Extremely basic question about namespaces in c++

Question

using namespace X;
    cout << var;
    using Y::var;
    cout << var;

So say I have a namespace X and a namespace Y that both contain a variable of type int called var. When I say using namespace X; what I imagine happening is if I use some variable that isn't in the global namescope what basically happens is it goes okay I'm gonna look for var in namespace X but now that I also use Y::var what does this exactly mean? Does that just say var is the same as Y::var? But then in that case what's happening with using namespace X does it not even look for var in there because I said I'm using Y::var?

Just create an example and check what compiler will do. This is best way to learn. If we serve you an answer you will forget it in 5 minutes. — Marek R, Oct 19 '21 at 13:56
Depends on context. Create a [mcve]. Instead of describing how the program is, show us the program. — eerorika, Oct 19 '21 at 13:59
There's a program that can answer these types of questions. [You can ask it](https://gcc.godbolt.org/z/Tca744nnK) — Raymond Chen, Oct 19 '21 at 14:01
I already did check what the compiler did and I got what I expected which is using namespace X I printed var and I got the var inside that namespace then when I used Y::var and printed that out I got the var inside the Y namespace I just want to ask to make sure is what I said before right? Me saying I'm using Y:::var kind of just overwrites using namespace X kind of? At least for var sorry if I'm not explaining this well but when I say I'm using Y::var is that basically saying ok ignore that other namespace I'm using if you see var in the code I'm talking about Y::var? — jaylse2, Oct 19 '21 at 14:01
"_I'm using Y:::var kind of just overwrites using namespace X kind of?_" - It only overrides it for `X::var`. The rest of `X` is still used. — Ted Lyngmo, Oct 19 '21 at 14:10
Yeah I said at least for var right after that thanks I basically understood this already I just wanted to make sure I was thinking of it the right way thanks — jaylse2, Oct 19 '21 at 14:12

Vlad from Moscow · Accepted Answer · 2021-10-19T14:28:10.307

After the using directive

using namespace X;

the compiler uses the unqualified name lookup to find the name var used in the following statement

cout << var;

And due to the using directive it will find the variable var in the namespace X.

This using declaration

using Y::var;

introduces the variable var from the namespace Y in the current scope and the next statement

cout << var;

will use the variable var from the namespace Y.

Here is a demonstration program.

#include <iostream>

namespace X
{
    int var = 1;
}

namespace Y
{
    int var = 2;
}

int main() 
{
    using namespace X;

    std::cout  << "var = " << var << '\n';

    using Y::var;

    std::cout  << "var = " << var << '\n';
}

The program output is

var = 1
var = 2

That is the using declaration that introduces the variable var in the block scope of the function main hides the declaration of the variable var declared in the namespace X.

In fact the below simplified demonstration program in essence behaves similarly as the above program relative to the name lookup.

#include <iostream>

int var = 1;

int main() 
{
    std::cout  << "var = " << var << '\n';

    int var = 2;

    std::cout  << "var = " << var << '\n';
}

score 2 · Answer 2 · answered Oct 19 '21 at 14:20

Yes, using Y::var; does hide X::var when you ask for var. The former is as-if Y::var were declared in this scope, the latter only influences unqualified lookup.

using ns_name::name;

using-declaration: makes the symbol name from the namespace ns_name accessible for unqualified lookup as if declared in the same class scope, block scope, or namespace as where this using-declaration appears.

Names introduced into a namespace scope by a using-declaration can be used just like any other names, including qualified lookup from other scopes

using namespace ns_name;

using-directive: From the point of view of unqualified name lookup of any name after a using-directive and until the end of the scope in which it appears, every name from ns_name is visible as if it were declared in the nearest enclosing namespace which contains both the using-directive and ns_name.

Extremely basic question about namespaces in c++

2 Answers2