没有范围的范围解析运算符

Question

In C++, what is the purpose of the scope resolution operator when used without a scope? For instance:

::foo();

Answer 1

It means global scope. You might need to use this operator when you have conflicting functions or variables in the same scope and you need to use a global one. You might have something like:

void bar();    // this is a global function

class foo {
    void some_func() { ::bar(); }    // this function is calling the global bar() and not the class version
    void bar();                      // this is a class member
};

If you need to call the global bar() function from within a class member function, you should use ::bar() to get to the global version of the function.

Answer 2

Also you should note, that name resolution happens before overload resolution. So if there is something with the same name in your current scope then it will stop looking for other names and try to use them.

void bar() {};
class foo {
    void bar(int) {};
    void foobar() { bar(); } // won't compile needs ::bar()
    void foobar(int i) { bar(i); } // ok
}

Answer 3

A name that begins with the scope resolution operator ( :: ) is looked up in the global namespace. We can see this by looking at the draft C++ standard section 3.4.3 Qualified name lookup paragraph 4 which says ( emphasis mine ):

A name prefixed by the unary scope operator :: (5.1) is looked up in global scope , in the translation unit where it is used. The name shall be declared in global namespace scope or shall be a name whose declaration is visible in global scope because of a using-directive (3.4.3.2). The use of :: allows a global name to be referred to even if its identifier has been hidden (3.3.10).

As the standard states this allows us to use names from the global namespace that would otherwise be hidden , the example from the linked document is as follows:

int count = 0;

int main(void) {
  int count = 0;
  ::count = 1;  // set global count to 1
  count = 2;    // set local count to 2
  return 0;
}

The wording is very similar going back to N1804 which is the earliest draft standard available.

Answer 4

如果在本地作用域中已经有一个名为foo（）的函数，但是您需要在全局作用域中访问该函数。

Answer 5

My c++ is rusty but I believe if you have a function declared in the local scope, such as foo() and one at global scope, foo() refers to the local one. ::foo() will refer to the global one.

Answer 6

指全球范围