Doubt in arrays

[psandeepnair1985]

Hi friends,

I have a major doubt in array concept.
For eg i declare
int a[10];
int b[10][10];

I am confused how &a and a are equal (&a = a)
and
how &b=b=*b..................

I would be thankful to any one who could explain me the logic behind this....

Please help friends.......

[wje_lf]

Pointers and arrays work similarly, but not quite identically.

If you declare something as an array, then using "something" by itself in a C program acts as a pointer to the first element of that something.

If you declare something as an array, then &something is the address of the first element of that something. So in this case, something and &something mean the same thing.

That takes care of your "a" case.

Your "b" case is similar. But b is an array of arrays, right? So b[0] is the first array within the array of arrays; b[1] is the second array; and so on.

So *b is the first array within the array of arrays, just as *a is the first element of a.

A simple reference to an array works like a pointer to the first element of that array. Since *b is an array, *b is equivalent to b[0], which is an array, so it works like a pointer to the first element of that array, which means a pointer to b[0][0].

Try this simple program and watch the magic:

#include <stdio.h>

int main(void)
{
  int a[10];
  int b[10][10];

  a[0]=9;
  b[0][0]=8;

  printf("0x%08X 0x%08X 0x%08X\n",
         &a,
         a,
         *a
        );

  printf("0x%08X 0x%08X 0x%08X 0x%08X\n",
         &b,
         b,
         *b,
         **b
        );

  return 0;

} /* main() */

If this explanation is insufficient, you may wish to become more grounded in how arrays and pointers work in C. To do so, google this:

C pointer array

and you'll get a whole bouquet of tutorials on the subject (along with other stuff that doesn't apply).

Hope this helps.