Re: [Boost-users] How to deduce (possibly void) return type of member function?

Terry Golubiewski schrieb:

How? I read the docs you referenced (previously), but I did not see a concrete example.

i thought it was straightforward, but looking into details it was not the right reference. =-O However, below is a solution which relies on BOOST_TYPEOF. BOOST_TYPEOF will have a problem if the return_type is void. So i replace it with mpl::void_.

typedef typename boost::result_of< ...what goes here???... (double)>::type f_double_return_type;

terry

HTH Kim # include # include # include # include # include struct X { int f(int) { return 1; } double f(double) { return 1.0; } void f(char) {} }; template typename boost::mpl::if_< boost::is_same , boost::mpl::void_ , return_type

::type deduce_type(return_type (X::*)(Param));

typedef BOOST_TYPEOF(deduce_type<char>(&X::f)) void_; typedef BOOST_TYPEOF(deduce_type<int>(&X::f)) int_; typedef BOOST_TYPEOF(deduce_type<double>(&X::f)) double_; int main() { BOOST_MPL_ASSERT((boost::is_same)); BOOST_MPL_ASSERT((boost::is_same)); BOOST_MPL_ASSERT((boost::is_same)); return 0; }