按位C $ C $Ç四舍五入问题 [英] Rounding issues with bitwise C code

问题描述

我有以下按位code这蒙上了浮点值（封装在一个int）为int值。

问：有四舍五入问题，所以它的情况下失败的地方输入0x80000001为例。我该如何处理呢？

下面是code：

 如果（X == 0）返回X;  unsigned int类型signBit = 0;
  unsigned int类型absX =（unsigned int类型）×;
  如果（X℃，）
  {
      signBit = 0x80000000u;
      absX =（unsigned int类型）-x;
  }  无符号整型指数= 158;
  而（（absX＆安培;为0x80000000）== 0）
  {
      指数 -​​ ;
      absX＆所述;＆下; = 1;
  }  unsigned int类型尾数= absX＆GT;＆GT; 8;  无符号整型结果= signBit | （指数＆LT;＆LT; 23）| （尾数＆安培; 0x7fffff）;
  的printf（\\ n对于X：为％x，结果是：％X，X，结果）;
  返回结果;
 

解决方案

这是因为precision 0x80000001 超出的一个浮动 。阅读链接的文章，一个浮动的precision为24位，所以任何对彩车的差异（说明X - Y ）小于的最高位两个＆GT;＆GT; 24 根本无法被检测到。
  GDB 同意你的施法：

main.c中：

 的#include＆LT;＆stdio.h中GT;诠释主（）{
    浮动X = 0x80000001;
    的printf（％F \\ N，X）;
    返回0;
}
 

GDB：

 断点1，主要的（）在test.c的：4
4浮动X = 0x80000001;
（GDB）N
5的printf（％F \\ N，X）;
（GDB）P X [
$ 1 = 2.14748365e + 09
（GDB）P（INT）x
$ 2 = -2147483648
（GDB）P / X（INT）x
$ 3 = 0x80000000的
（GDB）
 

这IM precision的限制：

 （GDB）芘为0x80000000 ==（浮点）0x80000080
$ 21 = 1
（GDB）p为0x80000000 ==（浮点）0x80000081
$ 20 = 0
 

实际按位重新presentation：

 （GDB）P / X（INT）（无效*）（浮点）为0x80000000
$ 27 = 0x4f000000
（GDB）P / X（INT）（无效*）（浮点）0x80000080
$ 28 = 0x4f000000
（GDB）P / X（INT）（无效*）（浮点）0x80000081
$ 29 = 0x4f000001
 

双击就做有足够的precision做出区分：

 （GDB）芘为0x80000000 ==（浮点）0x80000001
$ 1 = 1
（GDB）p为0x80000000 ==（双）0x80000001
$ 2 = 0
 

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;

解决方案

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

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