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

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在做leetcode 第2題時使用C語言編寫鏈表時報錯

• 錯誤復現
• 錯誤原因
• 解決辦法

錯誤復現

報錯時的代碼如下

/**
 * 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;
}

運行后報錯

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

錯誤原因

在程序倒數第6行處申請了一個tmpN指向的結構體ListNode空間，而該結構體中包含next指針，若該節點作為整個鏈表的最后一個節點，如l1和l2分別指向[5]和[6]時，此時tmpN指向的空間並沒有初始化next指針，因此報錯。

解決辦法

使用NULL初始化next指針指向的內容，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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