Calling a function overloaded in several namespaces from inside one namespace

Question

I have the following code snippet:

void foo(double a) {}

namespace bar_space
{
  struct Bar {};

  void foo(Bar a) {}
}

foo(double) is a g

Answer 1

In C++, there is a concept called name hiding. Basically, a function or class name is "hidden" if there is a function/class of the same name in a nested scope. This prevents the compiler from "seeing" the hidden name.

Section 3.3.7 of the C++ standard reads:

A name can be hidden by an explicit declaration of that same name in a nested declarative region or derived class (10.2)

So, to answer your question: in your example void foo(double a); is hidden by void bar_space::foo(Bar a); So you need to use the :: scoping operator to invoke the outer function.

Answer 2

However, in your sample code you could use something like that

namespace bar_space 
{
    using ::foo;
    void baz()
    {
       Bar bar;
       foo(5.0);
       foo(bar);
    }
}

Answer 3

Yes, bar_space is hiding the original function and no, you can't make foo(5.0) callable from whithin bar_space without explicit scoping if foo(double) is defined in the global namespace.