为什么输出在代码中不会是这个? [英] Why the output won't be this in the code?

问题描述

#include <stdio.h>

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

我试图了解C语言中操作符的关联性和优先级.在这里，输出结果为-2 3 0 1，但我认为输出应该为-2 3 1 1，因为k也已预先递增.为什么那不是答案?谢谢！

I am trying to learn about associativity and precedence of operators in C. Here, The output comes out to be -2 3 0 1, but I think the output should be -2 3 1 1 because k is also pre-incremented. Why that won't be the answer? Thanks!

推荐答案

||具有短路评估功能，这意味着仅当左侧为false时，右侧才被评估.在您的情况下，不会发生这种情况，因为i和j的值在递增后都不同于0，因此++k不会执行

the || has short-circuit evaluation, which means that the right hand side gets evaluated only if the left hand side is false. In your case this doesn't happen since both i and j have values different than 0after being incremented, so the ++k doesn't get executed

当您有一个&&且LHS表达式的计算结果为false

The same behavior occurs when you have a && in which the LHS expressions evaluates to false

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