为什么std :: is_const< const int&> :: value evaluate为false？ [英] Why does std::is_const<const int&>::value evaluate to false?

问题描述

这是一个问题的旋转如何检查对象是否为const？

This is a spin off of the question How to check if object is const or not?.

我惊讶地看到以下程序

#include <iostream>
#include <type_traits>

int main() 
{
   std::cout << std::boolalpha;
   std::cout << std::is_const<const int&>::value << "\n";
}

生成此输出

false

在上下文中，考虑 const int& ;

In what context does it make sense to think of const int& as a non-const type?

推荐答案

这可能会更容易理解与这个示例

Perhaps it'll be easier to understand with this example

std::cout << std::is_const<int const *>::value << "\n";  // pointer to const int
std::cout << std::is_const<int * const>::value << "\n";  // const pointer to int

输出：

false
true

一个指向 const int 的指针，而第二个 int * 本身是 const 。因此，它导致 true ，而前者是 false 。同样，你有一个引用 const int 。如果 int& const 有效，会导致 true 。

The first type is a pointer to a const int, while in the second the int * itself is const. Hence it results in true while the former is false. Similarly, what you have a reference to a const int. If int& const were valid it'd result in true.

这篇关于为什么std :: is_const< const int&> :: value evaluate为false？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！