混合增量运营商逻辑运算符 [英] Mixed increment operators with logical operators

问题描述

我有关于与逻辑运算符pre和后增量的问题
如果我有这样的code

I have a question concerning pre and post increments with logical operators if I have this code

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

知道递增和递减运营商有更高的precedence比＆安培;＆安培;和||
因此，他们会先执行然后和放大器;＆安培;比结果高
装置-2 || 3和;＆放大器; 1，它给出的值-2 3 1 1为printf的

knowing that the increment and the decrement operators have higher precedence than && and || So they'll be executed first Then the && is higher than
means -2||3&&1 which gives the values -2 3 1 1 for the printf

但输出上VS2010努力时，我得到的是-2 2 0 1

but the output I get when trying on VS2010 is -2 2 0 1

有没有人有任何的解释？
问候,,

Does anyone have any explanation for that ? Regards,,

推荐答案

这是你从短路得到什么。 + I 为-2，其余不必进行评估（并且不按照标准）。左侧 || 是真实的，因为-2不为0，因此整个前pression是真实的。

This is what you get from short circuiting. ++i is -2, and the rest doesn't have to be evaluated (and isn't according to the standard). The left side of || is true because -2 is not 0, so the whole expression is true.

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