On Tue, Apr 20, 2010 at 06:54:33PM -0700, Bill Broadley wrote:
> On 04/20/2010 06:37 PM, Brian Lavender wrote:
> > Our new guy (forget his name, doh!) and I figured out the problem with
> > my loop that would count down, but not terminate. Turns out I was using
> > an unsigned integer for my counter in my for loop and it is always
> > greater than zero (Example 1).
>
> No, it's not always greater than zero. Your test says i>=0 so if it's
> greater than or equal to zero it continues. Seems like you want i>0.
Sorry, I meant to say it's always greater than or equal than zero. zero
minus 1 is 4294967295 (0xffffffff) on a 32 bit machine. It can never go
negative because it is unsigned and the loop will never terminate. Thus,
I am thinking that the compiler could catch this due to the fact that i
is unsigned. I wanted to print out the reverse of an array.
If you run the following, the loop will never terminate.
#include <stdio.h>
int main() {
int a[] = {5,6,8,3,4};
unsigned int i;
for (i= (sizeof(a) -1)/sizeof(int) ; i >= 0; i--) {
printf("%d\n",a[i]);
}
return 0;
}
>
> > Funny thing is that -Wall didn't catch this. Seems that -Wall could
> > catch this assuming that we want to loop to terminate. Any thoughts?
>
> Seems strange, but legal to do what you wanted.
>
> > Say the compiler gave a warning, would that mess up the "for (;;)"
> > construct shown in Example 2?
> >
> > brian
> >
> > // Example 1
> > // Loop never terminates
> > #include<stdio.h>
> >
> > int main() {
> > unsigned int i, num=50;
> >
> >
> > for (i= num ; i>= 0; i--) {
> > printf("%u\n",i);
> > }
> >
> > return 0;
> > }
> >
> > // Example 2
> > // Purposely never terminates
> > #include<stdio.h>
> >
> > int main() {
> > for (;;) {
> > printf("Hello forever\n");
> > }
> > return 0;
> > }
--
Brian Lavender
http://www.brie.com/brian/
"For every complex problem there is an answer that is clear, simple, and wrong."
- H. L. Mencken
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