Question

我必须遵循将浮点值（打包在int中）转换为int值的按位代码。

问题：存在舍入问题，因此在输入为0x80000001的情况下失败。我该如何处理？

以下是代码：

  if(x == 0) return x;

  unsigned int signBit = 0;
  unsigned int absX = (unsigned int)x;
  if (x < 0)
  {
      signBit = 0x80000000u;
      absX = (unsigned int)-x;
  }

  unsigned int exponent = 158;
  while ((absX & 0x80000000) == 0)
  {
      exponent--;
      absX <<= 1;
  }

  unsigned int mantissa = absX >> 8;

  unsigned int result = signBit | (exponent << 23) | (mantissa & 0x7fffff);
  printf("\nfor x: %x, result: %x",x,result);
  return result;

Answer 1

那是因为0x80000001的精确度超过that of a float。阅读链接文章，浮点数的精度为24位，因此根本无法检测到差异（x - y）小于两个>> 24的最高位的任何一对浮点数。 gdb同意你的演员：

main.c中：

#include <stdio.h>

int main() {
    float x = 0x80000001;
    printf("%f\n",x);
    return 0;
}

GDB：

Breakpoint 1, main () at test.c:4
4       float x = 0x80000001;
(gdb) n
5       printf("%f\n",x);
(gdb) p x
$1 = 2.14748365e+09
(gdb) p (int)x
$2 = -2147483648
(gdb) p/x (int)x
$3 = 0x80000000
(gdb)

这种不精确的极限：

(gdb) p 0x80000000 == (float)0x80000080 
$21 = 1
(gdb) p 0x80000000 == (float)0x80000081
$20 = 0

实际按位表示：

(gdb) p/x (int)(void*)(float)0x80000000
$27 = 0x4f000000
(gdb) p/x (int)(void*)(float)0x80000080
$28 = 0x4f000000
(gdb) p/x (int)(void*)(float)0x80000081
$29 = 0x4f000001

double确实有足够的精确度来区分：

(gdb) p 0x80000000 == (float)0x80000001
$1 = 1
(gdb) p 0x80000000 == (double)0x80000001
$2 = 0

按位C代码舍入问题

1 个答案: