Multiple instances of operator of scope in namespace using statement

Sam_Gregson · #1 11-Nov-2009, 08:52

Hi guys - just a quick one (I think!) -

I am new to C++ programming, but have done some programming in Fortran, C and VBA.

I am currently undertaking a particle physics PhD at Cambridge University in the UK and have been given a large and complex LHC/CERN program to understand.

I am unclear on the following code section:

CPP / C++ / C Code:

using namespace Rich::Rec::MC;

Does this code line simply mean that I can use variables defined in all namespaces i.e Rich, Rec and MC without constantly using the operator of scope :: (e.g. Rich::a or Rec::a) ???

Isn't this programming style open to complications? e.g. if there is a variable called "a" in the Rich and Rec namespaces then the compiler will not know which "a" I am referring to!

Thanks,

Sam

davekw7x · #2 11-Nov-2009, 12:39

Quote:

Originally Posted by Sam_Gregson

...

CPP / C++ / C Code:

using namespace Rich::Rec::MC;

Does this code line simply mean that I can use variables defined in all namespaces i.e Rich, Rec and MC without constantly using the operator of scope :: (e.g. Rich::a or Rec::a) ???

No. It means that you can use stuff defined in the Rich::Rec::MC namespace without the scope operator.

CPP / C++ / C Code:

namespace n1 {
    const int i = 1;
    const int i1 = 11;
    namespace n2 {
        const int i = 2;
        const int i2 = 22;
        namespace n3{
            const int i = 3;
            const int i3 = 33;
        }
    }
}
using namespace n1::n2::n3;

#include <iostream>
using namespace std;  // No conflicts since "i" and "i3" are not in the std namespace

int main()
{
    cout << "        n1::i = " << n1::i << endl;
    cout << "    n1::n2::i = " << n1::n2::i << endl;
    cout << "n1::n2::n3::i = " << n1::n2::n3::i << endl;
    cout << "(Ta-daa)    i = " << i << endl;
    //cout << "i1 = " << i1 << endl; //won't work without scope
    //cout << "i2 = " << i2 << endl; //won't work without scope

    return 0;
}

Output:

Code:

        n1::i = 1
    n1::n2::i = 2
n1::n2::n3::i = 3
(Ta-daa)    i = 3

Now, uncomment the i2 or i3 print statements and try to compile.

Regards,

Dave

Sam_Gregson · #3 12-Nov-2009, 18:49

I will have a look at that in the morning - thanks Dave-

Sam

Sam_Gregson · #4 13-Nov-2009, 05:03

Super - cheers Dave

Sam

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Multiple instances of operator of scope in namespace using statement

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