From: Patrick Mézard (pmezard_at_[hidden])
Date: 2005-03-03 03:12:12
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.