C Programming - Pointers - Discussion

Thank you for explaining the answer.

Here is the coding part;


#include<stdio.h>
int main ()
{
int arr[2][3]={5, 10, 15, 20, 25, 30};
int (*ptr) [2][3]=&arr;
printf ("%d\t", ***ptr);
printf ("%d\t", ***(ptr+1));
printf ("%d\t", **(*ptr+1));
printf ("%d\t", *(*(*ptr+1)+2));
return 0;
}

Here is the coding part;


#include<stdio.h>
int main ()
{
int arr[2][3]={5, 10, 15, 20, 25, 30};
int (*ptr) [2][3]=&arr;
printf ("%d\t", ***ptr) ;
printf ("%d\t", ***(ptr+1)) ;
printf ("%d\t", **(*ptr+1)) ;
printf ("%d\t", *(*(*ptr+1)+2)) ;
return 0;
}

Thanks @Pradeep and @Ramanaji.

Thank you @Pradeep.

Thanks @Ramanaji.

Thanks, @Pradeep.

Here In the above problem "a" is an Array having $ dimensions.

Which indicates that a is a pointer and i , j , k , l are offset values.

Let Us Consider An Example : Of An Array;

int ptr = a + i;
where a is Base address and i is Offset,
now it gives address of a[i] which is in integer form
Hence we will it into address as *(a+i) ;
and then add it with the 2-nd offset i.e. j -> *(a+i)+j.

Similarly The Process Continues as Follows:
Address = *(a+i)+j Offset = k
new Address = (*(*(a+i)+j)+k)
Address = *(*(a+i)+j)+k Offset = l
new Address = *(*(*(a+i)+j)+k)+l

And Then Finally The Address Of The Element a[i][j][k][l] = *(*(*(*(a+i)+j)+k)+l).

Such a wonderful explanation and I can easily understand that. Thank you @Ramanaji & @Pradeep.

a[i] =*(a+i) .

I can't understand this, please explain.