Question

Is it possible to deduce a non-type function pointer type template argument (function pointer) from a function argument?

template <void(*fptr)()>
  void test(void(*fp)()) { fp(); }

to call this function I must explicitly declare the function template parameter:

test<somefunc>(somefunc);

I know I could also do it this way:

template <void(*fptr)()>
  void test() { fp(); }

test<somefunc>();

But I am just wondering if It is possible to do this way:

template <void(*fptr)()>
  void test() { fp(); }

test(somefunc);

Is it possible to declare in such a way that the compilier (GCC 4.7) will deduce from the function arguments?

Thanks alot in advance, been really wondering how to do this.
-Bryan

Answer 1

Bryan, that seems to be quite eccentric mix of low-level C and C++. Why do you need that? Why not to use functors?

struct clean
{
    void operator() () 
    {
        // do something here        
    }
};

template <typename FuncType> void call_func(FuncType func)
{
    func();
}

// here is how to pass 'clean' to be called
call_func(clean());

More on functors, for example, here: http://www.cprogramming.com/tutorial/functors-function-objects-in-c++.html