Question

If I do not to want to create a new container in order to do so?

Answer 1

I’ve written a snippet to do that for debugging. For example:

std::stack<int> s; // works with std::queue also!
s.push(1);
s.push(2);

std::cout << s; // [ 1, 2 ]

Please forgive me for this hackish code! but this is what I’ve written months ago:

#include <stack>
#include <queue>
#include <ostream>

template <class Container, class Stream>
Stream& printOneValueContainer
    (Stream& outputstream, const Container& container)
{
    typename Container::const_iterator beg = container.begin();

    outputstream << "[";

    while(beg != container.end())
    {
        outputstream << " " << *beg++;
    }

    outputstream << " ]";

    return outputstream;
}

template < class Type, class Container >
const Container& container
    (const std::stack<Type, Container>& stack)
{
    struct HackedStack : private std::stack<Type, Container>
    {
        static const Container& container
            (const std::stack<Type, Container>& stack)
        {
            return stack.*&HackedStack::c;
        }
    };

    return HackedStack::container(stack);
}

template < class Type, class Container >
const Container& container
    (const std::queue<Type, Container>& queue)
{
    struct HackedQueue : private std::queue<Type, Container>
    {
        static const Container& container
            (const std::queue<Type, Container>& queue)
        {
            return queue.*&HackedQueue::c;
        }
    };

    return HackedQueue::container(queue);
}

template
    < class Type
    , template <class Type, class Container = std::deque<Type> > class Adapter
    , class Stream
    >
Stream& operator<<
    (Stream& outputstream, const Adapter<Type>& adapter)
{
    return printOneValueContainer(outputstream, container(adapter));
}

You can stream std::stack and std::queue just like any other supported type!