[Info-vax] decrementing & for loops in C

[Bill Gunshannon]

In article <176uZD2KcidF-pn2-d8d17hV5X6v3 at rikki.tavi.co.uk>,
	"Bob Eager" <rde42 at spamcop.net> writes:
> On Thu, 8 Jan 2009 13:03:32 UTC, vaxinf at chemie.uni-konstanz.de wrote:
> 
>> int main (void) {
>> 
>> int i,sum=0,n=10;
>> 
>>     (void)printf ("Starting countdown loop...\n");
>>     (void)fflush (stdout);
>>     for (i = n; i = 0; i--) {
>>         (void)printf (" i: %i\n",i);
>>     }
>>     (void)printf ("done.\n");
>> 
>>     (void)printf ("Starting normal loop...\n");
>>     (void)fflush (stdout);
>>     for (i = 1; i <= n; i++) {
>>         (void)printf (" i: %i\n",i);
>>     }
>>     (void)printf ("done.\n");
>> }
>> 
>> Here is the output:
>> 
>> r SYSWRITE_TST
>> Starting countdown loop...
>> done.
>> Starting normal loop...
>>  i: 1
>>  i: 2
>>  i: 3
>>  i: 4
>>  i: 5
>>  i: 6
>>  i: 7
>>  i: 8
>>  i: 9
>>  i: 10
>> done.
>> 
>> So the countdown loop is ignored. g++ under linux behaves the same.
> 
> There's a logical error in the code. The second part of if() is the 
> condition to be fulfilled to keep the loop going (see the normal loop, 
> where it's going to keep going as long as i is less than or equal to n).
> 
> So it says:
> 
>  for (i = n; i = 0; i--) {
> 
> (i.e. "keep going as long as i is zero", which it obviously isn't)
> 
> It should be:
> 
>  for (i = n; i != 0; i--) {
> 
> (i.e. "keep going as long as i is not zero")

See my previous post.  It does not say "keep going as long as i is zero"
because you used the assignment operator instead of the comparison operator.
It says "set i to 0 and return success" and that terminates the loop.

bill

-- 
Bill Gunshannon          |  de-moc-ra-cy (di mok' ra see) n.  Three wolves
billg999 at cs.scranton.edu |  and a sheep voting on what's for dinner.
University of Scranton   |
Scranton, Pennsylvania   |         #include <std.disclaimer.h>