Re: [boost] result_of usability

From: David Abrahams (dave_at_[hidden])
Date: 2008-04-29 12:29:32

on Tue Apr 29 2008, Anthony Williams <anthony_w.geo-AT-yahoo.com> wrote:

> David Abrahams <dave_at_[hidden]> writes:
>
>> Now I just ran into another issue, from the other side of the result_of
>> interface.  If I have a function template taking a function object
>> argument by reference, and I want to use result_of with it, I need to do
>> something special to account for the case where I've been passed a
>> function reference.  To wit:
>>
>>          template <class F>
>>          typename result_of<F()>::type 
>>          call(F const& f)
>>          {
>>              return f();
>>          }
>>
>>          int f() { return 0; }
>>
>>          int x = call(f);
>>
>> The problem is that in the above case, F is deduced as int(), and we
>> form an illegal function type (one that itself returns a function type)
>> in result_of<F()>::type.
>>
>> Is there an easier way?
>
> In cases like this, I've often just overloaded call():
>
>     template<typename R>
>     R call(R(*f)())
>     {
>         return f();
>     }
>
> If you've got function call arguments to deal with, it's not so easy, as you
> can end up with a lot of overloads.

Yeah, and it doesn't work if you're really defining class templates
instead of simple functions because there you won't get the usual decay
to function pointer types.

    
-- 
Dave Abrahams
Boost Consulting
http://boost-consulting.com