Reversing a linked list in place

prev, cur and next walk down the list. Each pass saves cur->next, flips cur->next to point backward at prev, then moves all three one node along.

Watch the heap arrows turn around one at a time. At the end, prev is the new head.

#include <stdio.h>
#include <stdlib.h>

struct node {
    int value;
    struct node *next;
};

int main(void) {
    struct node *head = NULL;
    for (int v = 3; v >= 1; v--) {
        struct node *n = malloc(sizeof *n);
        if (!n) return 1;
        n->value = v;
        n->next = head;
        head = n;
    }
    struct node *prev = NULL;
    struct node *cur = head;
    while (cur) {
        struct node *next = cur->next;
        cur->next = prev;
        prev = cur;
        cur = next;
    }
    head = prev;
    printf("%d %d %d\n", head->value, head->next->value, head->next->next->value);
    while (head) {
        struct node *next = head->next;
        free(head);
        head = next;
    }
    return 0;
}

Output:

3 2 1

From the lesson: Linked data structures