Re: linux C array simplified.

[Andrej Gelenberg <andrej.gelenberg@udo.edu>]

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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