Loading an array using pointers

I am trying to use pointers to preload an array. I know how to do it as an array but the pointer method has my head twisted around.

The Array is defined as Array[Size]

If I used a function void load (int *loadmem) , how would I load the entire array with a single ( non-zero ) interger value. Would a FOR loop work from 0 to Size. But how do I tie the Array to a pointer and then increment the location in the pointer?

Code:

---

/*fills an array with 0's*/
char array[SIZE], *p;
for(p = &array[0]; p < &array[SIZE - 1]; )
    *p++ = 0;

---

-Prelude

Does this mean that if I wanted to load every element of the array with the number 8 it would be :
--------------------------------------------
/*fills an array with 8's*/
char array[SIZE], *p;
for(p = &array[0]; p < &array[SIZE - 1]; )
*p++ = 8;

and why a char and not an int if I'm loading it with integers?

>Does this mean that if I wanted to load every element of the array with the number 8 it would be
Actually, if you put in just 8 it would place the ASCII character represented by 8, for a character you would have to write '8', inside the single quotes.

>and why a char and not an int if I'm loading it with integers?
The array can be of any legal type as long as you're also loading it with a legal type. It really doesn't matter since the pointer only goes by the address of each element.

-Prelude

Sorry - maybe my question wasn't clear.

I did want to put 8 in every element. I realize that the computer will store this as an ascii number but my program will require number of + to - 10000. I will be adding number to specific elements but I wanted to preload the entire array with a value.

Just to be clear - Did I modify your lines of code properly to cause 8 to be loaded in each element?

Cheesehead

>Did I modify your lines of code properly to cause 8 to be loaded in each element?
No, to do that you need to change the array declaration from char array[SIZE] to int array[SIZE].

-Prelude