Using :: (scope resolution operator) in C++

Question

I am learning C++ and I can never tell when I need to use :: . I do know that I need to use std:: in front of cout and cin. Does this mean that inside of the iostream file the developers that created it made a namespace called std and put the functions cin and cout into the namespace called std? When I created a new class that isn't in the same file as main() for some reason I must add :: .

For example, if I create a class called A , why do I need to put A:: in front of a function that I make, even though I didn't put it into a namespace? For example void A::printStuff(){} . If I create a function in main, why don't I have to put main::printStuff{} ?

I know that my question is probably confusing, but could someone help me?

Answer 1

You're pretty much right about cout and cin. They are objects (not functions) defined inside the std namespace. Here are their declarations as defined by the C++ standard:

Header <iostream> synopsis

#include <ios>
#include <streambuf>
#include <istream>
#include <ostream>

namespace std {
  extern istream cin;
  extern ostream cout;
  extern ostream cerr;
  extern ostream clog;

  extern wistream wcin;
  extern wostream wcout;
  extern wostream wcerr;
  extern wostream wclog;
}

:: is known as the scope resolution operator. The names cout and cin are defined within std, so we have to qualify their names with std::.

Classes behave a little like namespaces in that the names declared inside the class belong to the class. For example:

class foo
{
  public:
    foo();
    void bar();
};

The constructor named foo is a member of the class named foo. They have the same name because its the constructor. The function bar is also a member of foo.

Because they are members of foo, when referring to them from outside the class, we have to qualify their names. After all, they belong to that class. So if you're going to define the constructor and bar outside the class, you need to do it like so:

foo::foo()
{
  // Implement the constructor
}

void foo::bar()
{
  // Implement bar
}

This is because they are being defined outside the class. If you had not put the foo:: qualification on the names, you would be defining some new functions in the global scope, rather than as members of foo. For example, this is entirely different bar:

void bar()
{
  // Implement different bar
}

It's allowed to have the same name as the function in the foo class because it's in a different scope. This bar is in the global scope, whereas the other bar belonged to the foo class.

Answer 2

The :: are used to dereference scopes.

const int x = 5;

namespace foo {
  const int x = 0;
}

int bar() {
  int x = 1;
  return x;
}

struct Meh {
  static const int x = 2;
}

int main() {
  std::cout << x; // => 5
  {
    int x = 4;
    std::cout << x; // => 4
    std::cout << ::x; // => 5, this one looks for x outside the current scope
  }
  std::cout << Meh::x; // => 2, use the definition of x inside the scope of Meh
  std::cout << foo::x; // => 0, use the definition of x inside foo
  std::cout << bar(); // => 1, use the definition of x inside bar (returned by bar)
}

unrelated: cout and cin are not functions, but instances of stream objects.

Answer 3

The :: is called scope resolution operator. Can be used like this:

:: identifier
class-name :: identifier
namespace :: identifier

You can read about it here
https://docs.microsoft.com/en-us/cpp/cpp/scope-resolution-operator?view=vs-2017

Answer 4

One use for the 'Unary Scope Resolution Operator' or 'Colon Colon Operator' is for local and global variable selection of identical names:

    #include <iostream>
    using namespace std;
    
    int variable = 20;
    
    int main()
    {
    float variable = 30;
    
    cout << "This is local to the main function: " << variable << endl;
    cout << "This is global to the main function: " << ::variable << endl;
    
    return 0;
    }

The resulting output would be:

This is local to the main function: 30

This is global to the main function: 20

Other uses could be: Defining a function from outside of a class, to access a static variable within a class or to use multiple inheritance.

Answer 5

look at it is informative [Qualified identifiers

A qualified id-expression is an unqualified id-expression prepended by a scope resolution operator ::, and optionally, a sequence of enumeration, (since C++11)class or namespace names or decltype expressions (since C++11) separated by scope resolution operators. For example, the expression std::string::npos is an expression that names the static member npos in the class string in namespace std. The expression ::tolower names the function tolower in the global namespace. The expression ::std::cout names the global variable cout in namespace std, which is a top-level namespace. The expression boost::signals2::connection names the type connection declared in namespace signals2, which is declared in namespace boost.

The keyword template may appear in qualified identifiers as necessary to disambiguate dependent template names]1