将类模板实例化与其基类模板进行比较时 is_same 返回 false？

发布于 2024-11-07 16:59:13 字数 1044 浏览 4 评论 0原文

*编辑：不知何故，我认为编译器正在创建 B 就像 A 一样，导致我假设 is_same 应该如何评估它们，无论继承/派生。我的不好:(对后续的误解表示抱歉:\ *

制作一些元函数来检查我的自定义类型，并遇到了这个问题，但不确定我明白这里发生了什么。我想我可以解决它通过将已知类型的 this_t 成员与传递的任何参数的 this_t 进行比较，但我只想了解为什么第一个和第三个 is_same 测试失败：

template<typename... Args> struct A {
    typedef A<Args...> this_t;
};

struct B : A<int, int, string> {
};

//tests
std::is_same<A<int, int, string>, B>::value; //false
std::is_same<A<int, int, string>, typename B::this_t>::value; //true
std::is_same<B, typename B::this_t>::value; //false

//more tests for kicks
std::is_base_of<A<int, int, string>, B>::value; //true
std::is_base_of<A<int, int, string>, typename B::this_t>::value; //true
std::is_base_of<B, typename B::this_t>::value; //false

is_same 通过 A<...>A<...> 进行区分code> base？A 和 B 之间有什么明显的区别？

原文

*Edit: Somehow I thought the compiler was creating B just as A<int, int, string>, leading to my assumption about how is_same should evaluate them, regardless of inheritance/derivation. My bad :( Sorry for subsequent misunderstandings :\ *

Making some meta-functions to check for my custom types, and ran into this issue, but not sure I understand what's going on here. I think I can work around it by comparing this_t member of a known type to this_t of whatever parameter is passed, but I just want to understand why the 1st and 3rd is_same tests fail:

template<typename... Args> struct A {
    typedef A<Args...> this_t;
};

struct B : A<int, int, string> {
};

//tests
std::is_same<A<int, int, string>, B>::value; //false
std::is_same<A<int, int, string>, typename B::this_t>::value; //true
std::is_same<B, typename B::this_t>::value; //false

//more tests for kicks
std::is_base_of<A<int, int, string>, B>::value; //true
std::is_base_of<A<int, int, string>, typename B::this_t>::value; //true
std::is_base_of<B, typename B::this_t>::value; //false

Is is_same differentiating by way of the A<...> base? What's the appreciable difference between A<int, int, string> and B?

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梦在深巷 2024-11-14 16:59:13

is_same 基本上是一个具有专门化的模板

template<class T, class U>
struct is_same : false_type
{ };

template<class T>
struct is_same<T, T> : true_type
{ };

，它永远不会给你 true ，除非你有完全相同相同的类型。请注意，专业化中只有一个 T。它永远不能同时匹配 A 和 B。

The is_same is basically a template with a specialization

template<class T, class U>
struct is_same : false_type
{ };

template<class T>
struct is_same<T, T> : true_type
{ };

That will never give you true, unless you have exactly the same type. Note that there is only one T in the specialization. It can never match both A and B.

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