Re: Dummy streams: eating std::endl?

From:

Alberto Ganesh Barbati <AlbertoBarbati@libero.it>

Newsgroups:

comp.lang.c++.moderated

Date:

1 Oct 2006 05:57:26 -0400

Message-ID:

<8TDTg.130466$_J1.886996@twister2.libero.it>

lindahlb@hotmail.com ha scritto:

#include <iostream>
using namespace std;

struct Ignore_Stream
{
<snip>
template<typename Char, typename Traits>
Ignore_Stream operator<<(basic_ostream<Char,Traits> &
(*)(basic_ostream<Char,Traits> &)) const;
<snip>
};

<snip>

It should work, considering basic_ostream gets it to work (gcc STL uses
a signature almost identical to the signature in the alternate
definition).

In addition to what other posters have already correctly said, I would
like to notice that the signature you think it's "almost" identical is
in fact completely different! Compare the signature in your example
above with this one:

template <class charT, class traits = char_traits<charT> >
class basic_ostream : ...
{
    //...
    basic_ostream<charT,traits>& operator<<
      (basic_ostream<charT,traits>& (*pf)(basic_ostream<charT,traits>&));
    //...
};

Do you spot the difference? Yours is a member template, while ostream's
is not: it's a regular member (of a class template). That means the
template parameters in ostream's operator<< are not deduced: they *must*
match the object's char and traits. In the case of std::endl, which is a
function template, this fact allows the compiler to select one (and only
one) specialization of the std::endl.

In your case, the compiler needs to deduce the parameters and can't
because there are (potentially) infinite specializations of std::endl
that matches your signature.

Regards,

Ganesh

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