C防爆Ex pression评价 [英] Evaluation of C expression
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问题描述
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;
}
我认为这和放大器;＆安培;有更多的precedence该||按照这个逻辑 ++Ĵ应该执行，但它从来不和节目输出 -2 2 0 1 。这里发生了什么？什么是中间步骤？
i thought that && has more precedence that || as per this logic ++j should execute, but it never does and the program outputs -2 2 0 1. What is going on here? What are the intermediate steps?
推荐答案
＆放大器;＆安培; 确实有更高的precedence比 || ，这意味着 ++我|| ++ J＆安培;＆安培; ++氏/ code>解析为 ++我|| （++ J＆安培;＆安培; ++ K）
&& does have higher precedence than ||, which means that ++i || ++j && ++k parses as ++i || (++j && ++k).
然而，这并不能改变一个事实，即的RHS || 只执行，如果LHS返回 0 。
However this does not change the fact that the RHS of || only executes if the LHS returns 0.
precedence不会影响评价的顺序。
Precedence does not affect order of evaluation.
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