解浮点数方程: X+1=X

不是纯数学意义上的方程, 对应计算机的一个浮点数问题:

if((float)(x+1.0) == (float)(x)) { x = ? }

简单分析, ieee754中float采用23bit表示有效位, 再加省略的1, 共有24bit. 当结果超出24bit时, 小数部分被被丢失.

大于 2^24 = 16777216 的x, 满足 x+1.0==x.

测试程序:

#include <stdio.h>

// Little endian
union ieee754_float {
    float f;

    /* This is the IEEE 754 single-precision format.  */
    struct {
        unsigned int mantissa:23;
        unsigned int exponent:8;
        unsigned int negative:1;
    } ieee;

    /* This format makes it easier to see if a NaN is a signalling NaN.  */
    struct {
        unsigned int mantissa:22;
        unsigned int quiet_nan:1;
        unsigned int exponent:8;
        unsigned int negative:1;
    } ieee_nan;
};

int main()
{
    union ieee754_float f = { 0.0f };

    // Find a positive floating point value x, for which x+1.0=x.
    for(f.ieee.exponent = 127; f.ieee.exponent < 255; f.ieee.exponent++) {
        if((float)(f.f + 1.f) == f.f) {
            printf("%f: exponent = %d\n", f.f, f.ieee.exponent);
        }
    }
    return 0;
}

// output:
// 16777216.000000: exponent = 151
// 33554432.000000: exponent = 152
// 67108864.000000: exponent = 153
// 134217728.000000: exponent = 154
//
// ... ...
//
// 21267647932558654000000000000000000000.000000: exponent = 251
// 42535295865117308000000000000000000000.000000: exponent = 252
// 85070591730234616000000000000000000000.000000: exponent = 253
// 170141183460469230000000000000000000000.000000: exponent = 254

IEEE754相关资源:

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