Re: deducing the return type of a function call...

"Alf P. Steinbach" <>
Fri, 27 Nov 2009 00:34:55 +0100
* James:

"Frank Neuhaus" <> wrote in message


"James" <no@spam.invalid> schrieb im Newsbeitrag

I am struggling to find a way to get around having to explicitly pass
a return type for the following callback scheme I am playing around
with. Here is some sample code:

Maybe this can help you?

How in the heck does it determine all of those traits? Anyway, I don't
think it would work for me unless I could do something like:

int foo()
   return 0;

void blah()
   typedef function_traits<foo>::result_type return_type;

and have return_type be an int.

AFAICT, that's just not going to work here.

What am I missing?

Difficult to say since you haven't made your design requirements very clear.

But in general you can't portably deduce function result types in C++98 without
"registering" all relevant types first (you can however do that in C++0x).

Anyway, perhaps this helps:

#include <stdio.h>

template< typename Result >
class AbstractInvokable
     virtual Result operator()() const = 0;

template< typename Result, typename Arg >
class Invokable
     : public AbstractInvokable< Result >
     Arg myArg;
     Result (*myF)( Arg );
     Invokable( Result f( Arg ), Arg a )
         : myArg( a )
         , myF( f )

     virtual Result operator()() const
         return myF( myArg );

template< typename Result, typename Arg >
Invokable< Result, Arg > bind( Result f( Arg ), Arg a )
     return Invokable< Result, Arg >( f, a );

int foo( int x ) { return printf( "f(%d)\n", x ); }
void blah( char const* s ) { printf( "blah(\"%s\")\n", s ); }

int main()
     AbstractInvokable<int> const& f = bind( foo, 42 );
     AbstractInvokable<void> const& b = bind( blah, "whoopie doo!" );


Cheers & hth.,

- Alf

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