如何在可变参数模板类中将lambda表达作为参数传递给mermber函数

Question

我想设计一个带有元组的模板类，提供该类的apply（）成员函数，该函数接受返回元组的lambda函数。 lambda的返回元组的类型应与模板类的类型列表兼容。 但是它不能编译。

main.cpp: In function ‘int main()’:
warning: lambda templates are only available with -std=c++14 or -std=gnu++14
     auto getTuple = []<typename ... Param, typename ... Types>(const Param & ..
                   ^
main.cpp: In lambda function:
main.cpp:75:32: error: parameter packs not expanded with ‘...’:
         std::tuple<Types>  items = std::tuple<Types>(p...);
                            ^
main.cpp:75:32: note:         ‘Types’
main.cpp:19:43: error: invalid use of ‘auto’
     std::tuple<Types...>  t = getTuple(param...);    

#include <iostream>
#include <tuple>
#include <string>
#include <type_traits>
#include <array>
template <typename... Types>
class Data
{
public:
    template<typename ... Param, typename Functor>
    std::tuple<Types...> apply(Functor getTuple, const Param& ... param)
    {
        std::tuple<Types...>  t = getTuple(param...);
        return t;
    }
};

int main()
{
    Data<std::string, int,double> data1;
    auto getTuple = []<typename ... Param, typename ... Types>(const Param & ... p)
    {
        std::tuple<Types>  items = std::tuple<Types>(p...);
        return items;
    };
    const auto t = data1.apply(getTuple , "hello", 3,1.9);

    return 0;
}

Answer 1

您不能将模板提供给lambda：您可以简单地使用auto代替：

auto getTuple = [](const auto& ... p)
{
    auto items = std::make_tuple(p...);
    return items;
};