const 引用可以分配一个 int 吗?

问题描述

我遇到了一个代码片段

const int& reference_to_const_int = 20;
cout<<"
  reference_to_const_int = "<<reference_to_const_int<<endl;     

此代码编译 &执行输出:-

This code compiles & executes with output :-

reference_to_const_int = 20

这对我来说很奇怪.据我所知参考不占用内存 &它们是其他变量的别名.因此我们不能说

This is something strange in for me. As I know reference do not occupy memory & they are aliases to other variables. hence we cannot say

int& reference_to_int = 30;

以上语句不能编译给出错误:-

The above statement shall not compile giving error :-

 error: invalid initialization of non-const reference of type ‘int&’ from an rvalue of type ‘int’

const int&"究竟发生了什么案件?需要完整的解释.

What exactly is happening in the "const int&" case? A full explanation is desired.

请帮忙.

谢谢

推荐答案

创建了一个临时对象，将 const 引用绑定到它是合法的，但将它绑定到非 const 引用是非法的code>const 一.

A temporary is created, and it's legal to bind a const reference to it, but illegal to bind it to a non-const one.

就像:

const int& reference_to_const_int = int(20);  //LEGAL
      int& reference_to_const_int = int(20);  //ILLEGAL

const 引用延长了临时文件的生命周期，这就是它起作用的原因.这只是语言的规则.

A const reference extends the life of a temporary, that's why this works. It's just a rule of the language.

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