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

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<<.

Note these workaround do not work or cannot be applied if the type is declared by a typedef. In this case, I have to define again the operator in namespace boost.test_tools.tt_detail.

If the type is declared by a typedef: typedef foo::C C2; you can define the operator<< in the 'foo' namespace, and that should work. - Volodya