获取可变参数模板可变参数模板参数可变参数

Question

是.

假设我有一个包含typedef的简单可变结构:

template<typename... TArgs> struct TupleTypeHolder {
    using TupleType = std::tuple<TArgs*...>;
};

我想将TupleTypeHolder<something>模板参数传递给另一个类,并获取该typedef.

我的所有尝试都没有编译.

// None of these is valid
template<template<typename...> class TTupleTypeHolder> struct TupleMaker {
     using MyTupleType = TTupleTypeHolder::TupleType; // Not valid
     using MyTupleType = typename TTupleTypeHolder::TupleType; // Not valid
};

 template<template<typename... A> class TTupleTypeHolder> struct TupleMaker2 {
     // A is not a valid name here
     using MyTupleType = TTupleTypeHolder<A...>::TupleType; // Not valid
     using MyTupleType = typename TTupleTypeHolder<A...>::TupleType; // Not valid
};

有没有办法使用可变参数模板类的可变参数模板参数(在本例中为TupleTypeHolders TArgs...)来自使用上述类作为模板可变参数模板参数的类？

---

用法示例:

template<typename... TArgs> struct TupleTypeHolder {
    using TupleType = std::tuple<TArgs*...>;
};

template<typename... TArgs> static int getSomeValue() { ... }

template<??? T1, ??? T2> class TupleMaker 
{ 
    std::pair<int, int> someValues;

    using TupleType1 = T1::TupleType;
    using TupleType2 = T2::TupleType;

    TupleMaker() : someValues{getSomeValue<T1's TArgs...>(), 
                             getSomeValue<T2's TArgs...>()} { }
};

class MyTupleMaker : TupleMaker<TupleTypeHolder<int, char>, 
                                TupleTypeHolder<int, float>> 
{ };

MyTupleMaker::TupleType1 tuple1{new int(1), new char('a')};  
MyTupleMaker::TupleType2 tuple1{new int(35), new float(12.f)};  

Answer 1

工作用法示例:

#include <tuple>

template<typename... TArgs> struct TupleTypeHolder {
    using TupleType = std::tuple<TArgs*...>;
};

template<typename... TArgs> static int getSomeValue() { return 42; }

// primary template:
template<class T1, class T2>
struct TupleMaker;

// partial specialization:
template<template<class...> class TT1, template<class...> class TT2,
         class... T1, class... T2>
struct TupleMaker < TT1<T1...>, TT2<T2...> >
{ 
    std::pair<int, int> someValues;

    using TupleType1 = typename TT1<T1...>::TupleType;
    using TupleType2 = typename TT2<T2...>::TupleType;

    TupleMaker() : someValues{getSomeValue<T1...>(), 
                              getSomeValue<T2...>()} { }
};

struct MyTupleMaker : TupleMaker<TupleTypeHolder<int, char>, 
                                TupleTypeHolder<int, float>> 
{ };

MyTupleMaker::TupleType1 tuple1{new int(1), new char('a')};  
MyTupleMaker::TupleType2 tuple2{new int(35), new float(12.f)};  

int main() {}

当您传递类型时,主模板有两种类型.TupleTypeHolder<int, char>是一种类型,一种模板的特化,而不是一种模板本身.然而,模板模板参数将模板作为参数(而不是类型),例如:

template<template<class...> class Foo>
struct Bar
{
    using type = Foo<int, double, char>;
};

Bar< std::tuple > b; // note: no template arguments for `std::tuple`!

通过部分特化,您可以将模板特化分割为模板和参数,这就是上述方法的工作原理.