typedef resolution across namespaces

Question

I am currently confused with the way "using (namespace)" statements work in C++.

I have:

//somewhere in included headers
typedef unsigned int uint;

namespace mine {
    typedef unsigned int uint;
}
namespace other {
    using namespace mine;
    void foobar () {
        uint offender = i;
    }
}

Results in (paraphrased):
reference to 'uint' is ambiguous. candidates are
typedef unsigned int uint
and
typedef unsigned int mine::uint

Meanwhile, when I do

namespace other {
    using namespace mine;
    using mine::uint;

    void foobar () {
        uint offender = i;
    }
}

Everything works fine. It seems strange to me that "using identifier;" changes the visibility of other typedef definition (conceals the global one?). Can someone point me to what kind of rules in C++ govern resolution of typedefs across namespaces?

Answer 1

Your original code confuses the compiler, because the uint can be either

::uint

or

::mine::uint

therefore compiler throws that error message to you. In the "fix", the using mine::uint explicitly specified that ::mine::uint shall be preferred.

However, typedef conflict shall be avoided IMO. It makes the code so hard to maintain.

Answer 2

A name made visible by a using-directive appears in the nearest enclosing scope that contains [- directly or indirectly -] both the using-directive and the nominated namespace. (7.3.4 [namespace.udir])

This means that both uint declarations appear at the global namespace scope when looked up after the using-directive in other.

A using-declaration, like any other declaration, declares a name at the scope in which it appears. This is why, in the second example, using mine::uint; hides the uint introduced by using namespace mine; as the latter appears to come from the global scope.