Java整数转换成字符

来源：互联网发布：java线程应用场景编辑：程序博客网时间：2024/05/11 20:37

Java中提供了三种整数转换成字符串的方式，大家都很熟悉：

1. String s = String.valueOf(i);
2. String s = Integer.toString(i);
3. String s = "" + i;
注：Double, Float, Long 转成字串的方法大同小异。

然而自己实现呢？

其实也很简单，我们只要参考Java的源代码即可实现。那么我们来分析一下Java中将整数转换成字符串的函数源码。

<pre name="code" class="java">    final static char [] DigitTens = {        '0', '0', '0', '0', '0', '0', '0', '0', '0', '0',        '1', '1', '1', '1', '1', '1', '1', '1', '1', '1',        '2', '2', '2', '2', '2', '2', '2', '2', '2', '2',        '3', '3', '3', '3', '3', '3', '3', '3', '3', '3',        '4', '4', '4', '4', '4', '4', '4', '4', '4', '4',        '5', '5', '5', '5', '5', '5', '5', '5', '5', '5',        '6', '6', '6', '6', '6', '6', '6', '6', '6', '6',        '7', '7', '7', '7', '7', '7', '7', '7', '7', '7',        '8', '8', '8', '8', '8', '8', '8', '8', '8', '8',        '9', '9', '9', '9', '9', '9', '9', '9', '9', '9',        } ;    final static char [] DigitOnes = {        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',        } ;

public static String toString(int i) {        if (i == Integer.MIN_VALUE)            return "-2147483648";

<span style="white-space:pre"></span>//求整数的位数，当x为负数时，是以它的绝对值求的位数，因此需要给它加上一位负号位        int size = (i < 0) ? stringSize(-i) + 1 : stringSize(i);        char[] buf = new char[size];//定义一个缓冲字符数组        getChars(i, size, buf);//将整数分解为字符分别存入字符数组        return new String(buf, true);//利用字符数组构造字符串    }    /**     * Places characters representing the integer i into the     * character array buf. The characters are placed into     * the buffer backwards starting with the least significant     * digit at the specified index (exclusive), and working     * backwards from there.     *     * Will fail if i == Integer.MIN_VALUE     */    static void getChars(int i, int index, char[] buf) {        int q, r;        int charPos = index;        char sign = 0;        if (i < 0) {            sign = '-';            i = -i;        }        // Generate two digits per iteration        while (i >= 65536) {            q = i / 100;        // really: r = i - (q * 100);            r = i - ((q << 6) + (q << 5) + (q << 2));            i = q;            buf [--charPos] = DigitOnes[r];            buf [--charPos] = DigitTens[r];        }        // Fall thru to fast mode for smaller numbers        // assert(i <= 65536, i);        for (;;) {            q = (i * 52429) >>> (16+3);            r = i - ((q << 3) + (q << 1));  // r = i-(q*10) ...            buf [--charPos] = digits [r];            i = q;            if (i == 0) break;        }        if (sign != 0) {            buf [--charPos] = sign;        }    }    final static int [] sizeTable = { 9, 99, 999, 9999, 99999, 999999, 9999999,                                      99999999, 999999999, Integer.MAX_VALUE };    // Requires positive x 提供整数 X,返回x的位数    static int stringSize(int x) {        for (int i=0; ; i++)            if (x <= sizeTable[i])                return i+1;    }

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