1. When you are using %u, it is for unsigned printing.

2. looking at the loop, it will run for only values of i
for(i<=5 && i>=-1;++i;i>0)
^^ the first statement here is not the conditional one.

3. since int is of 4 bytes on your system, the max positive value it will
print is power(2, 32) - 1. (4 bytes = 32 bits)

if suppose you used unsigned int, then that would have been an infinite
loop. since the condition is i>=-1. each time you incremented the value of i
, it would have been between (0, 2^64-1) only.


On 2 August 2011 03:16, Dharmendra Modi <[email protected]>wrote:

> 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 <[email protected]> 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???????????????????????????
>
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