增量和逻辑运算符precedence [英] Increment and logical operators precedence

问题描述

在下面所示的程序，preFIX应首先评估，因为它具有更高的precedence，但答案是 -2，2，0，1 并且在书中解释说，因为||的LHS是真实的RHS不计算。结果
为什么会这样呢？所有单位应先进行，然后逻辑应该因为precedence进行检查。

 ＃包括LT＆;＆stdio.h中GT;诠释的main（）
{
    INT I = -3，J = 2，K = 0，米;
    M = ++我|| ++ J＆安培;＆安培; ++ K表;
    的printf（％D，％D，％D，％d个\\ N，I，J，K，M）;
    返回0;
}
 

解决方案

不要误会与评价的 precedence 和订单的

评估顺序逻辑的或的 || 为从左到右。

所以，如果左= TRUE 然后左||权将不会执行右键。在您的code完全相同发生。

如你所知，任何非零值视为真正在C，因此， ++我或 -2 是真正。因此，

  M = ++我|| ++ J＆安培;＆安培; ++ K表;
 M =真||唧唧歪歪; //权甚至没有检查！
 M =真
 m = 1时
 

和你预期的输出。

In the program shown below, prefix should be evaluated first because it has higher precedence, But answer is -2, 2, 0, 1 and it is explained in book "as LHS of || is true RHS is not evaluated."
Why is it so? All the increments should performed first and then logical should be checked because of precedence.

#include<stdio.h>

int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i || ++j && ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}

解决方案

Don't get confused with Precedence and Order of evaluation.

The order of evaluation of logical OR || is left to right.

So if left = true then left || right will never execute right. In your code exactly same happened.

As you know, any non zero value treated as true in C, hence, ++i or -2 is true. So,

 m = ++i || ++j && ++k;
 m = true || bla bla bla; //right not even checked!
 m = true 
 m = 1

And you get the output as expected.

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