You do not have to use the class member access syntax to refer to a static member; to access a static member s of class X, you could use the expression X::s. The following example demonstrates accessing a static member:
#include <iostream>
using namespace std;
struct A {
static void f() { cout << "In static function A::f()" << endl; }
};
int main() {
// no object required for static member
A::f();
A a;
A* ap = &a;
a.f();
ap->f();
}The three statements A::f(), a.f(), and ap->f() all call the same static member function A::f().
You can directly refer to a static member in the same scope of its class, or in the scope of a class derived from the static member's class. The following example demonstrates the latter case (directly referring to a static member in the scope of a class derived from the static member's class):
#include <iostream>
using namespace std;
int g() {
cout << "In function g()" << endl;
return 0;
}
class X {
public:
static int g() {
cout << "In static member function X::g()" << endl;
return 1;
}
};
class Y: public X {
public:
static int i;
};
int Y::i = g();
int main() { }The following is the output of the above code:
In static member function X::g()The initialization int Y::i = g() calls X::g(), not the function g() declared in the global namespace.
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