C++ Function Pointer as Default Template Argument

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开发者_开发百科 it possible?

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

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1. Yes

2. No, you don't need PF.

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

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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.

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