为什么C ++ const引用可以归类为非const引用 [英] Why can C++ const references be collasped into non-const references

问题描述

考虑以下C ++程序：

Consider the following C++ program:

#include <iostream>

template<typename T>
class A
{
public:
    explicit A(T& x) : x_(x){}
    const T& get() { return x_; }

private:
    T x_;
};

int main()
{
    int x = 42;
    A<int&>(x).get() = 43; // compiles fine, even though get() looks like it returns a const ref
    std::cout << x << '\n';
}

程序编译OK并输出43。这表明返回了看似const引用实际上，通过get（）获得的是非const 引用，因为它允许修改其引用的值。

The program compiles OK and outputs 43. This suggests that the seemingly const reference returned by get() is in fact a non-const reference, because it allows to modifies the value it refers to.

这是规则吗？

如何强制执行get（）返回的引用的行为类似于const引用，即不允许修改

How to enforce that the reference returned from get() behaves like a const reference, that is, it doesn't allow to modify the value it refers to?

推荐答案

引用规则折叠会导致这种行为吗？

Is it a rule of reference collapsing that causes this behaviour?

是。您有：

T = int&
const T& = const (int&) &

引用不能为 const （您仍然无法重新绑定它们，因此将其忽略），对引用的引用只是引用。

References can't be const (you can't rebind them anyways, so it's ignored) and a reference to a reference is just a reference.

所以您拥有

const T& = int&

要解决此问题，您需要应用 const 到基础类型，您可以通过删除引用来实现：

To fix this, you need to apply const to the underlying type, which you can do like this by removing the reference:

const std::remove_reference_t<T>& get() { return x_; }
//    ^^^^^^^^^^^^^^^^^^^^^^^

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