No!
p sizeof is 4 bytes, p is the frst byt of &x, and p + 1 is the 4th
byte. Casting is only on attribution of &x to p.
Realize, p[0] evals to 1 and p[1] evals to 2 as it should be. Only
problem relates to p[0] * p[1]. I believe it should (1 * 2), i.e., 2.
Not a random value.
On 3/19/07, Nick ! <[EMAIL PROTECTED]> wrote:
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
