Subject: Re: [boost] [functional] [overloaded_function] polymorphic functions?
From: Gordon Woodhull (gordon_at_[hidden])
Date: 2012-03-14 16:50:45
Hi Lorenzo,
On Feb 18, 2012, at 6:24 AM, John Bytheway wrote:
> On 18/02/12 03:16, Lorenzo Caminiti wrote:
>> I spent some time trying to make overloaded_function support
>> polymorphic functions but I had little success :(
Thank you for attempting it! This is something that came up as a possible better solution in the Conversion review in August. (For which my report is grievously late but not forgotten.)
>>
>> 1) I was able to do this (see code below):
>>
>> overloaded_function<
>> void (int)
>> , functor<poly, void (generic)>
>>> o(f, p);
>> o(321);
>> o(3.21);
>> o("cba");
>>
>> But I don't like it because I explicitly have to pass the functor type
>> poly. I wanted instead to only use generic to tag the template
>> parameters in the signature:
>>
>> overloaded_function<
>> void (int)
>> , void (generic)
>>> o(f, p);
>> o(321);
>> o(3.21);
>> o("cba");
>>
>> I'd consider this acceptable but I wasn't able to program this because
>> I loose p's type (poly) in base so base cannot hold a reference to the
>> functor p... Is there a way around this? (Some type-erasure magic
>> maybe??)
>
> If the type of o did not involve poly, then you could pass a reference
> to o to another translation unit where the definition of poly was not
> visible. In that other translation unit you could call it with an
> arbitrary type, which would have to instantiate poly::operator() without
> sight of its definition. Therefore, I conclude that this is clearly
> impossible.
I had neglected the fact that overloaded_function does type erasure (duh). I guess this would only be possible if the polymorphic result_of (or equivalent) is passed to overloaded_function. Probably not worth it.
Sorry if I sent you on a wild goose chase! I really appreciate the effort, and knowing that this technique could be used as long as type erasure is not required (as it isn't for Conversion).
Cheers,
Gordon