const引用可以赋值int？ [英] const reference can be assigned an int?

问题描述

我遇到了一个代码片段。

I came across a code snippet

const int& reference_to_const_int = 20;
cout<<"\n  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。因为我知道参考不占用记忆&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;

上面的语句不会编译给出错误： -

The above statement shall not compile giving error :-

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

发生在const int&案件？需要完整的说明。

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

请帮忙。

感谢

推荐答案

创建了一个临时的，绑定一个 const 引用是合法的，一个非 const 一个。

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

A const 生命的临时，这就是为什么这样工作。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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