Java整数转换成字符

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;
}