为什么提振:: is_same＆LT; INT常量和放大器;,提振:: add_const＆LT; INT和放大器;＆GT; ::值等于假的？ [英] Why does boost::is_same<int const&, boost::add_const<int &>::value equal false?

问题描述

我通过C ++模板元编程的亚伯拉罕＆放工作; Gurtovoy
这实际上不是在第二章，但是我尝试过，而在第一次练习中（2.10，2.0），这是混淆了我：

I'm working through "C++ Template Metaprogramming" by Abrahams & Gurtovoy" This isn't actually in chapter two but is something I tried whilst working on the first exercise (2.10, 2.0) which is confusing me:

#include <iostream>
#include <boost/type_traits.hpp>

std::string display(bool b)
{
  return (b ? "true" : "false");
}

int main()
{
   using namespace std;

   cout << display(boost::is_same<int const&, boost::add_const<int &>::type >::value) << "\n";

     return 0;
}

输出是假。
但是，如果我删除了引用，即'廉政常量和INT。输出是'真'。

The output is 'false'. However if I remove the references, i.e. 'int const' and 'int'. The output is 'true'.

推荐答案

如果你试图用三分球一样的东西，如

If you tried the same thing with pointers, as in

boost::is_same<int const *, boost::add_const<int *>::type>::value

你会发现这也是假的，因为的boost :: add_const＆LT; INT *＆GT; ::类型生成为int * const的类型，这显然是不一样的 INT常量* 。

you'd discover that it is also false, since boost::add_const<int *>::type generates int *const type, which is obviously not the same as int const *.

从本质上同样的事情发生与引用，即的boost :: add_const＆LT; INT和放大器;＆GT; ::类型是生成一个尝试 INT ＆安培;常量。从形式上看，键入 INT和放大器;常量在C ++中非法 - CV-资格不能应用于引用本身。因此，的boost :: add_const 的设计是一个空操作在这种情况下，这意味着的boost :: add_const＆LT; INT和放大器;计算值： ：键入生成 INT和放大器; 再次

Essentially the same thing happens with references, i.e. boost::add_const<int &>::type is an attempt to generate int &const. Formally, type int &const is illegal in C++ - cv-qualification cannot be applied to the reference itself. So, boost::add_const is designed to be a no-op in this case, meaning that boost::add_const<int &>::type generates int & again.

这篇关于为什么提振:: is_same＆LT; INT常量和放大器;,提振:: add_const＆LT; INT和放大器;＆GT; ::值等于假的？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！