Rounding issues with bitwise C code

问题

I have to following bitwise code which casts a floating point value (packaged in an int) to an int value.

Question: There are rounding issues so it fails in cases where input is 0x80000001 for example. How do I handle this?

Here is the code:

  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;

回答1:

That's because the precision of 0x80000001 exceeds that of a float. Read the linked article, the precision of a float is 24 bits, so any pair of floats whose difference (x - y) is less than the highest bit of the two >> 24 simply cannot be detected. gdb agrees with your cast:

main.c:

#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)

The limit of this imprecision:

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

The actual bitwise representation:

(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

doubles do have enough precision to make the distinction:

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

来源：https://stackoverflow.com/questions/12344294/rounding-issues-with-bitwise-c-code

标签

bit-manipulation