T :: *在模板的参数是什么意思？ [英] What does T::* mean in template's parameters?

问题描述

按照这里撰写的文章：

我遇到了这个代码（为了清晰起见，缩写和改变了）：

I came across this code (shortened and changed for clarity):

template <class T> struct hasSerialize
{
    // This helper struct permits us to check that serialize is truly a method.
    // The second argument must be of the type of the first.
    // For instance reallyHas<int, 10> would be substituted by reallyHas<int, int 10> and works!
    // reallyHas<int, &C::serialize> would be substituted by reallyHas<int, int &C::serialize> and fail!
    // Note: It only works with integral constants and pointers (so function pointers work).
    // In our case we check that &C::serialize has the same signature as the first argument!
    // reallyHas<std::string (C::*)(), &C::serialize> should be substituted by 
    // reallyHas<std::string (C::*)(), std::string (C::*)() &C::serialize> and work!
    template <typename U, U u> struct reallyHas;

    // We accept a pointer to our helper struct, in order to avoid to instantiate a real instance of this type.
    // std::string (C::*)() is function pointer declaration.
    template <typename C>
    static char&
    test(reallyHas<std::string (C::*)(), &C::serialize>* /*unused*/) { }
};

因此

std::string (C::*)()

任何人都可以向我解释 C :: * 部分？这是一个函数指针，返回 std :: string 但还有什么？

Can anyone explain me the C::* part? That is a function pointer that returns std::string but what more?

推荐答案

成员函数指向C类中返回 std :: string 的成员的成员函数指针。

A member function pointer to a member in class C that returns a std::string.

检查 isocpp.org 了解有关成员函数指针的更多信息。

Check isocpp.org for more info on pointers to member functions.

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