What does :: operator mean in C++ when it's used at the beginning

Question

  double timestamp = ::a::b::c::NowInSeconds();

Already know the use of :: in varible selection, but what's the meaning of it when it is used at the beginning and followed with a function (with namespace specified) ?

Does this answer your question? [:: scope resolution operator in front of a template function call in c++](https://stackoverflow.com/questions/8937824/scope-resolution-operator-in-front-of-a-template-function-call-in-c) — Ryan Zhang, Jan 03 '21 at 04:35
"Look up the name that follows via the qualified name lookup in the global namespace". — Igor Tandetnik, Jan 03 '21 at 04:35

score 3 · Answer 1 · answered Jan 03 '21 at 04:35

The initial :: is used to explicitly specify that the name following it is located in the global/default namespace, and not in any other namespace that happens to be in scope. Here's an example of how that might be useful:

int a;

namespace b {
   int a;
};

using namespace b;

int main(int, char **)
{
   a = 5;   // error, reference to 'a' is ambiguous

   ::a = 5;   // ok
}

What does :: operator mean in C++ when it's used at the beginning

1 Answers1