Use this for statement:
for ( ; i < (int) (sizeof(array)/sizeof(int)); )
--
Regards,
Fangshuai 'David.Ko' Hu
On Sep 3, 2011, at 1:13 PM, Amahdy wrote:
> But this code will work:
>
> #include <stdio.h>
>
> int main()
> {
> int i;
> int array[]={ 2, 4, 45, 3, 21};
>
> i = 9;
> for ( ;i < sizeof(array);){
> printf("%d\n",array[3]);
> i++;
> }
> return 0;
>
> }
>
> sizeof is a keyword BTW, and it's designed to return a value of type size_t.
> It must be a number greater than or equal zero because it represents a size
> of something, which could never be negative.
> In gcc compiler implementation, size_t is actually a long unsigned int.
>
> In the for loop, you are trying to compare an unsigned value with a negative
> (signed) value, it won't work and the signed value get transformed into a
> very big number (because of reading it without the -ve sign).
>
> A solution for this could be a simple casting to int for example:
>
> for ( ;i < (int) sizeof(array);){
>
>
> -- Amahdy
> www.amahdy.net
>
>
>
> On Tue, Aug 30, 2011 at 17:42, addytheboss khandalkar
> <[email protected]> wrote:
> hello friends,
>
> i am facing trouble in understanding how does this code works and how does
> sizeof( ) works? and why it is not entering in the loop ?
>
>
>
> #include <stdio.h>
>
> int main()
> {
> int i;
> int array[]={ 2, 4, 45, 3, 21};
>
> i = -9;
>
> for ( ;i < sizeof(array);){
> printf("%d\n",array[3]);
> i++;
> }
>
> }
>
>
>
> Thanks
> Addytheboss
>
>
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