Path: utzoo!attcan!uunet!lll-winken!xanth!ames!apple!rutgers!rochester!kodak!ektools!randolph
From: randolph@ektools.UUCP (Gary L. Randolph)
Newsgroups: comp.lang.c
Subject: Re: Problem with ()?():() as ending test in for-loop
Message-ID: <1918@ektools.UUCP>
Date: 25 May 89 14:02:25 GMT
References: <1200@liszt.kulesat.uucp> <20658@news.Think.COM> <1917@ektools.UUCP>
Sender: randolph@ektools (Gary L. Randolph)
Reply-To: randolph@ektools.UUCP (Gary L. Randolph)
Organization: Eastman Kodak, Dept. 47, Rochester NY
Lines: 52

In article <1917@ektools.UUCP> randolph@ektools.UUCP (Gary L. Randolph) writes:
>In article <20658@news.Think.COM> barmar@kulla.think.com.UUCP (Barry Margolin) writes:
>>In article <1200@liszt.kulesat.uucp> vanpetegem@liszt.kulesat.uucp writes:
>>>The code with which I am in trouble, goes as follows:
>>>
>>>	 for(n = 1;  ((n % 10) ? (k > 2) : (n < 100));  n++)
>>>                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>>>           {
>>>             ..... (in some expressions k is changed, depending on n)
>>>	   }
>>>What I want to do is simply perform the test (k > 2) as ending test for the
>>>"for-loop", except in the cases where n is dividable by 10. In the latter
>>>situation the test (n < 100) has to be done.
>
=The sense of the test clause of your :? expression is wrong.  When n
=is divisible by 10, (n % 10) is 0, and 0 represents "false" in C.  I
=would write your end test as
=	((n % 10) == 0) ? (k > 2) : (n < 100)
=i.e. they would just interchange the true and false clauses, taking
=advantage of the fact that the result of the % can be used as a
=boolean.

My original anwer has some problems - PLEASE ignore it...
Here is, I believe, a correct answer.

NOOOOO! I do NOT think your logic is wrong.  I am sure it is correct
according to  your explanation of what you want to do.  There *is* a
problem however...
   ...Stuff here is too wrong (embarrASSing) for me to repeat...
You don't mention whether variable k was initialized before
entering the for.  IT SHOULD BE.

I ran a short test as follows:
main()
{
int n;
float k = 6;
for(n=1;((n % 10) ? (k>2) : (n<100)); n++)
	{
	k -= 2;
	}
printf("n: %d\n k: %f\n",n,k);
}

This produced the expected result:
     n: 3
     k: 2.000000


I hope THIS helps...

                           --Gary Randolph