为什么++ x ++给出编译错误？

本文介绍了为什么++ x ++给出编译错误？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我与后递增和预递增运算符的返回值混淆。它返回r值或l值。

I am having confusion with the return value of post-increment and pre-increment operator.Whether it returns r-value or l-value.

#include<iostream> using namespace std; int main(){ int a=10; cout<<++a++<<"\n"; }

以下代码会产生编译错误。

The following code give a compile error.

error: lvalue required as increment operator

为什么会出现错误？

编译器如何评估表达式 ++ a ++ ？

How does the compiler evaluates the expression ++a++?

推荐答案

后递增表达式的结果 a ++ rvalue ;在递增之前具有 a 的值的临时变量。作为右值，您可以使用其值，但不能修改它。具体来说，你不能像编译器所说的那样应用预递增。

The result of the post-increment expression, a++ is an rvalue; a temporary with the value that a had before incrementing. As an rvalue, you can use its value, but you can't modify it. Specifically, you can't apply pre-increment it, as the compiler says.

如果你要改变优先级，先做递增：

If you were to change the precedence to do the pre-increment first:

(++a)++

那么这将编译：预增量的结果是一个表示被修改的对象的lvalue 。然而，这可能具有未定义的行为;我不确定这两个修改和值的各种用法是否排序。

then this would compile: the result of pre-increment is an lvalue denoting the object that's been modified. However, this might have undefined behaviour; I'm not sure whether the two modifications and the various uses of the value are sequenced.

摘要：不要尝试写多个副作用的棘手表达式。

Summary: don't try to write tricky expressions with multiple side-effects.