On Wed, 12 Jan 2011 16:40:30 +0100, ratheesh k <ratheesh.ksz@xxxxxxxxx>
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 */
No. This is platform dependent and what's more, in theory both can have
different sizes.
Also, IIRC, strictly speaking &ptr is not a pointer but (an expression
that yields)
the address.
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 */
Again, you assume sizeof(ptr) is 4 which does not need to be true.
printf("pointer address ptr=%p ptr+1=%p\n" , ptr ,ptr+1);
/*since ptr is an array pointer ; element are accessed like (*ptr[0])
*/
You meant (*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 */
Same as above.
printf("sizeof(abc) = %ld\n" ,sizeof(abc));
/* pointer increment will point to next element. pointes is of type
^^^^^^^ -- typo
(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));
No return statement.
}
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
*/
Typos plus you meant functionB.
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
*/
sizeof(int) does not have to be 4.
printf("Sizeof(arr) = %ld\n", sizeof(arr));
You should probably write "sizeof" not "Sizeof".
/* address of array is &arr and its size is of 32bit (4bytes) */
Again, assumptions about type size.
printf("sizeof(&arr)=%ld\n" , sizeof(&arr));
/* sizeof(int) is 4 and arr[0] contains and integer */
See above.
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
*/
See above.
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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