Re: How to convert a pointer to member function type to a const member one

Hi

mcostalba wrote:

<snip>
>          template<class U> static
>          yes_type detect(helper<&U::operator()>*);
> 
>          template<class U> static
>          yes_type detect(helper_const<&U::operator()>*);

Not sure if the previously discussed solutions does sup****t it but this
version at least seems to me broken in regard to overloaded operator().
Supose you have a static_visitor functor (a functor with many overloaded
operator()(T&) each taking a different T, usually used with a static
visitor aproach). It would be nice if is_compatible<> would determine or
not if there is a suitable overload for the signature you want. The
expression "&U::operator()" would give an amibuity error in case of
overloading (which is usually solved either wraping it with a static_cast
to the exact function type wanted from the overload set or used as
initialization expression for a pointer/reference to function variable).
So
a solution might be "helper<static_cast<ptm>(&U::operator())>" and
something similar for the const version.

I'm also a little puzzled that "&U::operator()" does not give compile
error
in case of not giving a functor to is_compatible<> because I don't think
SFINAE helps here, you have an invalid "value expression" not an
invalid "type expression", but quite possible I misunderstand this.

-- 
Dizzy


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