Question

What exactly does this code fragment do?

#include <stdio.h>
 List *makeList(int n) { 
    List *l, *l1 = NULL; 
    for (int i = 0; i < n; i++) {
        l = malloc(sizeof(List)); 
        l->val  = n-i;
        l->next = l1;
        l1 = l;
    }
    return l;
}

My notes say that “Given a number n,
build a list of length n where
the ith element of the list
contains i”

But I don’t get that…

Answer 1

The tricky thing here is that the list is built backwards, note that each element’s value is set to n - i, and i counts from 0 to n - 1. So, the first element will get value n, the next one will get n - 1, and so on.

This is probably done in order to save on a variable (!); otherwise it would be required to have another pointer to remember the first node, in order to have something to return.

Also, it doesn’t check the return value of malloc(), which is always scary.

For n = 0, it will return an undefined value (the value of l), which is really scary.