const 引用的字面初始化

问题描述

以下代码如何在 C++ 中工作?合乎逻辑吗?

How does the following code work in C++? Is it logical?

const int &ref = 9;
const int &another_ref = ref + 6;

为什么 C++ 允许常量引用的字面初始化，而非常量引用不允许?例如:

Why does C++ allow literal initialization for const references when the same is not permitted for non-const references? E.g.:

const int days_of_week = 7;
int &dof = days_of_week; //error: non const reference to a const object

这可以通过一个事实来解释，非常量引用可用于更改它所引用的变量的值.因此，C++ 不允许对 const 变量的非常量引用.

This can be explained by the fact that, a non-const reference can be used to change the value of the variable it is referring to. Hence, C++ does not permit a non-const reference to a const variable.

这是一个可能的解释吗?C++ 不允许:

Could this be a possible explanation? C++ does not allow:

int &ref = 7;

因为这不合逻辑，但是:

Because that is not logical, but:

const int &ref = 7;

几乎相当于:

const int val = 7;

所以常量变量允许字面初始化.

So literal initialization is permitted for const variables.

P.S.:我目前正在学习 Lippman 的 C++ Primer.

P.S.: I'm currently studying Lippman's C++ Primer.

推荐答案

所以你可以这样写代码:

So you can write code like this:

void f( const string & s ) {
}

f( "foobar" );

尽管严格来说，这里实际发生的并不是将文字绑定到 const 引用——而是创建了一个临时字符串对象:

Although strictly speaking what is actually happening here is not the literal being bound to a const reference - instead a temprary string object is created:

string( "foobar" );

并且这个无名字符串绑定到引用.

and this nameless string is bound to the reference.

请注意，像您这样做时创建非参数引用变量实际上很不寻常 - 引用的主要目的是用作函数参数和返回值.

Note that it is actually quite unusual to create non-parameter reference variables as you are doing - the main purpose of references is to serve as function parameters and return values.

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