Why is calling a static member function with . or -> syntax legal? [duplicate]

Question

Possible Duplicate:
C++ Static member method call on class instance

Today I discovered that something I had long (and I mean long—like, for twenty years), thought illegal in C++ is actually legal. Namely, calling a static member function as if it belonged to an individual object. For example:

struct Foo
{
    static void bar() { cout << "Whatever."; }
};

void caller()
{
    Foo foo;
    foo.bar();    // Legal -- what?
}

I normally see static member functions being called strictly with "scope resolution syntax," thus:

Foo::bar();

This makes sense, because a static member function is not associated with any particular instance of the class, and therefore we wouldn't expect a particular instance to be syntactically "attached" to the function call.

Yet I discovered today that GCC 4.2, GCC 4.7.1, and Clang 3.1 (as a random sampling of compilers) accept the former syntax, as well as:

Foo* foo = new Foo;
foo->bar();

In my particular case, the legality of this expression led to a runtime error, which convinced me that the peculiarity of this syntax is of more than academic interest—it has practical consequences.

Why does C++ allow static member functions to be called as if they were direct members of individual objects—that is, by using the . or -> syntax attached to an object instance?

Answer 1

In The Design and Evolution of C++ at page 288, Bjarne Stroustrup mentions that in the days before static member functions, programmers used hacks like ((X*)0)->f() to call member functions that didn't need an object. My guess is that when static member functions were added to the language, access through -> was allowed so that programmers with code like that could change f to static without having to hunt down and change every use of it.

Answer 2

Presumably so you can call it in places where you may not know the class type of something but the compiler does.

Say I had a bunch of classes that each has a static member that returned the class name:

class Foo
{
    static const char* ClassName() { return "Foo"; }
};

class Bar
{
    static const char* ClassName() { return "Bar"; }
};

Then all over my code I could do things like:

Foo foo;

printf( "This is a %s\n", foo.ClassName() );    

Without having to worry about knowing the class of my objects all the time. This would be very convenient when writing templates for example.

Answer 3

It's like this because the standard says that's how it works. n3290 § 9.4 states:

A static member s of class X may be referred to using the qualified-id expression X::s; it is not necessary to use the class member access syntax (5.2.5) to refer to a static member. A static member may be referred to using the class member access syntax, in which case the object expression is evaluated. [ Example:

struct process { 
  static void reschedule(); 
}; 

process& g();

void f() { 
  process::reschedule(); // OK: no object necessary
  g().reschedule(); // g() is called 
} 

end example ]