member access within misaligned address 0x0000002c3931 for type 'struct ListNode‘

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在做leetcode 第2题时使用C语言编写链表时报错

• 错误复现
• 错误原因
• 解决办法

错误复现

报错时的代码如下

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {

    if(l1==NULL) return l2;
    if(l2==NULL) return l1;

    struct ListNode* result=(struct ListNode*)malloc(sizeof(struct ListNode));
    int tmp=l1->val+l2->val;
    if(tmp<10){
        result->val = tmp;
        result->next = addTwoNumbers(l1->next,l2->next);
    }else{
        result->val = tmp%10;
        struct ListNode* tmpN=(struct ListNode*)malloc(sizeof(struct ListNode));
        tmpN->val=tmp/10;
        result->next = addTwoNumbers(addTwoNumbers(l1->next,l2->next),tmpN);
    }
    return result;
}

运行后报错

member access within misaligned address 0x000000000e91 for type 'struct ListNode', which requires 8 byte alignment (ListNode.c)
0x000000000e91: note: pointer points here

错误原因

在程序倒数第6行处申请了一个tmpN指向的结构体ListNode空间，而该结构体中包含next指针，若该节点作为整个链表的最后一个节点，如l1和l2分别指向[5]和[6]时，此时tmpN指向的空间并没有初始化next指针，因此报错。

解决办法

使用NULL初始化next指针指向的内容，如

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {

    if(l1==NULL) return l2;
    if(l2==NULL) return l1;
    struct ListNode* result=(struct ListNode*)malloc(sizeof(struct ListNode));
    int tmp=l1->val+l2->val;
    if(tmp<10){
        result->val = tmp;
        result->next = addTwoNumbers(l1->next,l2->next);
    }else{
        result->val = tmp%10;
        struct ListNode* tmpN=(struct ListNode*)malloc(sizeof(struct ListNode));
        tmpN->next=NULL;
        tmpN->val=tmp/10;
        result->next = addTwoNumbers(addTwoNumbers(l1->next,l2->next),tmpN);
    }
    return result;
}

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By JZ
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