Re: [boost] [function] why do I function1 for type-of emulation?

AMDG On 04/10/2012 12:01 PM, lcaminiti wrote:

Hello all,

I was trying to use Boost.Function together with Boost.Typeof. It seems that if I run in type-of emulation mode I have to register boost::functionN instead of boost::function even if I only use boost::function... do you know why? <snip>

Because the type of &boost::function<int(int)>::operator() is int (boost::function1<int, int>::*)(int) The type of a member function pointer is determined by the class that it is declared in, not the class that it is accessed through. In Christ, Steven Watanabe