Do you see the bug here?

Help with Code Tags

C Syntax (Toggle Plain Text)

#include <stdio.h>
 
 #define N 10
 
 int main()
 {
    int i, array[N] = {1,2,3,4,5,6,7,8,9,10};
    for ( i = 1; i <= N; ++i )
    {
 	  printf("array[%d] = %d\n", i, array[i]);
    }
    return 0;
 }

#include <stdio.h> #define N 10 int main() { int i, array[N] = {1,2,3,4,5,6,7,8,9,10}; for ( i = 1; i <= N; ++i ) { printf("array[%d] = %d\n", i, array[i]); } return 0; }

Hehe. It should be like this:

Help with Code Tags

C Syntax (Toggle Plain Text)

    for(i=1;i<=n;i++)

for(i=1;i<=n;i++)

NOT

Help with Code Tags

C Syntax (Toggle Plain Text)

    for(i=1;i<=n;++i)

for(i=1;i<=n;++i)

The second one is wrong because i increases with 1 BEFORE entering the for .. so i practically starts with the value 2. While in my example , i starts from 1 and ends at 10 , meaning exactly what it has to do. :eek:

No, the second one is not wrong, and your explanation is incorrect. I take it you didn't even try it.:rolleyes:

And I take it that you still don't see the bug. This is exactly why you shouldn't be doing this.

Indeed , I haven't tested them until now. It seems like both versions give the same result ...

My guess is: i should start from 0? :p

--------------------------------------------------------------------------------
hi
look if u want array and every case on array it's name look:
#include<stdio.h>
int main()
{char *a[5]={"sami","rami","wael","dani","nami"};
for (i=0;i<5;i++)
printf("%s",a[i]);
return 0;
}

Help with Code Tags

C Syntax (Toggle Plain Text)

char *a[5];

char *a[5];

An array of uninitialized pointers. These are not automagically strings, you need to have them actually point to some memory you can write to before you try to write to it. Or actually declare an array for strings.

Help with Code Tags

C Syntax (Toggle Plain Text)

char a[5][80];

char a[5][80];