On 4/1/2011 12:33 AM, Joel Falcou wrote:

On 01/04/11 01:32, Noah Roberts wrote:

...

I had to write an invoke metafunction because I couldn't seem to find one that looked like it was what I needed:

template < typename MF > struct invoke { typedef typename MF::type type; };

I've looked through the documentation TOC: http://www.boost.org/doc/libs/1_42_0/libs/mpl/doc/refmanual/refmanual_toc.ht...

Is there something I'm missing?

http://www.boost.org/doc/libs/1_46_1/libs/mpl/doc/refmanual/apply.html

apply invokes a metafunction *class*.

On 4/1/2011 9:28 AM, Krzysztof Czainski wrote:

2011/4/1 Noah Roberts mailto:roberts.noah@gmail.com>

I had to write an invoke metafunction because I couldn't seem to find one that looked like it was what I needed:

template < typename MF > struct invoke { typedef typename MF::type type; };

I've looked through the documentation TOC: http://www.boost.org/doc/libs/1_42_0/libs/mpl/doc/refmanual/refmanual_toc.ht...

Is there something I'm missing?

How do you intend to use invoke?

The way I understand it, invoke<MF> would be equivalent to MF itself.

No, it would be the 'type' definition inside of MF.

2011/4/1 Noah Roberts

How do you intend to use invoke? template < typename Sq > // Sq is full of metafunctions struct do { typedef typename fold < Sq , start_state , funky,_1>

...

::type };

I would like to empasize here, that I'm no expert -- on the contrary, I'm trying to learn bases of mpl ;-) I don't understand your example above. I think 'do' cannot be a name of a struct, and typedef typename fold<...>::type seems to be missing something... Did you mean: struct do_ { typedef typename fold < Sq , start_state , funky,_1>

::type type; };

Could you please explain, what funky does, and start_state too? I would like to see an example I can compile, and if my understanding is correct, make a simple modification to avoid the need for invoke. Regards, Kris