如何求一个 int length 多长？

引言

没错，又是看到 Integer 里的代码，

如何求一个 int length多长🤔🤔

final static int [] sizeTable = { 9, 99, 999, 9999, 99999, 999999, 9999999,
                                      99999999, 999999999, Integer.MAX_VALUE };

// Requires positive x
static int stringSize(int x) {
    for (int i=0; ; i++)
        if (x <= sizeTable[i])
            return i+1;
}		

没想到有时候开发 jdk 大神们的思路又是如此的奇清😂😂

32进制内求进制字符串

 /**
     * All possible chars for representing a number as a String
     */
    final static char[] digits = {
        '0' , '1' , '2' , '3' , '4' , '5' ,
        '6' , '7' , '8' , '9' , 'a' , 'b' ,
        'c' , 'd' , 'e' , 'f' , 'g' , 'h' ,
        'i' , 'j' , 'k' , 'l' , 'm' , 'n' ,
        'o' , 'p' , 'q' , 'r' , 's' , 't' ,
        'u' , 'v' , 'w' , 'x' , 'y' , 'z'
    };

public static String toString(int i, int radix) {
    if (radix < Character.MIN_RADIX || radix > Character.MAX_RADIX)
        radix = 10;

    /* Use the faster version */
    if (radix == 10) {
        return toString(i);
    }

    char buf[] = new char[33];
    boolean negative = (i < 0);
    int charPos = 32;

    if (!negative) {
        i = -i;
    }

    while (i <= -radix) {
        buf[charPos--] = digits[-(i % radix)];
        i = i / radix;
    }
    buf[charPos] = digits[-i];

    if (negative) {
        buf[--charPos] = '-';
    }

    return new String(buf, charPos, (33 - charPos));
}

和之前求 int 长度的方法有异曲同工之妙，哈哈哈哈