按位补运算 [英] Bitwise complement operator

问题描述

你们可以解释一下下面的程序

Can you guys please explain the below program

int main() 
{   
 int max = ~0; 
 printf("%d\n",max);    
 return 0; 
}

据我所知〜会翻转位。在这种情况下，即0〜将设置所有位为1，那么最大变量应该包含MAX的价值，但我得到O / P为-1。因此，可以在这里任何人都请告诉我为什么我收到O / P为-1。

AFAIK ~ will flip the bits. In this case i.e ~0 will set all the bits into 1. So max variable should contain MAX value but I am getting o/p as -1. So can anyone here please tell me why I am getting o/p as -1.

推荐答案

为什么你期望获得最大值？在二进制补码重新签订presentation全1位模式代表 1 。这只是事情是这样的。

Why did you expect to obtain the "max value"? In 2's-complement signed representation all-1 bit pattern stands for -1. It is just the way it is.

在二进制补码的最大值重新签订presentation重新由 01111 ... 1 psented $ P $位模式（即第一位为 0 ）。你得到了什么是 1111 ... 1 ，这显然是否定的，因为第一个位 - 符号位 - 为 1 。

Maximum value in 2's-complement signed representation is represented by 01111...1 bit pattern (i.e the first bit is 0). What you got is 1111...1, which is obviously negative since the very first bit - sign bit - is 1.

如果你想在这里补充为零产生最大值，用一个例子的符号的再presentation

If you want an example where complemented zero produces a "max value", use unsigned representation

int main() {   
  unsigned max = ~0u; 
  printf("%u\n", max);    
}

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