leetcode第二题及 pointer points here memory cannot be printed

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    struct ListNode *p = l1,*q = l2,*r;
    struct ListNode *p1 = l1,*q1 = l2;
    struct ListNode* l = (struct ListNode*)malloc(sizeof(struct ListNode));
    l->next = 0;
    int length1 = 0,length2 = 0,add = 0;
    while(p1 != 0){
        length1++;
        p1 = p1->next;
    }
     while(q1 != 0){
        length2++;
        q1 = q1->next;
    }
    if(length1 >=length2){
        while(q != 0){
            add += p->val + q->val;
            if(add >= 10){
                p->val =add-10;
                add = 1;
            }
            else{
                p->val =add;
                add = 0;
            }
            r = p;
            p = p->next;
            q = q->next;
        }
        while(p !=0){
            add +=p->val;
            if(add >= 10){
                p->val =add-10;
                add = 1;
            }
            else{
                p->val =add;
                add = 0;
            }
            r = p;
            p = p->next;
        }
        if(add){
            l->val = 1;
            r->next = l;
            l->next == 0;
        }
        return l1;
    }
    else{
        while(p != 0){
            add += q->val + p->val;
            if(add >= 10){
                q->val =add-10;
                add = 1;
            }
            else{
                q->val =add;
                add = 0;
            }
            r = q;
            q = q->next;
            p = p->next;
        }
        while(q !=0){
            add +=q->val;
            if(add >= 10){
                q->val =add-10;
                add = 1;
            }
            else{
                q->val =add;
                add = 0;
            }
            r = q;
            q = q->next;
        }
        if(add){
            l->val = 1;
            r->next = l;
            l->next == 0;
        }
        return l2;
    }
}

一、第二题先求两个链表的长度length1和length2，在长的链表中做加法

二、长链表与短链表第一个数相加，如果相加之和小于10，则把值赋给长的链表；若大于等于10则把个位数赋给长链表，并记进位符add为1.

三、长短链表第二个数相加并加上进位符，以此类推。若短链表所有值加完，长链表中还有数，长链表中值加add（例：8->5->1->1 + 2->4 = 0-> 0->2->1，短链表完成，进位符为1，第三位为2，此时进位符为0，第四位不变为1）

四、若两链表相加后，长度大于长的链表，则再开辟一个节点“l”储存进位符（例9->9->9 + 1 = 0-> 0->0->1）。注意：新开辟的“l”中next指针未指向任何空间，可能出现runtime error: member access within misaligned address 0xbebebebebebebebe for type ‘struct ListNode’, which requires 8 byte alignment (ListNode.c) 0xbebebebebebebebe: note: pointer points here 错误，应该添加l->next = NULL;