[Boost-users] Re: BOOST_CHECK_EQUAL and operator<<

Vladimir Prus writes:

Patrick M?zard wrote:

...

namespace testns {

struct Test { .... };

namespace {

std::ostream& operator<<(std::ostream& o, Test const& t) { return o<

...

...

Does not compile and gives me a: C2679: binary 'operator' : no operator defined which takes a right-hand operand of type 'const testns::Test' (or there is no acceptable conversion), the failure being issued by "void boost::test_tools::tt_detail::print_log_value<T>::operator ()(std::ostream &,const T &)"

However, if I define operator<< as a public member of testns::Test it compiles. Same thing if I move it outside the anonymous namespace, with or without static linkage.

I think defining operator<< in the same namespace as the class 'Test' is the only way. Otherwise, it won't be found by ADL and ADL is the only way to boost.test to find your operator<<.

Thank you for your answer. I did not know that ADL treats normal and anonymous namespaces differently (and cannot find anything related in google.groups or other mailing lists). However, it does not work with typedefed types. Here is a new example : //********************* #include using boost::unit_test::test_suite; namespace testns { typedef std::pair Test; std::ostream& operator<<(std::ostream& o, Test const& t) { return o<add( BOOST_TEST_CASE( &testns::free_test_function )); return test; } //********************* I get the same error (C2679) again with the same function instantiation "void boost::test_tools::tt_detail::print_log_value<T>::operator ()(std::ostream &,const T &)". Until now, to make it works I moved operator<< into "namespace boost { namespace test_tools { namespace tt_detail" but that's clearly not the way to go. Can you think of any workaround for this one (workaround or just a correct way to write that :-). I do not pretend to understand anything about ADL...). Patrick Mézard