使用malloc时提示 Assigning: msgNode.data = storage returned from malloc(msgHead.msgLength + 44UL).
时间: 2024-05-16 14:17:04 浏览: 16
这个提示是在使用malloc分配内存时出现的,其中msgHead.msgLength + 44UL是要分配的内存大小。提示的意思是告诉你将malloc返回的内存地址赋值给msgNode.data。
为了避免出现这个提示,你可以将赋值操作拆分成两行代码。具体做法是先将malloc返回的内存地址赋给一个临时变量,然后再将临时变量赋给msgNode.data,如下所示:
```
void *temp = malloc(msgHead.msgLength + 44UL);
msgNode.data = temp;
```
这样做可以避免编译器对于malloc返回值的警告提示,同时使代码更加易读。
相关问题
使用malloc时提示Storage is returned from allocation function malloc.
这个提示意味着你使用了 malloc 分配了内存,但是没有将其赋值给任何指针变量。你需要将 malloc 返回的指针保存在一个指针变量中,以便稍后可以使用该指针访问分配的内存。
例如,以下代码段会导致这个警告:
```c
malloc(10);
```
应该将其修改为:
```c
char *ptr = malloc(10);
```
这样,你就可以使用指针变量 `ptr` 访问分配的 10 字节内存,而且不会收到警告。
用c语言写class ListNode: def __init__(self, x): self.val = x self.next = Nonedef mergeTwoLists(l1: ListNode, l2: ListNode) -> ListNode: if not l1: return l2 if not l2: return l1 if l1.val <= l2.val: head = l1 l1 = l1.next else: head = l2 l2 = l2.next cur = head while l1 and l2: if l1.val <= l2.val: cur.next = l1 l1 = l1.next else: cur.next = l2 l2 = l2.next cur = cur.next if l1: cur.next = l1 else: cur.next = l2 return head# 测试l1 = ListNode(1)l1.next = ListNode(3)l1.next.next = ListNode(5)l2 = ListNode(2)l2.next = ListNode(4)l2.next.next = ListNode(6)head = mergeTwoLists(l1, l2)while head: print(head.val, end=' ') head = head.next
以下是使用C语言实现的代码:
```c
#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int val;
struct ListNode *next;
};
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
if (!l1) return l2;
if (!l2) return l1;
struct ListNode *head, *cur;
if (l1->val <= l2->val) {
head = l1;
l1 = l1->next;
} else {
head = l2;
l2 = l2->next;
}
cur = head;
while (l1 && l2) {
if (l1->val <= l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if (l1) {
cur->next = l1;
} else {
cur->next = l2;
}
return head;
}
int main() {
struct ListNode *l1, *l2, *head;
l1 = (struct ListNode *)malloc(sizeof(struct ListNode));
l1->val = 1;
l1->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l1->next->val = 3;
l1->next->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l1->next->next->val = 5;
l1->next->next->next = NULL;
l2 = (struct ListNode *)malloc(sizeof(struct ListNode));
l2->val = 2;
l2->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l2->next->val = 4;
l2->next->next = (struct ListNode *)malloc(sizeof(struct ListNode));
l2->next->next->val = 6;
l2->next->next->next = NULL;
head = mergeTwoLists(l1, l2);
while (head) {
printf("%d ", head->val);
head = head->next;
}
printf("\n");
return 0;
}
```
输出结果为:1 2 3 4 5 6
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