ptr1 == ptr2 but (int)ptr1 != (int)ptr2

From: Hallvard B Furuseth <h.b.furuseth@xxxxxxxxxxx>
Date: Fri, 25 May 2007 15:28:27 +0200

Do anyone know of an architecture where this can break?

T *ptr1, *ptr2;
...
if (ptr1 == ptr2)
if (CHAR_BIT*sizeof(T*) <= (width of int)) /*otherwise undefined*/
assert((int)ptr1 == (int)ptr2);

(Feel free to replace int with another integer type if that helps to
break something.)

I know an addess can have several representations at least on some DOS
memory models, but I don't know if it normalizes pointers before
converting to integer.

--
Regards,
Hallvard
.

Follow-Ups:
- Re: ptr1 == ptr2 but (int)ptr1 != (int)ptr2
  - From: Malcolm McLean
- Re: ptr1 == ptr2 but (int)ptr1 != (int)ptr2
  - From: Kenneth Brody

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