What is the syntax for a function-pointer typedef?

Question

I am working on a thread pool and to avoid long qualifier names I would like to use typedef declaration.

But it is not as easy as it seems to be:

typedef unsigned ( __stdcall *start_address )( void * ) task;

When I tried it that way I got:

error C3646: 'task' : unknown override specifier

error, after playing for a little while with this declaration I'm stuck and can't find any reasonable solution to use to declare such type of typedef.

Task is a word which I want to use instead of this long declaration. But nevermind, the solution is found. Thanks to all, specially pb2q :) — unresolved_external
– unresolved_external, Commented Aug 30, 2012 at 13:49
Be warned: Implementig a reasonably good thread pool is far from trivial. Did you consider to use an existing implementation? — Paul Michalik
– Paul Michalik, Commented Aug 30, 2012 at 14:05
Well it is in educational purposes, so I think I will take a risk :) — unresolved_external
– unresolved_external, Commented Aug 30, 2012 at 14:11
@unresolved_external: and you didn't wonder what start_address was for? — PlasmaHH
– PlasmaHH, Commented Aug 30, 2012 at 14:34

pb2q · Accepted Answer · 2012-08-31 20:18:08Z

15

When creating a typedef alias for a function pointer, the alias is in the function name position, so use:

typedef unsigned (__stdcall *task )(void *);

task is now a type alias for: pointer to a function taking a void pointer and returning unsigned.

edited Aug 31, 2012 at 20:18

answered Aug 30, 2012 at 13:21

pb2q

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3 Comments

Christian Rau Over a year ago

@unresolved_external Don't forget to accept this answer if is the best solution to your problem (which from your comment it seems it is).

Ed Swangren Over a year ago

@unresolved_external: Forgetting function pointer syntax in C is nothing to be ashamed of :D

Youda008 Over a year ago

that sucks ... i like C++, but the syntax is really retarded, when it comes to pointers and constants

Matthieu M. · Accepted Answer · 2012-08-30 13:48:59Z

Since hmjd's answer has been deleted...

In C++11 a whole newsome alias syntax has been developed, to make such things much easier:

using task = unsigned (__stdcall*)(void*);

is equivalent the to typedef unsigned (__stdcall* task)(void*); (note the position of the alias in the middle of the function signature...).

It can also be used for templates:

template <typename T>
using Vec = std::vector<T>;

int main() {
    Vec<int> x;
}

This syntax is quite nicer than the old one (and for templates, actually makes template aliasing possible) but does require a quite newish compiler.

Dude · Accepted Answer · 2012-08-30 14:21:34Z

0

Sidenote: The __stdcall part might break your code under different compilers / compiler settings (unless the function is explicitly declared as __stdcall, too). I'd stick to the default calling convention only using proprietary compiler extensions having good reaons.

answered Aug 30, 2012 at 14:21

Dude

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1 Comment

Adam Rosenfield Over a year ago

Or use a macro that is #define'd to either __stdcall or something else depending on the compiler. Although typically you only explicitly declare the calling convention when interacting with a (usually platform-specific) library that has a defined calling convention.

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