Scope of using namespace

After running the following code segment, the output is

Outer.

Inner.

Inner.

I know this is about the usage of "namespace", but do not understand why the call of "Inner::message()" print out "Inner". Thanks for explanation.

#include <iostream>
using namespace std;
namespace Outer
{ 
    void message( );
    namespace Inner
    {  
        void  message( );
    }
}
int main( )
{ 
    Outer::message( );
    Outer::Inner::message( );

    using namespace Outer;
    Inner::message( );

    return 0;
}

namespace Outer
{   

    void message( )
    { 
        cout<< "Outer.\n";
    }
    namespace Inner
    { 开发者_运维技巧
        void message( )
        {
            cout << "Inner.\n";
        }
    }
}

---

This makes perfect sense. Your are using namespace Outer. Inside of namespace Outer you have two members...

• void message();
• void Inner::message();

You explicitly scoped into Inner and called message there. Why would you expect otherwise? Had you not explicitly scoped into Inner, then it would have called void Outer::message();.

---

What other function could Inner::message() possibly call?

---

They key is the line using namespace Outer; - this basically merges the namespace Outer with the global namespace, which means anything - even other namespaces inside that one - is now directly accessible from the global namespace, as if you were both in global and in the Outer namespace at once.

By the way: The only other alternative, if you assumed it would not print "Inner.", were for the code not to compile at all.

---

Inner::message() prints out "Inner". because of this

namespace Inner  
{   
    void message( )  
    {  
        cout << "Inner.\n";  
    }  
}