From: daniel.kruegler@googlemail.com
On 2012-04-13 04:46, James K. Lowden wrote:
> I feel like I've asked this question before, but if so I can't find
> it and still don't get it, so here goes....
Yes, this issue occurs once in a while.
> I want to copy a std::map to std::cout. I define operator<< for the
> map's value_type. I can use that operator by defererencing a the
> map's iterator. Why can't I compile std::copy with
> ostream_iterator?
>
> The program that follows compiles and runs as is. Change it to #if
> 0 instead, and get "no match for 'operator<<' ". g++ says there are
> many candidates, all standard, none mine.
>
> The whole "no match" error is:
> /usr/include/g++/bits/stream_iterator.h:196: error: no match for
> 'operator<<' in '*((std::ostream_iterator float>, char, std::char_traits >*)
> this)->std::ostream_iterator, char,
> std::char_traits >::_M_stream<< __value'
>
> Many thanks for your insight.
ADL name-lookup wont find your operator<< overload, because it only
looks in namespace std, but your operator<< overload for std::pair is
not part of this namespace. According to the library requirements,
user-code shall not add any components (except template
specializations with at least one user-defined type) to namespace std.
The only portable way to realize this use-case is to
a) Provide an io wrapper for the wanted library-component(s)
b) Use std::transform instead of std::copy.
Put it all together, this may look like so:
template
struct pair_io {
const std::pair& p;
pair_io(const std::pair& p) : p(p) {}
};
template
inline
pair_io io(const std::pair& p) { return pair_io(p); }
template
inline
ostream& operator<<( ostream& os, const pair_io& value )
{
return os << "(" << value.p.first << ", " << value.p.second << ")";
}
and at the call point:
std::transform( m.begin(), m.end(),
ostream_iterator >(cout, "\n"),
io);
HTH & Greetings from Bremen,
Daniel Krügler
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