The first part initialization of for loop (i<=5 && i>=-1) is just an
initialization and nothing else. It is of no use in this case.
The second part condition holds the condition which is in this case +
+i specifying to run loop till i is non zero. You are using prefix
operator hence it will first increment i and then test.
If you have used i++ the loop wont even have executed single time.
The condition ++i becomes zero when there is round of short and you
get zero at that time the loop terminates.
The format specifier for unsigned int is used hence positive values
are printed.
If the format specifier for %d is used then the results would be more
informative.
Hope that helps.
On Aug 1, 11:51 pm, jagrati verma <jagrativermamn...@gmail.com> wrote:
> #include<stdio.h>
> main()
> {
> short int i=0;
> for(i<=5 && i>=-1;++i;i>0)
> printf("%u\n",i);
> printf("\n");
> return 0;}
>
> o/p is 1.....
> 4294967295 hw???????????????????????????
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
