Re: cout << vector<string>

Maxim Yegorushkin <>
Tue, 11 Nov 2008 07:19:06 -0800 (PST)
On Nov 11, 2:40 pm, Hendrik Schober <> wrote:

Maxim Yegorushkin wrote:

On Nov 11, 1:54 pm, Maxim Yegorushkin <>

On Nov 11, 11:20 am, Hendrik Schober <> wrote:


  Having followed this whole battle of words, I wonder what's
  wrong with putting this operator into the global namespace
  and altogether avoiding the hassle of having to know about
  things you're not supposed to know about?

Nothing wrong and this is indeed the correct way to do so. I confused
this case with another one, sincere apologies.

This what I was confusing it with:

    #include <map>
    #include <iostream>
    #include <iterator>

    // should be in namespace std::
    template<class T, class U>
    std::ostream& operator<<(std::ostream& s, std::pair<T, U> const=

& p)

        return s << p.first << ' ' << p.second;

    int main()
        typedef std::map<int, int> Map;
        Map m;
            , m.end()
            , std::ostream_iterator<Map::value_type>(std::c=



It won't compile unless operator<<(std::ostream& s, std::pair<T, U>
const& p) is in namespace std.

  I would have asked the same question for this code. :)
  I don't understand why it doesn't compile. It comes down
  to this
    ostr << val;
  with 'ostr' being an 'std::basic_ostream<>' and 'val'
  being an 'std::pair<>'. Why doesn't this find the global

Because expression "ostr << val" is template argument dependent and
thus is bound at the second phase of the two-phase name lookup. At the
second phase it uses ADL only to search for functions within
namespaces associated with ostr and val. ostr is std::basic_ostream
and val is std::pair<int, int>, thus one associated namespace is std.
int has no associated namespaces. So, the only namespace considered
for expression "ostr << val" is std, which lacks a suitable


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