Possible to wrap or merge separate namespaces?

Question

I seem to recall seeing notes somewhere on a way to combine multiple namespaces into one.

Now, looking for said notes I am not finding them -- even searching using search terms combing, grouping, merging and wrapping I'm not coming up with anything. Maybe I misunderstood what I saw before. I don't have a specific application for this, it's just a curiosity and it's a bit contrived.

But, starting with two name spaces...

namespace a {int func() {return 1;}}
namespace b {int func() {return 2;}}

I was looking for syntax to either simply wrap them in another name -- after the fact -- (yes, I know I can rewrite it in a nested way) or merge them into one new space. But, I did find that I if I add to one of the namespaces that much works.

namespace c {namespace a{ int func2() {return 3;}} }


int main(int argc, char **argv)
{
    int a = a::func();          // normal case
    int c = c::a::func2();      // wrapped and added to

    //int c = c::func2();       // doesn't work
    //int d = a::func2();       // doesn't work
}

The question are:

1) is there syntax that just combines the two spaces into one new one?

2) is there a syntax to wrap the spaces without adding more to the subspaces?

Answer 1

You can do this:

namespace c
{
    using namespace a;
    using namespace b;
}

But if a and b have elements with the same names, you won't be able to use them from namespace c.

Answer 2

The best solution since C++11 is:

namespace c
{
    inline namespace a { using namespace ::a; }
    inline namespace b { using namespace ::b; }
}

This way for names that not conflict you can qualify only by c and you can resolve conflicts by qualifing c::a or c::b.

e.g.:

namespace a
{
    auto foo_a() { cout << "a::foo_a" << endl; }
    auto foo() { cout << "a::foo" << endl; }
}
namespace b
{
    auto foo_b() { cout << "b::foo_b" << endl; }
    auto foo() { cout << "b::foo" << endl; }
}

namespace c
{
    inline namespace a { using namespace ::a; }
    inline namespace b { using namespace ::b; }
}

int main()
{
  c::foo_a();
  c::foo_b();

  c::a::foo();
  c::b::foo();

  return 0;
}