[vox-tech] loop never exits!

Brian Lavender brian at brie.com
Wed Apr 21 09:37:21 PDT 2010


I told the compiler to go through a loop while a certain expression
evaluates to true. When the expression evalates to false, the loop
should terminate. If I had provided no expression, or I had provided
true, or some value that would always evaluate to true, then I would
expect the compiler to do as I tell it.

I have provided an expression for the loop to terminate that could
evaluate to false, yet will never evaluate to false.

This determination can be done through static analysis. Hence,
I am still convinced that the compiler should issue a warning. Yet,
I am still looking for a contradiction to my theory.




On Tue, Apr 20, 2010 at 08:30:42PM -0700, Matthew Holland wrote:
> It's not the compiler's job to tell you what you want, it's your job
> to tell it!  The -Wall flag doesn't mean, "warn me about every
> possible stupid construct I can come up with."
> 
> In all seriousness, if you don't get over using a "greater than or
> equal" construct for loop termination, you're going to be in a world
> of hurt if STL containers ever cross your transom.  Just don't do
> that, and everything will be fine.
> 
> Matt
> 
> On Tue, Apr 20, 2010 at 7:44 PM, Brian Lavender <brian at brie.com> wrote:
> > 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
> > _______________________________________________
> > vox-tech mailing list
> > vox-tech at lists.lugod.org
> > http://lists.lugod.org/mailman/listinfo/vox-tech
> >
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-- 
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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