nishantjr nishantjr - 1 month ago 6
C++ Question

BOOST_CHECK_EQUAL with pair<int, int> and custom operator <<

When attempting to do a BOOST_CHECK_EQUAL(pair, pair),
gcc doesnt find the stream operator for pair, inspite of declaring it.
The funny thing is that std::out finds the operator.

ostream& operator<<(ostream& s, const pair<int,int>& p) {
s << '<' << p.first << ',' << p.second << '>';
return s;
}


BOOST_AUTO_TEST_CASE(works)
{
pair<int,int> expected(5, 5);
pair<int,int> actual (5, 5);
std::cout << expected << std::endl;
std::cout << actual << std::endl;
BOOST_CHECK(actual == expected);
}

BOOST_AUTO_TEST_CASE(no_work)
{
pair<int,int> expected(5, 5);
pair<int,int> actual (5, 5);
BOOST_CHECK_EQUAL(actual, expected);
}


This doesnt compile with the error:

... instantiated from here
../boost-atp/release/include/boost/test/test_tools.hpp:326:9: error: no match for ‘operator<<’ in ‘ostr << t’

Answer

Putting operator<< in std like Remus's answer is undefined behavior in the C++ 14 draft (N4296 section:17.6.4.2.1). Boost provides a hook (used by this answer) and you can write:

namespace boost
{
    namespace test_tools
    {
        template<typename T,typename U>
        struct print_log_value<std::pair<T, U> >
        {
            void operator()(std::ostream& os, std::pair<T, U> const& pr)
            {
                os << "<" << std::get<0>(pr) << "," << std::get<1>(pr) << ">";
            }
        };
    }
}

print_log_value is a template so if you are not declaring a templated value like pair<T,U>, you will need to write something like:

template<>
struct print_log_value<MyType>{ /* implementation here*/ };

Edit

If you are using boost 1.59 or later you need to use namespace boost::test_tools::tt_detail instead. That is, the code needs to start:

namespace boost
{
    namespace test_tools
    {
        namespace tt_detail
        {
Comments