分��n代码�p�d��——vlist

changedi — Sun, 15 Apr 2012 04:29:00 GMT

摘要: vlist是一�U�列表的实现。结构如下图�Q? �Q�图来源wikipedia�Q?�c�M��链接表的�l�构�Q�但是，不是�U�性的。它的结构基于一�U?的幂�ơ扩展，�W�一个链接节点包含了列表的前一半数据，�W�二个包含了剩下一半的一半，依次递归。节点的基本�l�构不像LinkedList的节�Ҏ��双端队列�Q�每个VListCell包含了下个节点的指针mNext和前一个节点的指针mPrev�Q�同时内�|�一个数�l�mElems用来... 阅读全文

changedi 2012-04-15 12:29 发表评论

changedi — Fri, 06 Apr 2012 04:20:00 GMT

jdk中的Integer�c�L��int对象的包装类�Q�正常的Integer占用内存开销要比int大，比例大概�?:4 。今天分享的代码是Integer�c�M��的静态方法parseInt(String, int)。这个方法众所周知�Q�甚臛_��我们一开始学习编�E�时��尝试的写过�q�样的代码，一个正常的思�\�Q�遍历输入的字符数组�Q�java的字�W�串��是一个字�W�数�l�）�Q�然后parse每个char�Q�依据参数给定的�q�制�Q�判断每个char是否满��Q�满��_��l�箋�Q�否则抛出异常或中断�Q�直到处理完毕所有字�W�，�q�回�l�果�?/p>

那么我们看看jdk�l�出的实玎ͼ�

    public static int parseInt(String s, int radix)
        throws NumberFormatException
    {
        if (s == null) {
            throw new NumberFormatException("null");
        }

    if (radix < Character.MIN_RADIX) {
        throw new NumberFormatException("radix " + radix +
                        " less than Character.MIN_RADIX");
    }

    if (radix > Character.MAX_RADIX) {
        throw new NumberFormatException("radix " + radix +
                        " greater than Character.MAX_RADIX");
    }

    int result = 0;
    boolean negative = false;
    int i = 0, max = s.length();
    int limit;
    int multmin;
    int digit;

    if (max > 0) {
        if (s.charAt(0) == '-') {
        negative = true;
        limit = Integer.MIN_VALUE;
        i++;
        } else {
        limit = -Integer.MAX_VALUE;
        }
        multmin = limit / radix;
        if (i < max) {
        digit = Character.digit(s.charAt(i++),radix);
        if (digit < 0) {
            throw NumberFormatException.forInputString(s);
        } else {
            result = -digit;
        }
        }
        while (i < max) {
        // Accumulating negatively avoids surprises near MAX_VALUE
        digit = Character.digit(s.charAt(i++),radix);
        if (digit < 0) {
            throw NumberFormatException.forInputString(s);
        }
        if (result < multmin) {
            throw NumberFormatException.forInputString(s);
        }
        result *= radix;
        if (result < limit + digit) {
            throw NumberFormatException.forInputString(s);
        }
        result -= digit;
        }
    } else {
        throw NumberFormatException.forInputString(s);
    }
    if (negative) {
        if (i > 1) {
        return result;
        } else {    /* Only got "-" */
        throw NumberFormatException.forInputString(s);
        }
    } else {
        return -result;
    }
    }

�q�程��是按照思�\来的�Q�但是更全面一些，首先做一些参数检查，然后定义了局部变量用于计��：result是对应的int�l�果�Q�negative对应是否是负数的判断�Q�i是遍历用的烦引指针，max代表字符串的长度�Q�limit是合法数字的上限�Q�下限）�Q�digit是当前扫描到的字�W�对应的数字�Q�multmin是在做乘法计��时能走到的合法下限�?/p>

严�}是这�D늨�序最大的特点�Q�因为有�W�号int的上下限�?2147483648~2147483647�Q�可见负数表辄��范围比正数多一个，�q�样��好理解��Z��么在开头要把limit全部表达��敎ͼ�下限�Q�，�q�样的操作减��了后箋的判断，可以一步到位，相当于二者选择取其大一��P��大的包含了小的。同理，那么multmin也就是负��C��Q�而且可以认�ؓ是只和进制参数radix有关�p�R��接着每个char的扫描计��digit利用��C��Character.digit(char,int) �Ҏ��Q�这个方法就是在调用CharacterDataLatin1.digit(codePoint, radix) �Ҏ��Q�而这个新的方法其实只是去静态数�l�中取个映射而已。最后当��利的执行完while循环后，result�l�果也就计算好了�?/p>

作�ؓ�E�序设计人员�Q�我最初接触的语言是C++�Q�当初用到的库函数是atoi�Q�那么我们看看atoi的库标准实现�Q?/p>

int
atoi(str)
	const char *str;
{
	_DIAGASSERT(str != NULL);

	return((int)strtol(str, (char **)NULL, 10));
}

其中调用了strtol�Ҏ���Q�参��C��递的radix�?0�Q�也���是说我们常用的atoi是默认�{化字�W�串�?0�q�制的。其中开始时�q�进行了一个trim的操作，而且支持16�q�制�?x开��_��可谓完全的尽善尽���啊�?/font>

strtol�Ҏ���Q?/font>

#define	_FUNCNAME	strtol
#define	__INT		long
#define	__INT_MIN	LONG_MIN
#define	__INT_MAX	LONG_MAX

__INT
_FUNCNAME(const char *nptr, char **endptr, int base)
{
	const char *s;
	__INT acc, cutoff;
	char c;
	int i, neg, any, cutlim;

	_DIAGASSERT(nptr != NULL);
	/* endptr may be NULL */

	/* check base value */
	if (base && (base < 2 || base > 36)) {
		errno = EINVAL;
		return(0);
	}

	/*
	 * Skip white space and pick up leading +/- sign if any.
	 * If base is 0, allow 0x for hex and 0 for octal, else
	 * assume decimal; if base is already 16, allow 0x.
	 */
	s = nptr;
	do {
		c = *s++;
	} while (isspace(c));
	if (c == '-') {
		neg = 1;
		c = *s++;
	} else {
		neg = 0;
		if (c == '+')
			c = *s++;
	}
	if ((base == 0 || base == 16) &&
	    c == '0' && (*s == 'x' || *s == 'X')) {
		c = s[1];
		s += 2;
		base = 16;
	}
	if (base == 0)
		base = c == '0' ? 8 : 10;

	/*
	 * Compute the cutoff value between legal numbers and illegal
	 * numbers.  That is the largest legal value, divided by the
	 * base.  An input number that is greater than this value, if
	 * followed by a legal input character, is too big.  One that
	 * is equal to this value may be valid or not; the limit
	 * between valid and invalid numbers is then based on the last
	 * digit.  For instance, if the range for longs is
	 * [-2147483648..2147483647] and the input base is 10,
	 * cutoff will be set to 214748364 and cutlim to either
	 * 7 (neg==0) or 8 (neg==1), meaning that if we have accumulated
	 * a value > 214748364, or equal but the next digit is > 7 (or 8),
	 * the number is too big, and we will return a range error.
	 *
	 * Set any if any `digits' consumed; make it negative to indicate
	 * overflow.
	 */
	cutoff = neg ? __INT_MIN : __INT_MAX;
	cutlim = (int)(cutoff % base);
	cutoff /= base;
	if (neg) {
		if (cutlim > 0) {
			cutlim -= base;
			cutoff += 1;
		}
		cutlim = -cutlim;
	}
	for (acc = 0, any = 0;; c = *s++) {
		if (!isascii(c))
			break;
		if (isdigit(c))
			i = c - '0';
		else if (isalpha(c))
			i = c - (isupper(c) ? 'A' - 10 : 'a' - 10);
		else
			break;
		if (i >= base)
			break;
		if (any < 0)
			continue;
		if (neg) {
			if (acc < cutoff || (acc == cutoff && i > cutlim)) {
				any = -1;
				acc = __INT_MIN;
				errno = ERANGE;
			} else {
				any = 1;
				acc *= base;
				acc -= i;
			}
		} else {
			if (acc > cutoff || (acc == cutoff && i > cutlim)) {
				any = -1;
				acc = __INT_MAX;
				errno = ERANGE;
			} else {
				any = 1;
				acc *= base;
				acc += i;
			}
		}
	}
	if (endptr != 0)
		/* LINTED interface specification */
		*endptr = __DECONST(char *, (any ? s - 1 : nptr));
	return(acc);
}

当然�Q�类似的代码�q�有很多�Q�这里只列出了两大语�a�的库实现�Q���M��思�\是一致的�Q�当我们设计api�Ӟ���q�种�~�程思�\和风��g��及功能的考虑是我们需要学习的�?/font>

下面�q�两���stackoverflow的问�{�给��Z��一些比较全面的c风格代码�Q�可以参考，�q�里不脓全文只给link�Q?/font>

http://stackoverflow.com/questions/194465/how-to-parse-a-string-to-an-int-in-c

http://stackoverflow.com/questions/4442658/c-parse-int-from-string

参考文献：

jdk文档及源�?/font>

c库函数源码及文档

changedi 2012-04-06 12:20 发表评论

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分��n代码�p�d��——vlist

分��n一个ThreadMonitor