为什么我不能向前声明静态函数？

Question

I need to forward-declare a class in my header file, like this:

class MyStaticClass;

I understand why one cannot forward-declare data members of this class. I used to think that you could however forward-declare functions. I would like to declare a static function of this class, like this:

class MyStaticClass;
static int MyStaticClass::AddTwoNumbers(const int a, const int b);

This gives me a compile error though:

error C2027: use of undefined type 'MyStaticClass'

Why can this not be done? Or is there a secret way of doing this after all?

Answer 1

This is not allowed because it would allow others to add member functions to a class without even editing the class itself.

Consider this,

struct X
{
    static void f(float a) { std::cout << a << std::endl; }
 private:
    static int _data; //inaccessible to non-member
};

X::f(0); //convert 0 (an int) to float, and call X::f().

Now imagine someone came and forward-declare the following function, just before including header which defines the above class:

static void X::f(int);

Now the previous call X::f(0) would give linker error ( unresolved name ) because now 0 wouldn't convert to float , because it does not need to as there is a declared function which accepts an int , though it is not defined — worse, if it is defined, then you wont even get the linker error and you would probably not easily know that a different function is being called.

Morever. f(int) can now access the private member _data as well — in this way, anyone can access any private/protected members just by adding functions at whim.