Pointers in C Language (Code Explanation) - c

Question

Could someone please explain to me step by step what this code does?
int main(int argc, const char* argv[]) {
int numbers [2] [3] = {{1,2,3},{4,5,6}};
int (*num_ptr) [3]= numbers;
printf ("value is : %d\n",*(*(num_ptr+1)));
printf ("value is : %d\n",**num_ptr);
printf ("value is : %d\n",*(*num_ptr)+1);
printf ("value is : %d\n",*(*(num_ptr+1))+1);
//return (EXIT_SUCCESS);
}

Answer 1

It is easier to understand the code if you use array index notation. A[B] is the same as *(A+B), and vice versa, in all cases; with a special case being that A[0] is the same as *(A + 0), which is the same as *A if A is a valid pointer.
Applying this transformation, your code becomes:
printf ("value is : %d\n", num_ptr[1][0]);
printf ("value is : %d\n", num_ptr[0][0]);
printf ("value is : %d\n", num_ptr[0][1]);
printf ("value is : %d\n", num_ptr[1][0] + 1);
Hopefully you can take it from here.
The definition of num_ptr ensures that num_ptr[A][B] is the same as numbers[A][B]. Your program would be unchanged if the printf statements used numbers instead of num_ptr, because the array numbers decays to the same pointer value that num_ptr has.

Answer 2

*(*(num_ptr+1))
^^^^^^^^^^
skip the first row
No columns skipped after that
Ans = 4
**num_ptr
No rows or columns skipped
Ans = 1
*(*num_ptr)+1
^^^^^^^^^^^
no rows or column skipped
add 1 to the ans
Ans = 2
*(*(num_ptr+1))+1
^^^^^^^^^^^^^^^
skip first row, no columns skipped
add 1 to answer
Ans = 5

Answer 3

int (*num_ptr) [3]= numbers; declares num_ptr as a pointer to an array of 3 int and initializes it to the first element of array numbers. It is equivalent to
int (*num_ptr) [3]= &numbers[0];
num_ptr points to array numbers[0] and num_ptr+1 points to numbers[1]. *(num_ptr+1) dereferences num_ptr+1 and gives an array. As array decays to pointer to its first element in most context, *(num_ptr+1) will finally give &number[1][0]. So, *(*(num_ptr+1)) is equivalent to *(*(&number[1])) == *(&number[1][0]) == 4.
Similarly, **num_ptr will give 1, *(*num_ptr)+1) will give 1+1 =2 and *(*(num_ptr+1))+1 will give 4+1 =5.

Answer 4

First of all let's remove redundant parenthese inside the printf statement.
It will look like
printf ("value is : %d\n", **(num_ptr + 1) );
printf ("value is : %d\n", **num_ptr );
printf ("value is : %d\n", **num_ptr + 1 );
printf ("value is : %d\n",**( num_ptr + 1 ) + 1 );
num_ptr is declared like pointer to the first element of array numbers.
The elements of the array numbers are two on-dimensional arrays { 1, 2, 3 } and { 4, 5, 6 }
So expression *num_ptr yields the first one-dimensional array. In expressions with rare exceptions arrays are implicitly converted to pointers to their first elements. Thus this expression *num_ptr is converted to pointer to first element that is to 1 And **num_ptr gives the value.
As result this statement
printf ("value is : %d\n", **num_ptr );
outputs 1.
This statement
printf ("value is : %d\n", **num_ptr + 1 );
just adds 1 to the value of the prvious expression and outputs it that is 2.
As num_ptr is pointer to the first one-dimensioanle array then num_ptr + 1 is pointer to the second on-dimensional array { 4, 5, 6 }. Other calculations are the same as described above.
Thus statement
printf ("value is : %d\n", **(num_ptr + 1) );
outputs 4 and statement
printf ("value is : %d\n",**( num_ptr + 1 ) + 1 );
outputs 4 + 1 that is 5.