Gennadiy Rozental <gennadiy.rozental <at> thomson.com> writes:
You have several choices
1. You could define operator<< in namespace foo
It does not work because of the typedef (see below). Please see my first answer to Vladimir. It works if it is defined in namespace std:: however.
2. You could use BOOST_TEST_DONT_PRINT_LOG_VALUE(your type) on global scope to prevent values of this type being printed at all
Okay, but since I am the one writing the tests, I would simply use BOOST_CHECK().
3. You could write struct C2 : foo::C {}; And unless you use non-default constructor it will work.
Maybe, but it seems much more work just to use BOOST_CHECK_EQUAL(). I could understand the anonymous namespace and typedefs have issues with ADL and there are workaround to be found. For those interested with ADL rules, I finally got an old post from llewely on comp.lang.c+.moderated on a similar problem stating that (look for "Re: stream iterators and operator<< " in google.groups): <quote> A typedef name is not a type. The occurance of a typedef name in a scope does not add that scope to consideration in argument-dependent lookup. [3.4.2/2] Unqualified lookup does not occur because inside the definition of the function called by '*it = *v.begin()', operator<< is used for a function call. Instead, argument-dependent lookup is used. [3.4.2/1] Argument dependent lookup will lookup names in the classes and namespaces associated with the argument types. pair<int,int> is a template-id, and defined at namespace scope in namespace std, with arguments int,int, which have no associated namespaces or classes, so the only associated namespace is std. [3.4.2/2, bullet 8] I believe both compilers are conforming in this respect. Some notes: If you move the operator<< into namespace std, it will be found during argument-dependent lookup (I tested with gcc 3.2.1), except for the caveat that one is not allowed to add overloads to namespace std. If you change 'typedef pair<int,int> Type' to 'typedef pair<int,Foo>', where Foo is in the global namespace, the operator<< will also be found. </quote> Personnaly, I would stick with declaring the operator<< in std:: for the moment. That's ugly but that's just tests after all. Maybe you could just add a documentation note on this topic. Thank you for your answer. Patrick Mézard.