模板派生的演绎指南中断 class

Question

这是最简单的例子：

#include <cstddef>
struct base_pod
{
};
template<typename T, std::size_t N>
struct der_pod : public base_pod
{
    T k[N];
};
template<typename T, typename... U>
    der_pod(T, U...)
        ->der_pod<std::enable_if_t<(std::is_same_v<T, U> and ...), T>, 1 + sizeof...(U)>;

int main()
{
    der_pod dp {{}, {3, 3} };
}

live demo

prog.cc:16:9: error: no viable constructor or deduction guide for deduction of template arguments of 'der_pod'
der_pod dp {{}, {3, 3} };
        ^
prog.cc:11:2: note: candidate template ignored: couldn't infer template argument 'T'
        der_pod(T, U...)
        ^
prog.cc:6:8: note: candidate function template not viable: requires 0 arguments, but 2 were provided
struct der_pod : public base_pod
       ^
prog.cc:6:8: note: candidate function template not viable: requires 1 argument, but 2 were provided
1 error generated.

I have to add explicit template argumentder_pod<int, 2> to pass compilation.

但是推导指南在没有基础的情况下工作正常class:

template<typename T, std::size_t N>
struct der_pod
{
    T k[N];
};
template<typename T, typename... U>
    der_pod(T, U...)
        ->der_pod<std::enable_if_t<(std::is_same_v<T, U> and ...), T>, 1 + sizeof...(U)>;

int main()
{
    der_pod dp {3, 3};
}

live demo

1。后面是什么？

2。如何修复

Answer 1

{}、{3, 3}没有类型，不允许推导T和U.

相反，您可以使用：

template <typename T, std::size_t N>
der_pod(base_pod, const T (&)[N]) -> der_pod<T, N>;

Demo

模板派生的演绎指南中断 class

deduction guide breaks for templated derived class

c++

inheritance

template-argument-deduction

1。后面是什么？

2。如何修复