下面前pression评价 [英] Evaluation of the following expression

问题描述

以下code片断：

int i=-3,j=2,k=0,m;

m=++i && ++j || ++k;

可以用两个概念来评价，我相信：

can be evaluated using two concepts,I believe:

1.Since ++运算符具有更大的precedence比逻辑运算符，所以首先增量运营商将evaluted，然后和放大器;＆安培;具有高于precedence ||将computed.In此过程中，k会增加。

1.Since ++ operator has greater precedence than the logical operators,so first all increment operators will be evaluted,then && having higher precedence than || will be computed.In this process,k will be incremented.

2.First＆放大器;＆放大器;运营商将evaluated.For此++ i和++ j应该的＆放大器computed.Since结果;＆放大器;运营商是1，无需评估++ k.So K将不起递增。

2.First && operator will be evaluated.For this ++ i and ++j will be computed.Since the result of the && operator is 1,no need to evaluate the ++k.So k will not be incremented.

当我尝试它的系统上，结果证明推理2是正确的，1是错误的。为什么会这样呢？

When I try it on a system, the result proves reasoning 2 to be correct and 1 to be wrong. Why is it so?

推荐答案

奥利是正确的......你混淆precedence与评估顺序。

Oli is right... You're confusing precedence with evaluation order.

precedence意味着前pression是PTED为跨$ P $：

Precedence means that the expression is interpreted as:

m = ((((++i) && (++j)) || (++k));

至于反对，说：

m = (++(i && ++(j || (++k)))

precedence并不能改变一个事实，即 || 运营商的LHS总会在RHS前进行评估。

Precedence doesn't change the fact that the LHS of the || operator will always be evaluated before the RHS.

这篇关于下面前pression评价的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！