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

10 Apr 2012

      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?

For example:

#include <boost/function.hpp>
#include <boost/typeof/typeof.hpp>
#include BOOST_TYPEOF_INCREMENT_REGISTRATION_GROUP()

//BOOST_TYPEOF_REGISTER_TEMPLATE(boost::function, 1) // (1)
BOOST_TYPEOF_REGISTER_TEMPLATE(boost::function1, 2) // (2)

int main() {
    typedef BOOST_TYPEOF(&boost::function<int (int)>::operator()) f;
    return 0;
}

In order to compile this (on MSVC 8.0 and GCC 4.5.3) with
BOOST_TYPEOF_EMULATION #defined, I have to register boost::function1 as in
line (2) but I'd expect it to work with the registration of boost::function
of line (1) instead.

Why do I have to register boost::function1 (2) instead of boost::function
(1) for type-of emulation?

Thanks a lot.
--Lorenzo

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

lcaminiti