在做leetcode 第2题时使用C语言编写链表时报错
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;
}
By JZ
Less is more
