[vox-tech] loop never exits!

[Matthew Holland]

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
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