Why doesn't deleting my pointer delete my pointer?

Question

So to understand new/delete better (really to prove to myself with small examples why virtual destructors are needed for interfaces), I want to understand memory leaks, so that I may live in fear of them. But I am having a hard time getting my leak on, so to speak; actually, I am have a hard time with new/delete too.

Here's my simplest version:

int* P1 = new int(43);

cout<<"P1 = "<<P1<<endl;
cout<<"*P1 = "<<*P1<<endl;

delete P1;

cout<<"P1 = "<<P1<<endl;
cout<<"*P1 = "<<*P1<<endl;

This prints:

P1 = 0xcc0340
*P1 = 43
P1 = 0xcc0340
*P1 = 43

I had something more complicated inside of a class, but this example illustrates my fail. I thought delete takes a pointer and frees it's memory, thereby invalidating the pointer or at least what it points to? I must be doing something very simple very wrong.

Answer 1

You are causing undefined behaviour. This means that anything can happen. Since something did indeed happen, everything behaves as documented. (Sometimes "something" looks very similar to something else that you might erroneously expect. Doing exactly what you think you were trying to achieve is one of the possible allowed instances of "undefined behaviour".)

Note also that a "memory leak" is sort of the opposite of what you're trying to do - in a memory leak you forget to free memory, whereas you already freed the memory and are now accessing invalid memory.

Here's a real memory leak, which also does not cause undefined behaviour -- don't confuse "bad but correct" with "incorrect" programming!

int * factorial(int * n)
{
  if (*n == 0) return new int(1);
  else return new int(*n * *factorial(*n - 1));
}