linux C array simplified.

[ratheesh k <ratheesh.ksz@gmail.com>]

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);
   }