Why different behavior for “TYPE* const” pointers?

问题:

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?

回答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.

回答2:

Is doing such typecasting is undefined behavior ?

Yes.

(D*&)p = new D;

It invokes undefined behavior, as it tries to change the const pointer.

Recall that D* const p declares a variable p which is a const pointer to non-const D.

回答3:

D* const p = 0;

This declaration says that p is a pointer to D that is constant, that is it will never, ever change. It is always 0.

cout<<"p = "<<p<<endl;

Here you display the value of p, which you earlier said would always be 0. Guess why a 0 is displayed!