类模板参数推导因派生类而失败 [英] Class template argument deduction failed with derived class

问题描述

#include <utility>

template<class T1, class T2>
struct mypair : std::pair<T1, T2>
{ using std::pair<T1, T2>::pair; };

int main()
{
    (void)std::pair(2, 3); // It works
    (void)mypair(2, 3);    // It doesn't work
}

以上格式是否正确？

如果构造函数被继承，是否可以在第二种情况下推断类模板参数？ std :: pair 的构造函数是否正在参与 mypair 的隐式扣除指南的创建？

Is it possible deduce the class template arguments in the second case if the constructors are being inherited? Are the constructors of std::pair participating in the creation of implicit deduction guides for mypair?

我的编译器是g ++ 7.2.0。

My compiler is g++ 7.2.0.

推荐答案

请参见 Richard Smith的答案。

A此答案的先前版本已说明以下应该应该起作用

A previous version of this answer had stated that the following should work

template <class T> struct B { B(T ) { } };
template <class T> struct D : B<T> { using B<T>::B; };

B b = 4; // okay, obviously
D d = 4; // expected: okay

但这并不是真正可行的方法，甚至都不是一个好方法像我想的那样工作的想法（我们继承了构造函数，但没有继承推论指南？）

But this isn't really viable, and wouldn't even be a good idea to work as I thought it would (we inherit the constructors but not the deduction guides?)

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