如何翻转链表中的相邻两个元素？

gistfile1.c

/*
 * Q: 如何翻转链表中的相邻两个元素？ 
 * Link: http://www.foreach.cn/question/16
 * 比如原链表是：
 * a->b->c->d->e->f
 * 我们只翻转相邻两个，即1和2翻转，3和4翻转，翻转后如下：
 * b->a->d->c->f->e
 */
#include <stdio.h>

typedef struct nodeStruct {
    int data;
    struct nodeStruct *next;
} node; 

//print list
void P(node *head){for(; head; printf("%d", head->data), head = head->next);}

void foo(node *head){
    // add one node before head
    node x = {0, head}, *p=head, *m=&x, *n, *q;

    while(p){
        // .. m->p->n->q ..
        n = p->next;
        if(n) q = n->next;
        else break;
        // turn to: m->n->p->q
        m->next = n;
        n->next = p;
        p->next = q;
        //update m and p for next iter
        m = p;
        p = q;
    }
}

int main()
{
    node f={6, 0}, e = {5, &f}, d = {4, &e}, c = {3, &d}, b = {2, &c}, a = {1, &b};
    P(&a);
    printf("\n");
    foo(&a);
    P(&b);
    return 0;
}