如何在c语言中清除内存

Question

我有一个结构体如下:

struct Node{
int *arr;
int *sol;
struct Node *Next;
};

我用这种方式创建Node:

Node* MyNode = (Node *)malloc(sizeof (struct Node));
MyNode->arr = malloc(sizeof(int)*N);
MyNode->sol=  malloc(sizeof(int)*N);

然后将MyNode添加到链表中。如何为列表中的元素释放内存?
这是正确的吗?

pop(){
   free(first->arr);
   free(first->sol);
   first=first->Next; 
}

Answer 1

要使struct成为linked-list中的节点，则需要将self-referential structure variable声明为struct Node *next;

struct Node{
    int *arr;
    int *sol;
    struct Node *next;
}

要为链表的一个节点分配内存，您需要以下配置:

/* allocate memory for a node */
struct Node * MyNode = (struct Node *)malloc((int)sizeof(struct Node));
if (MyNode == NULL) {
    printf("ERROR: unable to allocate memory n");
    return 1;
}
/* allocate memory for the content of a node */
MyNode->arr = (int *) malloc((int)sizeof(int) * N);
if (MyNode->arr == NULL) {
    printf("ERROR: unable to allocate memory n");
    return 1;
}
MyNode->sol = (int *) malloc((int)sizeof(int) * N);
if (MyNode->sol == NULL) {
    printf("ERROR: unable to allocate memory n");
    return 1;
}
/* add the new node to a list by updating the next variable */
MyNode->next = ... 

如果您不确定需要执行哪些操作来删除链表中的节点，您可以使用temp变量以更简单的方式完成相同的操作。

pop()
{
    struct Node * temp = first;
    first = first->next;
    free(temp->arr);
    free(temp->sol);
    free(temp);
}

free的经验法则——每个malloc()都应该有一个free()

OTOH，要了解在链表中删除节点的各种场景，请参考此链接。

Answer 2

几乎，您需要释放节点本身:

pop(){
   Node* old_first = first;
   free(first->arr);
   free(first->sol);
   first=first->Next;
   free(old_first); 
}

Answer 3

pop(){
   free(first->arr);
   free(first->sol);
   Node* temp = first; //<========
   first=first->Next;
   free (temp);  //<=======
}

Answer 4

这很接近但不正确-你应该有尽可能多的free和malloc。你忘了释放Node本身。

要修复它，添加一个临时的:

Node *next = first->next;
free(first->arr);
free(first->sol);
free(first); 
first = next;