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

Gennadiy Rozental 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 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 Type' to 'typedef pair', 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.