Template argument deduction for member function pointers

Question

It is known that template arguments can be pointers to member functions.

So I can write:

struct Bar
{
    int fun(float x);
};

template <int (Bar::*FUN)(float)>
struct Foo
{ /*...*/ };

typedef Foo<&Bar::fun> FooBar;

But what if I want the the Bar type itself to be a template argument:

template <typename B, int (B::*FUN)(float)>
struct Foo
{ /*...*/ };

typedef Foo<Bar, &Bar::fun> FooBar;

Now, when I use it, I have to write Bar twice!

My question is: Is there a way to force the compiler to deduce the class type automatically?

The objective is for this to just work:

typedef Foo<&Bar::fun> FooBar;
typedef Foo<&Moo::fun> FooMoo;

Answer 1

Simple answer: no there isn't.

The problem is that for typedef Foo<&Bar::fun> FooBar; to work, the template would have to have a single non-type argument, but the type of that argument would be unknown when the template is being declared, which is not valid. On the other side, type deduction is never applied to the arguments of the template (only to the arguments to function templates, but those are the arguments to the function, not the template).