So, why when i printf p[1], it correctly prints 2? 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; > } > -Nick
