Re: operator<< for std::map

From:
Barry <dhb2000@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sun, 21 Oct 2007 10:43:16 CST
Message-ID:
<fffa69$nfq$1@news.cn99.com>
Richard Smith wrote:

On 19 Oct, 23:28, mojmir <svobod...@gmail.com> wrote:

hello,

i am trying to dump whole map container to ostream like this:

template<typename T, typename U>
inline
std::ostream & operator<< (std::ostream & s, std::pair<T,U> const & p)
{
     s << "<" << p.first << "," << p.second << ">";
     return s;

}

template<typename K, typename T>
inline
std::ostream & operator<< (std::ostream & s, std::map<K,T> const & m)
{
     s << "(";
     typedef typename std::map<K,T>::value_type t;
     std::copy(m.begin(), m.end(), std::ostream_iterator<t>(s, " "));
     s << ")";
     return s;
}

but the copy line fails to compile and i am unable to guess why.


I assume that you have the correct #includes?

   #include <algorithm> // For std::copy
   #include <iterator> // For std::ostream_iterator
   #include <map> // For std::map
   #include <ostream> // For std::ostream
   #include <utility> // For std::pair

Beyond that, whether the code works or not will depend exactly what
your test case looks like, but there definitely is scope to be bitten
by a fairly subtle problem. The following test case, for example,
should exhibit the problem:

   #include <iostream>
   #include <string>

   int main() {
     std::map< std::string, int > m;
     std::cout << m;
   }

The fundamental problem is to do with name lookup of operator<<, and
because you have put it in the global namespace it is not being
found. The short answer is that you can fix this by moving the two
operator<<s into namespace std -- technically this isn't legal as
you're not really allowed to put overloaded operators into namespace
std, but it's probably the least worst solution.


Actually, we only need to put the *first* operator<< into namespace std,
to make the code compiles.

Because within std::ostream_iterator, only namespace std is visible.

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