Subject: Re: [boost] [function] why do I function1 for type-of emulation?
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2012-04-10 16:32:05
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
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.
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