Question

static void just_traverse(sll **head_ref) {
    sll *first = *head_ref;
    sll *second = (*head_ref)->next;

    if(second == NULL) {
            return;
    }

    just_traverse(&(second));
    *head_ref = second;
    printf("%d \t",second->payload);
}

In above code if I remove *head_ref = second; I get output with respect to stack however If I put this line *head_ref = second; It always print last element say {4,3,2,1} then it always prints 1?Can anybody explain why?

Answer 1

If Your list contains {4,3,2,1}, then your output should be 1 1 1 right.

This because of the statement *head_ref = second;. In the 3rd call to the function just_traverse head_ref will points to 2 and second will points to 1. Now you are updating **head_ref to 1. Now you are printing second. So output will be value 1

And then it will return to the 2nd call, here head_ref is 3 and second is 1 (not 2 becuase we updated in 3rd all). Now againg assigning 1 to head and printing second. Again it will print the value 1 only.

And then it will return to the 1st call, here head_ref is 4 and second is 1 (not 3 because we updated in 2nd call). Now againg assigning 1 to head and printing second. Again it will print the value 1 only.

So the output will be three 1

1 1 1