How do you typedef a function pointer type with parameter pack arguments

Question

What's the syntax to typdef a parameter pack into a function pointer?

I want to be able to typedef a function pointer, but the compiler complains when I do something like this

template< class ...Args >
struct method { typedef typename void(*type)(void*, Args...); };

with a message along the lines of error: expected nested-name-specifier before 'void'

See also: http://stackoverflow.com/questions/610245/where-and-why-do-i-have-to-put-the-template-and-typename-keywords — Sam Varshavchik, Nov 29 '16 at 00:36
Shit, sorry guys. I misread an error from my compiler that said `error: need 'typename' before ...` but I put it there in the struct instead of the reference to the dependent type :/ — Stefan Sullivan, Nov 29 '16 at 01:23

score 2 · Answer 1 · answered Nov 29 '16 at 00:28

2

You shouldn't use typename here because there is no nested-name-specifier.

answered Nov 29 '16 at 00:28

Brian Bi

max66 · Answer 2 · 2016-11-29T00:49:24.680

1

I suppose that you should remove typename from the typedef line

template <typename ... Args>
struct method
 { typedef void(*type)(void*, Args...); };

Another solution could be using using instead of typedef (IMHO is a little clearer)

template <typename ... Args>
struct method
 { using type = void(*)(void*, Args...); };

edited Nov 29 '16 at 00:49

answered Nov 29 '16 at 00:28

max66

2 Answers2