可以从指向成员函数模板参数的指针推导出类类型

Question

Is it possible to deduce the type of the class T from its pointer to memmber T::*f as shown below.

struct Foo
{
    void func(){}
};

template<typename T, void (T::*f)()>
void bar()
{
}

int main()
{
    bar<Foo,Foo::func>();
    // bar<Foo::func>(); // Desired
}

Answer 1

In C++11/14 I would say no, unless you accept to deduce it by passing the pointer as a function argument:

template<typename T>
void bar(void(T::*f)())
{
}

int main()
{
    bar(&Foo::func);
}

In C++17 you can have a single parameter function template as shown by @Jarod42 , but you don't have the type T deduced anyway (if it was the purpose, as it seems to be from the question).

Answer 2

If you have access to C++17, you can use auto template parameter and decltype to inspect the class type:

struct Foo { void func(){} };

template<typename R, typename C, typename... Args>
C function_pointer_class(R (C::*)(Args...));

template<auto f>
std::enable_if_t<std::is_member_function_pointer_v<decltype(f)>>
bar() {
    using class_t = decltype(function_pointer_class(f));

    // stuff...
}

int main() {
    bar<&Foo::func>();
}

Answer 3

In C++17, you will be allowed to write

template<auto M>
void bar();

Which allows

bar<&Foo::func>();