基于模板参数成员函数

时间:2016-05-23 18:20:35

标签: c++ template-meta-programming

我可以运行此代码,但是当我启用3个已注释的行时,它不再编译并出现以下错误:

1>d:\git\testprojekt\testprojekt\testprojekt.cpp(41): warning C4346: 'first_argument<F>::type': dependent name is not a type
1>  d:\git\testprojekt\testprojekt\testprojekt.cpp(41): note: prefix with 'typename' to indicate a type
1>  d:\git\testprojekt\testprojekt\testprojekt.cpp(43): note: see reference to class template instantiation 'Bla<Type>' being compiled
1>d:\git\testprojekt\testprojekt\testprojekt.cpp(41): error C2923: 'DoStuff': 'first_argument<F>::type' is not a valid template type argument for parameter 'Arg'
1>  d:\git\testprojekt\testprojekt\testprojekt.cpp(22): note: see declaration of 'first_argument<F>::type'

我的想法为什么它不起作用是编译器想要确保Bla编译各种模板参数,但first_argument只能处理具有operator()定义的模板参数。 有谁知道如何让这个例子有效? 我需要它来选择一个类,这里​​是doStuff,基于模板参数operator()是否接受一个参数,在另一个模板化的类中,这里是Bla。

#include <iostream>


template<typename F, typename Ret>
void helper(Ret(F::*)());

template<typename F, typename Ret>
void helper(Ret(F::*)() const);


template<typename F, typename Ret, typename A, typename... Rest>
char helper(Ret(F::*)(A, Rest...));

template<typename F, typename Ret, typename A, typename... Rest>
char helper(Ret(F::*)(A, Rest...) const);

template<typename F>
struct first_argument {
    typedef decltype(helper(&F::operator())) type;
};

template <typename Functor, typename Arg = first_argument<Functor>::type>
struct DoStuff;

template <typename Functor>
struct DoStuff<Functor, char>
{
    void print() { std::cout << "has arg" << std::endl; };
};

template <typename Functor>
struct DoStuff<Functor, void>
{
    void print() { std::cout << "does not have arg" << std::endl; };
};


template <typename Type>
struct Bla
{
    //DoStuff<typename Type> doStuff;
    //void print() { doStuff.print(); };
};



int main()
{
    struct functorNoArg {
        void operator() () {};
    };

    struct functorArg {
        void operator()(int a) { std::cout << a; };
    };

    auto lambdaNoArg = []() {};
    auto lambdaArg = [](int a) {};


    std::cout << std::is_same<first_argument<functorArg>::type,int>::value <<std::endl;  // this works

    DoStuff<functorArg> doStuff;
    doStuff.print();

    DoStuff<functorNoArg> doStuff2;
    doStuff2.print();

    DoStuff<decltype(lambdaArg)> doStuff3;
    doStuff3.print();

    DoStuff<decltype(lambdaNoArg)> doStuff4;
    doStuff4.print();

    Bla<functorArg> bla;
    //bla.print();

    return 0;
}

感谢所有模板书呆子帮助:)

1 个答案:

答案 0 :(得分:1)

struct Bla中,您应该说DoStuff<Type> doStuff;(此处不需要或不允许使用typename)。

In(更正版):

template <typename Functor, typename Arg = typename first_argument<Functor>::type> struct DoStuff;

typename之前您遗失了first_argument<Functor>::type