'using namespace {some namespace nested inside of another}`

Question

Is it possible to access the type Foo under the namespace first by using a using declaration (or something similar) in the caller code?

namespace first {
    namespace second {
        struct Foo { int i; }; 
    } 
} 

int main() {
    using namespace first::second; 
    first::Foo foo { 123 }; 
    return 0; 
}

I get these error messages:

error: 'Foo' is not a member of 'first'
  first::Foo foo{ 123 };```

Have you tried yourself before asking? Having any issues with your code? By the way, why you declare as first::Foo while the struct Foo is inside second? — armagedescu
– armagedescu, Commented Jul 5, 2020 at 19:11
@ThomasSablik access the type Foo under namespace first. — user13865989
– user13865989, Commented Jul 5, 2020 at 19:17
@armagedescu yes I have tried, and that was part of the question. Sorry I didn't make it clear enough. — user13865989
– user13865989, Commented Jul 5, 2020 at 19:18
Then you should edit the question with the compiler error messages that you got. — cigien
– cigien, Commented Jul 5, 2020 at 19:18

HolyBlackCat · Accepted Answer · 2020-07-05 19:17:32Z

5

You have several options:

using namespace first::second; 
Foo foo{123};

namespace ns = first::second; 
ns::Foo foo{123};

I guess you could also do namespace first = first::second;. Then first::Foo foo{123}; would work, but to access the contents of the actual namespace first (other than those in namespace second) you'll have to use ::first.

answered Jul 5, 2020 at 19:17

HolyBlackCat

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Comments

Thomas Sablik · Accepted Answer · 2020-07-05 23:15:58Z

0

Namespaces can be expanded and you can create an alias to a token in a different namespace:

namespace first {
    namespace second {
        struct Foo { int i; }; 
    } 
}

namespace first {
    using second::Foo;
}

int main() {
    first::Foo foo { 123 }; 
    return 0; 
}

answered Jul 5, 2020 at 23:15

Thomas Sablik

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