18 Dec
2009
18 Dec
'09
5:58 p.m.
Jeremiah Willcock wrote:
OK. It actually should be "typename boost::result_of
::type" like what is in the specialization of property_traits.
Ok fixed it.
In case your operator()() in the functor is non-const (even though I said I was assuming it was const); switching to a reference is fine.
And to a const reference ? My operator() is const.
OK. Did it work otherwise?
I didn't test it on a real world test case yet but I added a std::cout in my functor's operator() and it is called, so I'd say yes. -- Maxime