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Home/ Questions/Q 4006732
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T08:32:58+00:00 2026-05-20T08:32:58+00:00

After running the following code segment, the output is Outer. Inner. Inner. I know

  • 0

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
    { 
        void message( )
        {
            cout << "Inner.\n";
        }
    }
}
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  1. Editorial Team
    Editorial Team
    2026-05-20T08:32:59+00:00Added an answer on May 20, 2026 at 8:32 am

    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();.

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