Namespaces qualified with :: in C++

What does it mean if namespace in C++ is qualified with ::? For e开发者_如何学JAVAxample ::testing::Test.

:: is the scope resolution operator. It always means "search the global namespace for the symbol on the right." For example:

namespace testing {
    int a = 1;
}

namespace foo {
    namespace testing {
        int a = 2;
    }

    int b = ::testing::a; // b has the value 1
    int c = testing::a; // c has the value 2
}

It means that the testing namespace being referred to is the one off the global namespace, rather than another nested namespace named testing.

Consider the following corner case, and probably an example of bad design:

namespace foo
{
    struct gizmo{int n_;};
    namespace bar
    {
        namespace foo
        {
            float f_;
        };
    };

};

int main()
{
    using namespace foo::bar;

    ::foo::gizmo g;
    g.n_;
}

There are 2 namespaces named foo. One is the top-level hanging off of the "global" namespace, and another one is nested within foo::bar. then we go on to using namespace foo::bar, meaning any unqualified reference to gizmo will pick up the one in foo::bar::foo. If we actually want the one in foo then we can use an explicit qualification to do so:

::foo::gizmo

If :: is used on the left side it means the global scope.

If you want an analogy with the filesystem, testing::Test would be a kind of relative path (with respect of König lookup) whereas ::testing::Test would be an absolute path. I hope this analogy makes it clearer and doesn't mess up your mind :-).

I if you precede a variable name with ::, it resolves the variable to the global scope. In this way, you can have both a local variable testing and global variable testing and differentiate between them.

#include <iostream>

using namespace std;
int testing = 37;

int main()
{
   int testing = 42;

   cout << ::testing << ' ' << testing << endl;
}

Output will be 37 42

This invokes something called the qualified name lookup:

$3.4.3/1 - "The name of a class or namespace member can be referred to after the :: scope resolution operator (5.1) applied to a nested-name-specifier that nominates its class or namespace. During the lookup for a name preceding the :: scope resolution operator, object, function, and enumerator names are ignored. If the name found is not a class-name (clause 9) or namespace-name (7.3.1), the program is ill-formed."