C++ Function Pointer as Default Template Argument

Question

how do I write a function pointer as default template argument, I'am guessing to write it like this:

template<typename R,
         typename A,
         typename F=R (*PF)(A)> 
class FunctionPtr { ...

my question is,

1.is it possible?

2.if it is and my guess code above is correct, what the purpose of PF here? do I need this?

Answer 1

1. Yes

2. No, you don't need PF.

    template<typename R, typename A, typename F=R (*)(A)> 

Answer 2

1. Yes, it is possible

2. The PF not only is useless but must be removed in this context. It would, for example, be necessary in the context of a function pointer declaration :

    int (*PF)(double) = &A::foo; // declares a 'PF' variable of type 'int (*)(double)' 

   but it is neither required nor legal here.

Answer 3

I suggest typedef ing pointer to function. typedef R (*PF)(A); then give PF as a default template argument.