std::copy to std::cout for std::pair

Question

I have next code:

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

//namespace std
//{

std::ostream& operator << ( std::ostream& out, 
                const std::pair< size_t, size_t >& rhs )
{
    out << rhs.first << ", " << rhs.second;
    return out;
}
//}

int main() 
{

    std::map < size_t, size_t > some_map;

    // fill  some_map with random values
    for ( size_t i = 0; i < 10; ++i )
    {
        some_map[ rand() % 10 ] = rand() % 100;
    }

    // now I want to output this map
    std::copy( 
        some_map.begin(), 
        some_map.end(), 
        std::ostream_iterator< 
              std::pair< size_t, size_t > >( std::cout, "\n" ) );

    return 0;
}

In this code I just want copy map to output stream. For do this I need define operator <<(..) - OK. But according names finding rules compiler can't find my operator<<().
Because std::cout, std::pair and std::copy which called my operator<< - all from namespace std.

Quick solution - add my oerator<< to std namespace - but it is ugly, imho.

What solutions or workaround for this problem do you know?

Answer 1

There is no standard way to cout a std::pair because, well, how you want it printed is probably different from the way the next guy wants it. This is a good use case for a custom functor or a lambda function. You can then pass that as an argument to std::for_each to do the work.

typedef std::map<size_t, size_t> MyMap;

template <class T>
struct PrintMyMap : public std::unary_function<T, void>
{
    std::ostream& os;
    PrintMyMap(std::ostream& strm) : os(strm) {}

    void operator()(const T& elem) const
    {
        os << elem.first << ", " << elem.second << "\n";
    }
}

To call this functor from your code:

std::for_each(some_map.begin(),
              some_map.end(),
              PrintMyMap<MyMap::value_type>(std::cout));