C++ Overload Static Function with Non-Static Function

No, it is directly prohibited by the standard:

ISO 14882:2003 C++ Standard 13.1/2 – Overloadable declarations

Certain function declarations cannot
be overloaded:

• Function declarations that differ only in the return type cannot be overloaded.
• Member function declarations with the same name and the same parameter types cannot be overloaded
  if any of them is a static member function declaration (9.4).

…

[Example:

class X {
    static void f();
    void f();                // ill-formed
    void f() const;          // ill-formed
    void f() const volatile; // ill-formed
    void g();
    void g() const;          // OK: no static g
    void g() const volatile; // OK: no static g
};

—end example]

…

Besides, it would be ambiguous anyway since it’s possible to call static functions on instances:

ISO 14882:2003 C++ Standard 9.4/2 – Static members

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:

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

—end example]

…

So there would be ambiguity with what you have:

class Foo
{
public:
    string bla;
    Foo() { bla = "nonstatic"; }
    void print() { cout << bla << endl; }
    static void print() { cout << "static" << endl; }
};

int main()
{
    Foo f;
    // Call the static or non-static member function?
    // C++ standard 9.4/2 says that static member
    // functions are callable via this syntax. But
    // since there's also a non-static function named
    // "print()", it is ambiguous.
    f.print();
}

---

To address your question about whether you can check what instance a member function is being called on, there is the this keyword. The this keyword points to the object for which function was invoked. However, the this keyword will always point to an object i.e. it will never be NULL. Therefore it’s not possible to check if a function is being called statically or not à la PHP.

ISO 14882:2003 C++ Standard 9.3.2/1 – The this pointer

In the body of a nonstatic (9.3)
member function, the keyword this is a
non-lvalue expression whose value is
the address of the object for which
the function is called.