使用const引用删除引用 [英] Remove reference with const references

问题描述

对于参数化类C，我想始终使用原始类型，而不考虑指针，const或引用修饰符。

For a parametric class C, I want to get always the "primitive" type irrespective of pointer, const or reference modifiers.

template<typename __T>
class C
{
public:
    typedef std::some_magic_remove_all<__T>::type T;
}

int main()
{
    C<some_type>::type a;
}

例如，对于 some_type 等于：

• int&


• int **


• int *&


• int const&&&


• int const * const


• 等等

• int&
• int**
• int*&
• int const &&
• int const * const
• and so on

我想要 a 始终为 int 类型。如何实现？

I want a is always of type int. How can I achieve it?

推荐答案

template<class T> struct remove_all { typedef T type; };
template<class T> struct remove_all<T*> : remove_all<T> {};
template<class T> struct remove_all<T&> : remove_all<T> {};
template<class T> struct remove_all<T&&> : remove_all<T> {};
template<class T> struct remove_all<T const> : remove_all<T> {};
template<class T> struct remove_all<T volatile> : remove_all<T> {};
template<class T> struct remove_all<T const volatile> : remove_all<T> {};
//template<class T> struct remove_all<T[]> : remove_all<T> {};
//template<class T, int n> struct remove_all<T[n]> : remove_all<T> {};

我最初也剥离了扩展区（数组），但Johannes注意到这会导致 const char [] ，问题没有提及。如果我们还想要删除数组（另见评论中提到的想法），下面的内容不会太复杂：

I originally also stripped extents (arrays), but Johannes noticed that this causes ambiguities for const char[], and the question doesn't mention them. If we also want to strip arrays (see also ideas mentioned in the comments), the following doesn't complicate things too much:

#include <type_traits>
template<class U, class T = typename std::remove_cv<U>::type>
struct remove_all { typedef T type; };
template<class U, class T> struct remove_all<U,T*> : remove_all<T> {};
template<class U, class T> struct remove_all<U,T&> : remove_all<T> {};
template<class U, class T> struct remove_all<U,T&&> : remove_all<T> {};
template<class U, class T> struct remove_all<U,T[]> : remove_all<T> {};
template<class U, class T, int n> struct remove_all<U,T[n]> : remove_all<T> {};

或使用辅助类但单个模板参数：

or with a helper class but a single template parameter:

#include <type_traits>
template<class T> struct remove_all_impl { typedef T type; };
template<class T> using remove_all =
  remove_all_impl<typename std::remove_cv<T>::type>;
template<class T> struct remove_all_impl<T*> : remove_all<T> {};
template<class T> struct remove_all_impl<T&> : remove_all<T> {};
template<class T> struct remove_all_impl<T&&> : remove_all<T> {};
template<class T> struct remove_all_impl<T[]> : remove_all<T> {};
template<class T, int n> struct remove_all_impl<T[n]> : remove_all<T> {};

这是正常的，如果所有的变种开始看起来相同; - ）

It is normal if all the variants start looking about the same ;-)

这篇关于使用const引用删除引用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！