Hi,
keep in mind, that on 64-bit systems pointer is 8 bytes long.
Regards,
Andrej
On 01/12/2011 03:40 PM, ratheesh k wrote:
> I wrote program to make c array simple. I have added comments in
> between so that my student could understand it easily.
> Could you please check whether this comments are correct. Could you
> correct me if i am wrong. Thanks in advance.
> ********************************************************************************************************************************************************
> # include <stdio.h>
>
> int functionA(int (*ptr)[10])
> {
> /* passing array address. */
>
> /* ptr pointer to an array and &ptr is the pointer to array pointer
> Both are of size 4 bytes */
> printf("sizeof(ptr) = %ld\n", sizeof(ptr));
> printf("sizeof(&ptr) = %ld\n", sizeof(&ptr));
>
> /* if ptr is incremented . ptr + 10*4 will be the answer */
> printf("pointer address ptr=%p ptr+1=%p\n" , ptr ,ptr+1);
>
> /*since ptr is an array pointer ; element are accessed like (*ptr[0])
> */
> printf("access memory ptr=%d\n" , (*ptr)[0] );
> /* we could do this by *(*ptr)+0); this is
> same as **ptr
> */
> printf("access memory **ptr=%d\n" , (**ptr) );
>
> /* access memory at *(*ptr+1)) */
> printf("access memory *(*ptr+1)=%d\n" , *(*ptr+1) );
>
> return 0;
> }
>
> int functionB(int *abc )
> {
> /* address of first element is passed */
>
> /* sizof pointer is 4 bytes */
> printf("sizeof(abc) = %ld\n" ,sizeof(abc));
>
> /* pointer increment will point to next element. pointes is of type (int *)
> */
> printf("pointer address abc=%p abc+1=%p\n", abc ,abc+1 );
> /* we could access memory locations by dereferencing each location */
>
> printf("access memory at *(abc)=%d, *(abc+1)=%d\n" , *abc ,*(abc+1));
> }
>
> int functionC(int pqr[])
> {
> /* prq[] is same as char *abc (in functionA ). eventhough we have
> pqr[] declaration
> compiler will treat this as simple char *pqr.
> Please refer funnctiuonA for all explanations
> */
> printf("sizeof(pqr) = %ld\n" ,sizeof(pqr));
> printf("pointer address pqr=%p pqr+1=%p\n", pqr ,pqr+1 );
> printf("access memory at *(pqr)=%d, *(pqr+1)=%d\n" , *pqr ,*(pqr+1));
> }
>
> int functionD(int stp[20])
> {
> /*stp[20] is also treated as simple char *stp */
> printf("sizeof(stp) = %ld\n" ,sizeof(stp));
> printf("pointer address stp=%p stp+1=%p\n", stp ,stp+1 );
> printf("access memory at *(stp)=%d, *(stp+1)=%d\n" , *stp ,*(stp+1));
> }
>
>
> int main()
> {
> /* integer array is defined and values are initialized */
> int arr[10]={1,2,3,4,5,6,7,8,9,0};
>
> /* sizeof(arr) gives totoal sizeof array. that is equal to
> no_of_elements * sizeof_an_element
> here it 10*4 = 40
> */
> printf("Sizeof(arr) = %ld\n", sizeof(arr));
> /* address of array is &arr and its size is of 32bit (4bytes) */
> printf("sizeof(&arr)=%ld\n" , sizeof(&arr));
> /* sizeof(int) is 4 and arr[0] contains and integer */
> printf("sizeof(arr[0])=%ld\n" , sizeof(arr[0]));
> /* arr contains pointer to first element of the array . so it is
> incremented
> it will point to next element
> */
> printf("pointer address arr=%p arr+1=%p \n" ,arr, arr+1 );
> /* &arr contains array address. so if it is incremented, it will
> point to (&arr) + (no.of.elements) * (sizof.each.element)
> eg: if &arr is 1000 ;then
> 1000 + 10*4 =1040
> */
> printf("pointer address &arr=%p &arr+1=%p\n" ,&arr ,&arr+1);
>
> /* arr points to first element of the array. and *arr dereference
> first element */
> printf("access element *arr=%d\n", *arr);
> /* normal array access */
> printf("access element arr[0]=%d\n", arr[0]);
>
> /* (arr+1) points to next element in the array */
> printf("access element *(arr+1)=%d\n", *(arr+1));
>
> functionA(&arr);
> functionB(arr);
> functionC(arr);
> functionD(arr);
> }
> --
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