关于C ++未命名的命名空间

Question

#include <iostream>

namespace
{
        int a=1;
}

int a=2,b=3;

int main(void)
{
        std::cout<<::a<<::b;
        return 0;
}

I complie it with my g++,but the output is 23, who can explain it? is that a way to get access to the <unnamed> namespace ::a ?

Answer 1

No, you can't. You can work around it thus:

namespace
{
    namespace xxx
    {
        int a = 1;
    }
}
...
std::cout << xxx::a << ::b;

Answer 2

:: in ::a refers to the global namespace. Anonymous namespace should be accessed via just a (or to be more specific, you shouldn't do like this at all)

Answer 3

Using unnamed namespaces, this is not possible. Refer the below article

http://publib.boulder.ibm.com/infocenter/lnxpcomp/v8v101/index.jsp?topic=/com.ibm.xlcpp8l.doc/language/ref/unnamed_namespaces.htm

You have to go for named namespaces.

Answer 4

You can access the global namespace, but don't redefine it.

#include <iostream>

namespace
{
        int a=1;
}


int b=3;

int main(void)
{
        std::cout<<::a<<::b;
    return 0;
}

here the out is 13.