People,

Thank you so much for all the great explanations. I now understand now
the actual problem. [In such that's not really a problem :-)]

Thankx..

Jason




On Sun, 21 Oct 2007 19:12:25 +0100, Malcolm McLean wrote:
> "jason" <jisis@notmal.com> wrote in message news:471b8fa0$0$14414
>> But just to make sure; what does `decay' exactly mean in this case ?
>> And what properties of the pointer, when passed to a function actually
>> change ?
>>
>>
> int main(void)
> {
> int array[10] = {1,2,3,4,5,6,7,8,910};
>
> printf("array is %d bytes\n", (int) sizeof(array)); foo(array, 10);
> }
>
> void foo(int *ptr, int N)
> {
> int i;
>
> printf("pointer is %d\n", (int) sizeof(ptr))' /* treat as array */
> for(i=0;i<N;i++)
> print("%d\n", ptr[i]);
> /* treat as pointer */
> for(i=0;i<N;i++)
> printf("%d\n", *ptr++);
> }
>
> in main your array is an array. When you pass it to foo it converts -
> decays - into a pointer. arrays are very nearly, but not quite, constant
> pointers. You can use the brackets notation on either an array or a
> pointer. However you can also increment a pointer. Whilst sizeof(ptr)
> gives the size of the memory item needed to hold the address, usually 4
> bytes, whilst sizeof(array) gives the size of the data in the array,
> probably either 20 or 40 bytes.