Priority of name hiding in C++ inheritance

Question

Why the call of print() from pointer to base class (Class A) is working and call to print() from Child class object(Class C) is not working?

Statement 1: Will give "print A" as output as since A::print() is a virtual function it will call print() function of Class C which is inherited from A::print(). This will prove that Class C indeed has print() function

Statement 2: This will give compiler error as B::print(int x) will hide A::print().

Statement 3: Will give compilation error. Why?

Probably because print() is still hidden in Class C as Class C is also inheriting B::print(int x). If this is the case then why call from a->print() worked?

Another question: Is there any rule which specifies that B::print(int x) will hide A::print() in the child classes?

class A{
    public:
    virtual void print();
};

void A::print(){
    cout<<"print A\n";
}

class B:public A{
    public:
        void print(int x);
};

void B::print(int x){
    cout<<"print B " << x <<"\n";
}

class C:public B{
};

void funca(A *a){
    a->print();//Statement 1
}

void funcb(B *b){
    //b->print(); Statement 2
}

void funcc(C *c){
    //c->print(); Statement 3
}

int main(){

    C d;
    funca(&d);
    funcb(&d);
    funcc(&d);
}

Answer 1

Probably because print() is still hidden in Class C as Class C is also inheriting B::print(int x). If this is the case then why call from a->print() worked?

A is base class so there is nothing to hide there, a->print() just works from base class context.

Both B and C hides the original print() function with a different prototype print(int) and so the error as the function is called with wrong prototype (there is no more print() in B or C class)

Answer 2

Statement 3: Will give compilation error. Why?

For the same reason that b->print() didn't work. This is the error when Statement 2 runs:

In function 'void funcb(B*)':
error: no matching function for call to 'B::print()'
     b->print(); //  Statement 2
              ^
note: candidate: 'void B::print(int)'
 void B::print(int x){
      ^
note:   candidate expects 1 argument, 0 provided

This is the error when Statement 3 runs:

In function 'void funcc(C*)':
error: no matching function for call to 'C::print()'
     c->print(); //  Statement 3
              ^
note: candidate: 'void B::print(int)'
 void B::print(int x){
      ^
note:   candidate expects 1 argument, 0 provided

It's practically the same error, since as you've guessed: C inherits B , whose print function hides A 's.

If this is the case then why call from a->print() worked?

Because your reference to C was casted into a pointer to A , as such, exposing A 's print function instead of B 's. If you do this explicitly, you'd get the same results:

static_cast<A*>(&d)->print();    //  print A
static_cast<B*>(&d)->print();    //  error: no matching function for call to 'B::print()'