Re: [boost] Overloading a function taking a boost::function

From: Paul Mensonides (pmenso57_at_[hidden])
Date: 2002-05-12 18:04:26

> > But that only says how many parameters there are not there types.  Using
> the
> > sizeof trick won't work for a complete function_traits class.
> >
> > Paul Mensonides
>
> Very true. But as was said to me earlier in this thread, for some function
> objects
> not all the parameter types can be determined. (Some results of a
> boost::bind for
> instance.)
>
> Dirk Gerrits

BTW, I already have my own full-fledge function_traits implementation, but it
uses my own preprocessor library to generate it, so I haven't post it.  However,
all of these things work:

typedef function_traits<void (int, int, double)> a;
std::cout
    << typeid(a::return_type).name() << '\n' // void
    << a::has_ellipsis << '\n' // false
    << a::arity << '\n' // 3
    << typeid(a::parameter<0>::type).name() << &std::endl; // int

typedef function_traits<void (*)(...)> b;
std::cout
    << typeid(b::return_type).name() << '\n' // void
    << b::has_ellipsis << '\n' // true
    << b::arity << '\n' // 0

typedef function_traits<double (&)(void)> c;
std::cout
    << typeid(c::return_type).name() << '\n' // double
    << c::has_ellipsis << '\n' // false
    << c::arity << '\n' // 0

typedef function_traits<int (X::*)(int, short) const> d;
std::cout
    << typeid(d::return_type).name() << '\n' // int
    << d::has_ellipsis << '\n' // false
    << typeid(d::member_of).name() << '\n' // X
    << d::is_const << '\n' // true
    << d::arity << '\n' // 2
    << typeid(d::parameter<1>::type).name() << &std::endl; // short

// etc.

I was just wondering what other things are required that I may have missed.

Paul Mensonides