[vox-tech] loop never exits!

[Brian Lavender]

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