Why should we use "::" operator on global functions / objects? - c++

Question

I have seen this used in various places. C++ programmers will often use the :: operator right before a global function call.
e.g
::glGenBuffers( 1, &id );
Why is this? Why not just use:
glGenBuffers( 1, &id );

Answer 1

To avoid accidental namespace clashing. For example if you current namespace would have glGenBuffers which does something different from the "good" glGenBuffers with :: you can specify to call the glGenBuffers which is in the global namespace.

Answer 2

The problem is 1) names in inner scopes hide names in outer scopes and 2) there can be ambiguous of function calls when using directive is used.
For example (ambiguity)
#include <algorithm>
using namespace std;
void swap( int &x, int &y )
{
int tmp = x;
x = y;
y = tmp;
}
int main()
{
int x = 5, y = 10;
//swap( x, y ); compiler error: what function to call?
::swap( x, y ); // the user defined function is called
std::swap( x, y ); // the standard function is called.
}
Another example of hidding names
#include <iostream>
void f() { std::cout << "::f()\n"; }
namespace MyNamespace
{
void f() { std::cout << "MyNamespace::f()\n"; }
void g() { f(); } // calls MyNamespace::f()
void h() { ::f(); } // calls ::f()
}
int main()
{
MyNamespace::g();
MyNamespace::h();
}

Answer 3

As you can see here C++ Scope Resolution Operator ::
The :: (scope resolution) operator is used to qualify hidden names so
that you can still use them. You can use the unary scope operator if a
namespace scope or global scope name is hidden by an explicit
declaration of the same name in a block or class.

Answer 4

::glGenBuffers to force choose the method in global namespace
void method()
{
std::cout << "method in global namespace";
}
class Test {
void method()
{
std::cout << "method in Test class";
}
void test()
{
method(); // method in Test class
::method(); //method in global namespace
}
}

Answer 5

Explicitly marking the scope may save you from unexpected matches through using declarations or argument dependent lookup. For example, consider the following:
namespace foo
{
class X {};
void bar(X*, int):
}
// ... much in between ...
foo::X some_object
// ... more in between ...
void bar(X*, long);
int main()
{
bar(&some_object, 42); // calls foo::bar, because it is a better match
::bar(&some_object, 42); // calls ::bar, because it is explicitly told to
}
If you were not aware that there is a bar in namespace foo, or that some_object has a type from namespace foo, the call to foo::bar instead of ::bar may catch you by surprise.

Answer 6

The scope operator (::) is used for accessing the global namespace on 2 occasions mainly:
a) importing a global method / variable / typedef (from global namespace) into a namespace:
int fun()
{
return 1;
}
namespace x
{
using ::fun;
// you cannot do `using fun`, as using requires a namespace after
// in this case, we are `using` the global namespace
}
int main()
{
std::cout << fun() + x::fun(); // prints 2
}
Edit: This is done by some STL implementations, for example: std::size_t:
typedef unsigned int size_t;
namespace std
{
using ::size_t;
}
size_t x = 1;
std::size_t y = 2;
b) To avoid clashing:
# include <algorithm>
using namespace std;
template<class T> inline
void swap(T &left, T &right)
{
T tmp = left;
left = right;
right = tmp;
}
int main()
{
int a = 1, b = 2;
// swap(a, b) - error: which 'swap'?
std::swap(a, b); // the one in <algorithm>
::swap(a, b); // the one i defined
}
Note: using namespace std; is not a reccomended practise.
Edit
MSVC compiler provides the _CSTD macro (#define _CSTD ::), so that you can use using _CSTD fun;, e.t.c instead if you prefer.