C language. Can't understand the output of the pointers

Question

I have the following piece of code.

    int a = 10, b = 20;
    int *x, *y, *z;
    int **pp;
    int arr[3] = { 5, 13, 29 };
    x = &b;
    pp = &y;
    y = arr + 1;
    *y = a;
    z = y;
    **pp = 999;
    y++;

    printf("%i %i %i %i %i %i %i %i %i", a, b, *x, *y, *z, **pp, arr[0], arr[1], arr[2]);

I expected the output would be 10 20 20 29 10 999 5 999 29 , however, the correct output appeared to be 10 20 20 29 999 29 5 999 29 . The differences are in *z and **pp . I can't understand why so, even though I tried to draw diagrams and analyze the result step by step. Why *z and **pp are 999 and 29, respectively, not 10 and 999? Thank you.

Answer 1

After this statement

y = arr + 1;

the pointer y points to the second element of the array arr ,

So the second element is set to 10 after the next statement

*y = a;

And we have the following content of the array arr

{ 5, 10, 29 }

The pointer z also points to this second element

 z = y;

However the second element was rewritten

 **pp = 999;

because the dereferenced pointer pp always yields the pointer y . The content of the array now is

{ 5, 999, 29 }

and the pointer y was incremented and after this statement

y++;

points to the third element of the array. Pay attention to that the pointer pp still points to the pointer y that stores now the address of the third element of the array.

The variables a and b were not changed> so the call of printf outputs

10 20

the pointer x points to the variable b so we have

10 20 20

The pointer y points to the third element of the array. So

10 20 20 29

The pointer z points to the second element of the array because it has the value of the pointer y before its increment.

10 20 20 29 999

The pointer pp points to the pointer y and as a result the expression **pp yields the same value as the expression *y

10 20 20 29 999 29

And these expressions arr[0], arr[1], arr[2] sequentially provide values of the array

10 20 20 29 999 29 5 999 29