On Tue, Jun 18, 2013 at 11:01 PM, vijay nag <vijunag@xxxxxxxxx> wrote:
>
> Consider the following expression
>
> char *foo = &bar;
> ((unsigned long*)foo)++
I think the correct way to get the same effect these days is not what
you suggested, but rather something like
foo = (char *) (((unsigned long*)foo) + 1);
Ian
