是否可以使用指向函数类型的指针来专门化模板，这样可以分别查看此指针中使用的所有类型？

Question

#include <iostream> 

template<typename T_function_type> 
struct pointer_wrapper { 

    T_function_type function_pointer; 

    explicit pointer_wrapper(T_function_type ptr) : function_pointer(ptr) { } 

    ~pointer_wrapper() { } 

}; 

template<typename T_return, typename T_arg1, typename T_arg2> 
pointer_wrapper<T_return (*)(T_arg1, T_arg2)> fabricate_some_trait(T_return (*ptr)(T_arg1, T_arg2)) { 

    return pointer_wrapper<T_return (*)(T_arg1, T_arg2)>(ptr); 

} 

void hello(std::string const& name, std::string const& surname) { 

    std::cout << "Hi " << name << " " << surname << "! "; 

    std::cout << "Was wondering if it\'s possible to modify "; 

    std::cout << "some_trait template in that way that it\'s "; 

    std::cout << "capable of deducing function (as well as "; 

    std::cout << "static and member one) type like the "; 

    std::cout << "template function fabricate_some_trait is. "; 

    std::cout << "So I can use these traits in typedef. "; 

    std::cout << "Will be perfect if no <functional>, "; 

    std::cout << "Boost.Function nor sigc::signal is going "; 

    std::cout << "to be used. Hope you can help and sorry "; 

    std::cout << "in advance about the form of this question, "; 

    std::cout << "just feeling good today." << std::endl; 

    std::cout << "Cheers!" << std::endl; 

} 

int main() { 

    // need 
    // some_trait<hello>::function_type hello_pointer = hello; 
    // hello_pointer("Stackoverflow", "User"); 

    fabricate_some_trait(hello).function_pointer("Stackoverflow", "User"); 

} 

/* 
template<typename T_return_type(*)(typename T_arg, typename T_arg2)> 
struct some_trait { 

    typedef T_return (*function_type)(T_arg1, T_arg2); 

    typedef T_return result_type; 

    typedef T_arg1 arg1_type; 

// (...) 

}; 
*/ 

Answer 1

你想提升的类型特征，具体功能特点：

http://www.boost.org/doc/libs/1_43_0/libs/type_traits/doc/html/boost_typetraits/reference/function_traits.html

它的工作原理只是普通的C函数指针细，无需使用boost ::为sigc ::信号的功能。通过函数特征，你可以单独提取返回类型和每个参数类型，不需要自己完成特殊化。

如果你根本不能使用boost，那么你仍然可以检查它们的实现以获得它的工作原理的一般概念。