On 3/19/07, Gustavo Rios <[EMAIL PROTECTED]> wrote:
On 3/19/07, Nick ! <[EMAIL PROTECTED]> wrote:
> On 3/19/07, Gustavo Rios <[EMAIL PROTECTED]> wrote:
> > I am writing a very simple program but the output change for the c
> > variable value change every time i run it. What would it be my mistake
> > on the source? Did i forget some thing?
> >
> > #include <stdio.h>
> >
> > int
> > main(int argc, char **argv)
> > {
> > unsigned long long x, c;
> > unsigned *p;
>
> ^ this is bad. always say your types in full.
>
> >
> > x = 1, x+= (unsigned long long)1 << 33 ;
>
> This sets *(&x) to 1, and then sets *(&x) (yes, the same one) to 1+(1<<33)
>
> > p = (void *)&x;
> > c = p[0] * p[1];
>
> That is, p[1] == *(&x+1) is never getting set to anything. Thus the
> reason the output is always changing is because p[1] is always
> pointing at a different, random location in memory that has some
> previous value.
>
> Further, p[1] is not your memory, and it's only by chance that you're
> not segfaulting.
>
> > fprintf(stdout, "x:%llu\n", x);
> > fprintf(stdout, "0,1:%u,%u\n", p[0], p[1]);
> > fprintf(stdout, "c:%llu\n", c);
> >
> > return 0;
> > }
> >
So, why when i printf p[1], it correctly prints 2?
Uhm. Hmm. Well, x is a long long which is 8 bytes right? A void (when
doing pointer arithmetic) is taken to only be 1 byte (right?). So p[0]
is the first byte of those 8, and p[1] is the second, and due to
coincidence and twos-complement encoding it happens to show you the
expected numbers. Maybe?
But that doesn't explain why the output is always changing.
Wait, how is * defined on two voids????? That shouldn't even compile
(unless it's autocasting to int?).
-Nick
