如何使用成员函数指针作为模板arg实例化类

Question

I have

template <void (*T)(Entity *), typename Caller>
class Updater 
{
public:
    Updater(Caller c):m_caller(c){}
    void process(Entity * e)
    {
        (m_caller->*T)(e);              //Is this right?
    }
private:
    Caller m_caller;
};

I understand I can instantiate it like

Foo f;
Updater<&Foo::Bar> updater(&f);

assuming that Foo has

void Foo::Bar(Entity *e);

but what if it has desired method tempated? Like this

template <typename T>
void Bar(T t);

how shoult I instanciate it? Like this:?

Foo f;
Updater<&Foo::Bar<Entity *>> updater(&f);

When I do this in my real code, I get

invalid template argument for ..., expected compile-time constant expression

So 2 questions:

1, is (m_caller->*T)(e); correct? If it is not, how shout i call it?

2, how can I instantiate it?

Answer 1

template <typename Caller, void (Caller::*Func)(Entity *)>
class Updater 
{
public:
    Updater(Caller *c):m_caller(c){}
    void process(Entity * e)
    {
        (m_caller->*Func)(e); // use pointer to member operator ->*
    }
private:
    Caller *m_caller;
};

// call like this
Foo f;
Updater<Foo, &Foo::Bar> updater(&f);

Answer 2

edit:

user2k5 edited his answer, so I accepted it.

my previous msg:

Thanks to user2k5 I figured out the right working code,

working sample follows: (Foo2 can be replaced by Foo)

#include <iostream>

struct Entity { int i; };

template < typename Caller, void (Caller::*Func)(Entity *)>
class Updater 
{
public:
    Updater(Caller *c):m_caller(c){}
    void process(Entity * e)
    {
        (m_caller->*Func)(e);
    }
private:
    Caller *m_caller;
};

struct Foo
{
    void bar(Entity * e) 
    {
        std::cout << e->i << std::endl;
    }
};

struct Foo2
{
    template <typename T>
    void bar(T t)
    {
        std::cout << t->i << std::endl;
    }
};

int main ()
{
    Foo2 f;
    Updater<Foo2, &Foo2::template bar<Entity *>> updater(&f);
    Entity e;
    e.i = 5;
    updater.process(&e);


    return 0;
}