Question

Below code is dealing with a TYPE* const pointer.

struct D {
  void Check ()
  {
    D* const p = new D; // 2nd test is "p = 0;"
    cout<<"p = "<<p<<endl;
    (D*&)p = new D;
    cout<<"p = "<<p<<endl; // prints 0, "p = 0;" at declaration
  }
};

int main ()
{
  D o;
  o.Check();
}

My questions are,

1. If you initialize with 0, then even though typecasting next time will not work. Is doing such typecasting is undefined behavior ?
2. this pointer is also of TYPE* const type, then why compiler doesn’t allow the same operation for this?

Answer 1

1. As others have said, this is undefined behaviour since it attempts to modify a const object. If you initialise it with zero then the compiler might treat it as a compile-time constant, and ignore any attempt to modify it. Or it might do something entirely different.

2. this is not an ordinary variable of type TYPE * const; it is an rvalue expression of type TYPE *. This means that it cannot be used as the target of an assignment expression, or bound to a non-constant reference, at all.