在C中使用++运算符对语句进行短路评估 [英] Short circuit evaluation of a statement with ++ operator in C

问题描述

我已经在Windows 7的Code :: Blocks 10.05中执行了以下代码.

I have executed the following code in Code::Blocks 10.05 on Windows 7.

int a=0,b=0,c;
c=a++&&b++;
printf("\na=%d\nb=%d\nc=%d\n\n",a,b,c);

我获得的输出如下，

a=1
b=0
c=0

由于具有短路评估功能，因此这很合理.

This makes perfect sense because of short circuit evaluation.

表达式a++是后递增，并且0返回到逻辑和(&&).因此，部分b++未被评估，因为0 && 0和 0 && 1评估为0.

The expression a++ is post increment and 0 is returned to the logical and (&&). Hence the part b++ is not evaluated since both 0 && 0 and 0 && 1 evaluates to 0.

但我对此表示怀疑.运算符的优先级值明确指出++的优先级高于&&.所以我的理解是这样的，对a++和b ++都进行了评估，然后&&仅检查表达式a++的结果来做出决定.但这并没有发生，仅在此处评估a++.

But here arises my doubt. The precedence value of operators clearly states that ++ is having higher precedence over &&. So my understanding was like this, both a++ and b++ are evaluated and then && only checks the result of expression a++ to come to a decision. But this has not happened only a++ is evaluated here.

此行为的原因是什么? &&作为序列点是否与此行为有关?如果是这样，为什么我们说&&的优先级低于++?

What is the reason for this behavior? Does && being a sequence point has something to do with this behavior? If so why we say that && is having lower precedence than ++?

推荐答案

您对优先级和评估顺序感到困惑.

优先级定义了运算符的分组方式，即

Precedence defines how the operators are grouped, i.e

c = a++ && b++;

等效于:

c = ((a++) && (b++));

求值顺序定义了表达式的求值方式，&&的短路表示a++首先求值，如果为零，则结束；如果它不为零，则对b++求值.

Order of evaluation defines how the expression is evaluated, the short circuit of && means a++ is evaluated first, if it's zero, the end; if it's not zero, b++ is then evaluated.

另一个例子:

c = (a++) + (b++);

在b++之前评估了a++吗?答案是我们不知道.大多数运算符不定义评估顺序. &&是很少定义的运算符之一. (其余为||，,和?:)

Is a++ evaluated before b++? The answer is we don't know. Most operators don't define the order of evaluation. && is one of the few operators that do define. (The rest are ||, , and ?:)

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