c – 使用位操作查找最小值

It is used to find the minimum of two numbers.

对于许多int组合来说,这个代码很不幸失败,因为它的价值非常低. @harold一个好的编译器会识别x<你？ x：y;并生成正确和快速的代码. @David C. Rankin

确定它是如何工作的并不像它失败那样有趣.

>未定义的行为：如果x-y溢出,兼容的编译器可能会生成任何输出 – 甚至崩溃.优化编译器利用这一点使新程序员感到懊恼.
>移动符号位是实现定义的行为,如some_negative_int>> (sizeof(int)* CHAR_BIT – 1))). int的算术右移是常见的,但未由C指定.
> some_int>> (sizeof(int)* CHAR_BIT – 1)))如果int包含填充,则可以超过允许的最大移位(这很少见).

对于x,y的许多组合,OP的代码失败 – 121个测试案例中的31个失败 – 见下文. “It accomplishes this by an arithmetic shift”是实现定义的行为. x-y的潜在溢出是未定义的行为.如果不解决这些问题,任何答案都不完整

拐角情况“it works for any other sizes”通常是正确的,但罕见的平台可能在int中使用填充,使sizeof(int)* CHAR_BIT – 1成问题.

#include <stdio.h>

int minz(int x,int y) {
  return  y + ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1)));
}

void testmin(int x,int y) {
  static unsigned count = 0;
  static unsigned fail = 0;
  int min0 = x < y ? x: y;
  int min1 = minz(x,y);
  count++;
  if (min0 != min1) {
    fail++;
    printf("%u/%u min(%d,%d)--> %d,should be %dn",fail,count,x,y,min1,min0);
  }
}
int main(void) {
  const int i[]={INT_MIN,INT_MIN+1,INT_MIN+2,-2,-1,1,2,INT_MAX-2,INT_MAX-1,INT_MAX};
  int x,y;
  for (x=0; x<sizeof i/sizeof i[0]; x++) {
    for (y=0; y<sizeof i/sizeof i[0]; y++) {
      testmin(i[x],i[y]);
    }
  }
}

输出(故障)

1/7 min(-2147483648,1)--> 1,should be -2147483648
2/8 min(-2147483648,2)--> 2,should be -2147483648
3/9 min(-2147483648,2147483645)--> 2147483645,should be -2147483648
4/10 min(-2147483648,2147483646)--> 2147483646,should be -2147483648
5/11 min(-2147483648,2147483647)--> 2147483647,should be -2147483648
6/19 min(-2147483647,should be -2147483647
7/20 min(-2147483647,should be -2147483647
8/21 min(-2147483647,should be -2147483647
9/22 min(-2147483647,should be -2147483647
10/31 min(-2147483646,should be -2147483646
11/32 min(-2147483646,should be -2147483646
12/33 min(-2147483646,should be -2147483646
13/44 min(-2,should be -2
14/56 min(0,-2147483648)--> 0,should be -2147483648
15/67 min(1,-2147483648)--> 1,should be -2147483648
16/68 min(1,-2147483647)--> 1,should be -2147483647
17/78 min(2,-2147483648)--> 2,should be -2147483648
18/79 min(2,-2147483647)--> 2,should be -2147483647
19/80 min(2,-2147483646)--> 2,should be -2147483646
20/89 min(2147483645,-2147483648)--> 2147483645,should be -2147483648
21/90 min(2147483645,-2147483647)--> 2147483645,should be -2147483647
22/91 min(2147483645,-2147483646)--> 2147483645,should be -2147483646
23/100 min(2147483646,-2147483648)--> 2147483646,should be -2147483648
24/101 min(2147483646,-2147483647)--> 2147483646,should be -2147483647
25/102 min(2147483646,-2147483646)--> 2147483646,should be -2147483646
26/103 min(2147483646,-2)--> 2147483646,should be -2
27/111 min(2147483647,-2147483648)--> 2147483647,should be -2147483648
28/112 min(2147483647,-2147483647)--> 2147483647,should be -2147483647
29/113 min(2147483647,-2147483646)--> 2147483647,should be -2147483646
30/114 min(2147483647,-2)--> 2147483647,should be -2
31/115 min(2147483647,-1)--> 2147483647,should be -1