVToArray:function(strData, strDelimiter){

        // Check to see if the delimiter is defined. If not,
        // then default to comma.
        strDelimiter = (strDelimiter || ",");
        // Create a regular expression to parse the CSV values.
        var objPattern = new RegExp(
            (
                // Delimiters.
                "(\\" + strDelimiter + "|\\r?\\n|\\r|^)" +

                // Quoted fields.
                "(?:\"([^\"]*(?:\"\"[^\"]*)*)\"|" +

                // Standard fields.
                "([^\"\\" + strDelimiter + "\\r\\n]*))"
            ),
            "gi"
            );

        // Create an array to hold our data. Give the array
        // a default empty first row.
        var arrData = [[]];

        // Create an array to hold our individual pattern
        // matching groups.
        var arrMatches = null;

        // Keep looping over the regular expression matches
        // until we can no longer find a match.
        while (arrMatches = objPattern.exec( strData )){

            // Get the delimiter that was found.
            var strMatchedDelimiter = arrMatches[ 1 ];

            // Check to see if the given delimiter has a length
            // (is not the start of string) and if it matches
            // field delimiter. If id does not, then we know
            // that this delimiter is a row delimiter.
            if (
                strMatchedDelimiter.length &&
                (strMatchedDelimiter != strDelimiter)
                ){

                // Since we have reached a new row of data,
                // add an empty row to our data array.
                arrData.push( [] );
            }

            // Now that we have our delimiter out of the way,
            // let's check to see which kind of value we
            // captured (quoted or unquoted).
            if (arrMatches[ 2 ]){

                // We found a quoted value. When we capture
                // this value, unescape any double quotes.
                var strMatchedValue = arrMatches[ 2 ].replace(
                    new RegExp( "\"\"", "g" ),
                    "\""
                    );

            } else {

                // We found a non-quoted value.
                var strMatchedValue = arrMatches[ 3 ];
            }

            // Now that we have our value string, let's add
            // it to the data array.
            arrData[ arrData.length - 1 ].push( strMatchedValue );
        }

        // Return the parsed data.
        return( arrData );
    }

yuxh 2010-12-03 09:06 发表评论

yuxh — Mon, 15 Nov 2010 11:59:00 GMT

31
while (i != 0)
    i >>>= 1; //无符号右�U�，不管正负左边都是�?
��Z��表达式合法，�q�里的i必须是整�?byte, char, short, int, or long)。谜题的关键在于>>>= 是一个复合赋值操作符�Q�不�q�的是复合赋值操作符�?x��)默默的�?span class="docEmphasis">narrowing primitive conversions�Q�即从一个数据类型�{换�ؓ(f��)一个更��的数据�c�d��?span class="docEmphStrong">Narrowing primitive conversions can lose information about the magnitude or precision of numeric values。�ؓ(f��)了��问题更具体，假设�q�样定义�Q?br>short i = -1; 因�ؓ(f��)初始值i ((short)0xffff) 非零�Q��@环执行。第一步位�U�M��(x��)把i提升为int�?tt>short, byte, or char�c�d��的操作数都会(x��)做这��L(f��ng)��操作。这�?span class="docEmphasis">widening primitive conversion�Q�没有信息丢失。这�U�提升有�W�号扩展�Q�因此结果是int�?tt>0xffffffff。无�W�号右移一位��生int�?tt>0x7fffffff。�ؓ(f��)了把int值存回short变量�Q�Java执行了可怕的narrowing primitive conversion,即简单去掉高十六位。这样又变回�?tt>(short)0xffff。如果定义类�?tt>short or byte型的负数�Q�都�?x��)得到类似结果。你如果定义的是char话，则不�?x��)无限��@环，因�ؓ(f��)char值非负，位移之前的宽扩展不会(x��)做符��h��展�?br>
�ȝ��Q?span style="color: red">不要�?tt>short, byte, or char变量上��用复合赋值操作符。这�U�表辑ּ��q�行混合�c�d��计算�Q�非常容易�؜淆。更�p�糕的是隐含的窄(ji��ng)映射�?x��)丢掉信息�?br>
32
while (i <= j && j <= i && i != j) {}
i <= j and j <= i, surely i must equal j�Q�对于实数来说是�q�样的。他非常重要�Q�有个名�U�ͼ�(x��)The �?relation on the real numbers is said to be antisymmetric。Java's <= operator used to be antisymmetric before release 5.0, but no longer.Java 5之前数据比较�W�号(<, <=, >, and >=) 需要两边的操作数必��Mؓ(f��)基础�c�d��(byte, char, short, int, long, float, or double).在Java 5变�ؓ(f��)两边操作��Cؓ(f��)凡是可�{变�ؓ(f��)基础�c�d��的类型。java 5引入autoboxing and auto-unboxing 。The boxed numeric types are Byte, Character, Short, Integer, Long, Float, and Double。具体点�Q�让上面�q�入无限循环�Q?br>Integer i = new Integer(0);
Integer j = new Integer(0);
(i <= j and j <= i) perform unboxing conversions on i and j and compare the resulting int values numerically。i和j表示0�Q�所以表辑ּ�为true�?tt>i != j比较的是对象引用�Q�也为true。很奇怪规范没有把�{�号改�ؓ(f��)比较倹{��原因很��单：(x��)兼容性。当一�U�语�a��q�泛应用的时候，不能破坏已存在的规范来改变程序的行�ؓ(f��)。System.out.println(new Integer(0) == new Integer(0));��L��输出false�Q�所以必��M��留。当一个是boxed numeric �c�d��Q�另一个是基本�c�d��的时候可以值比较。因为java 5之前�q�是非法的，具体点：(x��)
System.out.println(new Integer(0) == 0); //之前的版本非法，Java 5输出True
�ȝ��Q?span style="color: red">当两边的操作数是boxed numeric�c�d��的时候，数字比较�W�和�{�于�W�号是根本不同的�Q�数字比较符是值比较，�{�号比较的是对象引用�?br>
33
while (i != 0 && i == -i) {}
有负可��C�i一定是数字�Q�NaN不行�Q�因��Z��不等于�Q何数。事实上�Q�没有实数可以出现这�U�情��c(di��n)��但Java的数字类型�ƈ没有完美表达实数。��Q�Ҏ(gu��)��׃��个符号位�Q�一个有效数字（��数�Q�，一个指数构成。��Q�Ҏ(gu��)��只有0才会(x��)和自��q��负数相等�Q�所以i肯定是整数。有�W�号整数用的是二�q�制补码计算�Q�取反加一。补码的一大优势是用一个唯一的数来表�C?。然而有一个对应的�~�点�Q�本来可表达偶数个��|��现在用一个表达了0�Q�剩下奇��C��来表�C�正负数�Q�意味着正数和负数的数量不一栗��比如int��|��他的Integer.MIN_VALUE�Q?tt>-2³¹�Q�。十六进制表�?tt>0x8000000。通过补码计算可知他的负数仍然不变。对他取负数是溢��Z��的，不过Java在整数计��中忽略了溢出�?br>�ȝ��Q�Java使用二进制补码，是不对称的。有�W�号整数(int, long, byte, and short) 负数值比整数值多一个�?br>
34
        final int START = 2000000000;
        int count = 0;
        for (float f = START; f < START + 50; f++)
            count++;
        System.out.println(count);
注意循环变量是float。回忆谜�?8 �Q�明显f++没有��M��作用。f的初始化值接�q?tt>Integer.MAX_VALUE�Q�因此需�?1位来准确表达�Q�但是float�c�d��只提供了24位精度。增加这样大的一个float��g��?x��)改变倹{��看��h��?x��)死循环�Q�运行程序会(x��)发现�Q�输�?。��@环中f�?tt>(float)(START + 50)做比较。但int和float比较的时候，自动把int先提�C�Zؓ(f��)float。不�q�的是三个会(x��)引�v�_�ֺ�丢失的宽基本�c�d��转换的其中之一�Q�另外两个是long到float�Q�long到double�Q�。f的初始值巨大，�?0再�{换�ؓ(f��)float和直接把f转换为float是一��L(f��ng)��效果�Q�即(float)2000000000 == 2000000050�Q�所�?tt>f < START + 50 ��p�|。只要把float改�ؓ(f��)int卛_��修正�?br>     没有计算器，你怎么知道 2,000,000,050 和float表示2,000,000,000一��P��…�?br>The moral of this puzzle is simple: Do not use floating-point loop indices, because it can lead to unpredictable behavior. If you need a floating-point value in the body of a loop, take the int or long loop index and convert it to a float or double. You may lose precision when converting an int or long to a float or a long to a double, but at least it will not affect the loop itself. When you use floating-point, use double rather than float unless you are certain that float provides enough precision and you have a compelling performance need to use float. The times when it's appropriate to use float rather than double are few and far between�?br>

35
下面�E�序模拟一个简单的旉��
int minutes = 0;
for (int ms = 0; ms < 60*60*1000; ms++)
if (ms % 60*1000 == 0)
     minutes++;
System.out.println(minutes);
�l�果�?tt>60000�Q�问题在于布?y��u)��(d��ng)表辑ּ�ms % 60*1000 == 0�Q�最��单的修改�Ҏ(gu��)��是：(x��)if (ms % (60 * 1000) == 0)
更好的方法是用合适命名的帔R��代替��力数字�Q?/tt>

private static final int MS_PER_HOUR = 60 * 60 * 1000;
private static final int MS_PER_MINUTE = 60 * 1000;
public static void main(String[] args) {
   int minutes = 0;
   for (int ms = 0; ms < MS_PER_HOUR; ms++)
     if (ms % MS_PER_MINUTE == 0)
        minutes++;
    System.out.println(minutes);
}
�l�不要用�I�格来分�l�；使用括号来决定优先��

yuxh 2010-11-15 19:59 发表评论

yuxh — Mon, 15 Nov 2010 08:39:00 GMT
24
for (byte b = Byte.MIN_VALUE; b < Byte.MAX_VALUE; b++) {
        if (b == 0x90)
            System.out.print("Joy!");
    }
Ox90 ��过了byte的取��D��?font face="Courier New">-128�?27。byte和int比较是一�U��؜合类型比较。考虑个表辑ּ�((byte)0x90 == 0x90)得到的是false。byte和int做比较的时候，Java先对byte�q�行了widening primitive conversion 再比较两个int倹{��因为byte是有�W�号�c�d��Q��{变做了符��h��展，把负的byte��D�{换�ؓ(f��)相应的int倹{��这个例子中(byte)0x90被�{变�ؓ(f��)-112�Q�当然不�{�于int�?0x90或者说+144。�؜合比较总让��惑，因�ؓ(f��)��L��pȝ��L��升一个操作数来和另一�U�类型匹配。有几种方式可避免�؜合比较。可以把int映射为byte�Q�之后比较两个byte��|��(x��) if (b == (byte)0x90)     System.out.println("Joy!"); 另外�Q�可用mask抑制�W�号扩展�Q�把byte转换为int�Q�之后比较两个int��|��(x��) if ((b & 0xff) == 0x90)     System.out.println("Joy!"); 但最好的�Ҏ(gu��)��是把帔R��值移出��@环放到常量声明中�?br />     private static final byte TARGET = 0x90; // Broken!     public static void main(String[] args) {         for (byte b = Byte.MIN_VALUE; b < Byte.MAX_VALUE; b++)             if (b == TARGET)                 System.out.print("Joy!");     } 不幸的是�Q�上面编译通不�q�：(x��)0x90对于byte�c�d��来说不是一个有效倹{��这样修改即可：(x��) private static final byte TARGET = (byte)0x90; To summarize: Avoid mixed-type comparisons, because they are inherently confusing (Puzzle 5). To help achieve this goal, use declared constants in place of "magic numbers." You already knew that this was a good idea; it documents the meanings of constants, centralizes their definitions, and eliminates duplicate definitions.现在你知道他�q�可以强制你为每一个常量定义适用的类型，避免一�U��؜合类型比较的来源�?/p> 25 int j = 0; for (int i = 0; i < 100; i++) j = j++; System.out.println(j); //打印出的�?
问题出在 j = j++; �{�同于下列操作：(x��)

int tmp = j; j = j + 1; j = tmp;
�q�次的教训和��N��7一��P��(x��)在一个表辑ּ�中不要给同一个变量赋��D��q�一�ơ�?/font>

26

public static final int END = Integer.MAX_VALUE;
public static final int START = END - 100;
public static void main(String[] args) {
    int count = 0;
    for (int i = START; i <= END; i++)
       count++;
     System.out.println(count);
}
看�v来像100�Q�再仔细看��@环是��于�{�于�Q�应该是101�Q�结果是�E�序没有输出��M��|��陷入一个死循环。问题出�?tt>Integer.MAX_VALUE�Q�当�l�箋增加的时候，他�?zh��n)��?zh��n)�变�?tt>Integer.MIN_VALUE。如果你需要��@环int��D��界，最好用long变量做烦(ch��)引：(x��)
for (long i = START; i <= END; i++) //输出101
教训是：(x��)ints are not integers。无��Z��时用基本�c�d��Q�注意边界倹{��上溢或下溢�?x��)出��C��么情况？一般来说最好用大一点的�c�d��Q�基本类型是byte, char, short, int, and long�Q��?/font>也可以不用long�Q?/tt>

int i = START;
do {
count++;
} while (i++ != END);
考虑到清晰和��单，��L��用long索引�Q�除了一�U�特�D�情况：(x��)如果要遍历所有int��|��q�样用int索引的话�?x��)快两倍�?br /> 一个��@环四十亿int��|��调用�Ҏ(gu��)��的常规用法：(x��)

// Apply the function f to all four billion int values int i = Integer.MIN_VALUE; do { f(i); } while (i++ != Integer.MAX_VALUE);

27 位移
��C��java是��用二�q�制补码计算�Q�在��M��有符号基本类型中(byte, short, int, or long)都是用所有位�|?来表�C?1�?
        int i = 0;
        while (-1 << i != 0)       //左位�U?br />             i++;
        System.out.println(i);
int型的-1�?tt>0xffffffff 表示。不断左�U�，双��?补位。移�?2�ơ，变�ؓ(f��)�?�Q�蟩出��@环打�?2�Q�实际上�E�序�?x��)死循环。问题出�?1<<32不等�?而是�{�于-1�Q�因��Z��Uȝ��号只用右�Ҏ(gu��)��作数的低五位作�ؓ(f��)�U�d��距离�Q�如果左操作数是long的话用六位。三个位�U�L��作符�Q?lt;<,>>,>>>都是�q�样。移动距��L��L��0�?1�Q�左�Ҏ(gu��)��作数是long的话0�?3。位�U�距��ȝ��32取模�Q�左�Ҏ(gu��)��long则用64取模。给int��g��U?2位或�l�long��g��U?4位只�?x��)返回本�w�。所以不可能用位�U�d��全移除一个数据。幸�q�的是，有一个简单的办法解决�q�个问题。保存上一�ơ的位移�l�果�Q�每一�ơ�P代多�U�d��一位�?br />         int distance = 0;
        for (int val = -1; val != 0; val <<= 1)
            distance++;
        System.out.println(distance);
修改后的�E�序说明了一个原则：(x��)位移距离如果可能的话�Q�用帔R��?br /> 另外一个问题，许多�E�序员认为右�U�M��个负的移动距��，��和左移一��P��反之亦然。事实上不是�q�样�Q�左�U�L��左移�Q�右�U�d��是右�U�R��负数距��d��留下低五位（long留六位）�Q�其余的�|?��变��Z��正数距离。比如，左移一个int�?1的距��，实际上是左移31位�?br />
28 无穷的表�C?br /> for (int i = start; i <= start + 1; i++) {
}
看�v来��@环两�ơ就�?x��)结束，如果�q�样定义呢：(x��)
int start = Integer.MAX_VALUE - 1;//��d�@�?br /> while (i == i + 1) {
}
�q�个不可能死循环�Q�如果i是无�I�呢�Q�Java采用IEEE 754��点数算术，用double或float来表�C�无�I��所以可以用��M��点数计��表辑ּ�得出无穷来初始化i。比如：(x��)double i = 1.0 / 0.0;
更好的是可以利用标准库提供的帔R��:double i = Double.POSITIVE_INFINITY;
事实上，�Ҏ(gu��)��用不着用无�I�初始化i来引��h��循环。只要��够大的��Q�Ҏ(gu��)��够了：(x��)double i = 1.0e40;
�q�是因�ؓ(f��)��点数越大，他的值和下一个数的��D��M��p��大。distribution of floating-point values is a consequence of their representation with a fixed number of significant bits. �l��够大的��Q�Ҏ(gu��)��加一不会(x��)改变��|��因�ؓ(f��)他不能填充这个数和下一个数之间的距��R��Q�Ҏ(gu��)��操作�q�回最接近准确数学�l�果的��Q点倹{��一旦两个相��L�Q点��g��间的距离大于2�Q�加1��׃��?x��)有效果。float�c�d��来说�Q�超�q?²⁵�Q�或33,554,432�Q�再�?��无效；对double来说��过2⁵⁴�Q�接�q?.8 x 10¹⁶�Q�再�?��无效�?br />      盔R��点��C��间的距离�U�Cؓ(f��)ulp�Q�unit in the last place的羃写）。在Java 5�?Math.ulp�Ҏ(gu��)��被引入来计算float或double值的ulp�?/tt>
�ȝ��Q?font color="#ff0000">不可能用float或double来表�C�无�I��另外，在一个大的��Q�Ҏ(gu��)��上加一个小的��Q�Ҏ(gu��)��Q��g��?x��)改变。有点不合常理，实数�q�不是这栗��记住二�q�制��点数计��只是近��g��实数计算�?/font>29 NaN while (i != i) { } IEEE 754 ��点数计��保留了一个特�D�值来来表�C�Z��是数字的数量。NaN是��Q点计��不能很好定义的敎ͼ�比如0.0 / 0.0。规范定�?font color="#ff0000">NaN不等于�Q何数包括自己。因此double i = 0.0 / 0.0; 可让开始的�{�式不成立。也有标准库定义的常量：(x��)double i = Double.NaN; 如果一个或多个操作��Cؓ(f��)NaN�Q�那么��Q�Ҏ(gu��)��计算��׃��(x��)�{�于NaN�?/p> �ȝ��Q�float和doule存在�Ҏ(gu��)��的值NaN�Q�小心处理�?/p> 30 while (i != i + 0) { } �q�一�ơ不能��用float或者double�?br /> +操作�W�除了数字，��只能处理String�?操作�W�会(x��)被重载：(x��)对于String�c�d��Q�他做的是连接操作。如果操作数有非String�c�d��Q�会(x��)先做转换变�ؓ(f��)String之后再做�q�接。i一般用作数字，要是对String型变量这么命名容易引赯��解�?/tt> �ȝ��Q?font color="#ff0000">操作�W�重载非常容易误��g�h。好的变量名�Q�方法名�Q�类名和好的注释对于�E�序的可��L��一样重要�?/font> yuxh 2010-11-15 16:39 发表评论