Can class type be deduced from pointer to member function template parameter

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).