的＆放行为;＆安培;在C语言编程 [英] Behaviour of && in C programming language

问题描述

我在C语言编程初学者，最近我看了逻辑和＆放大器;＆安培; 运营商

I am beginner in C programming language, recently I have read about Logical AND && operator.

我也知道，在C语言所有非零值被当成 TRUE

I also know that, in C programming language all non-zero values are treated as TRUE.

NON-ZERO && NON-ZERO = 1  
NON-ZERO && ZERO = 0  
ZERO && NON-ZERO = 0
ZERO && ZERO = 0  

但是，当我处理下面的程序，然后我没有得到预期的答案。

But when I am dealing with the following program then I am not getting expected answer.

int main(){  
  int x, y, z;  
  x = y = z = -1;  
  y = ++x && ++y && ++z;  
  printf("x = %d, y = %d, z = %d, x, y, z);  
  return 0;  
} 

我期待

x = 0, y = 0, z = 0 

但得到的答复是

x = 0, y = 0, z = -1

任何人都可以请解释一下，为什么我得到这个答案？

Can anyone please explain, Why I am getting this answer?

推荐答案

，当 X 是 0 ，是和ž并不真正需要的，因为 0放大器进行评估;＆安培; ANYTHING 是 0 。

Because of Short-circuit evaluation, when x is 0, y and z don't really need to be evaluated since 0 && ANYTHING is 0.

在 X 递增到 0 ，结果是 0 ，这就是是得到。

Once x is incremented to 0, the result is 0, and that's what y gets.

以Z 保持不变（ 1 ）。

 x  | y  |  z 
----+----+-----
 -1 | -1 | -1   //x = y = z = -1;  
  0 | -1 | -1   //++x && ... Now the whole expression is evaluated to 0
  0 |  0 | -1   //y = ++x && ++y && ++z;

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