Question

以下是我的两个代码。一个使用结构和另一个使用指针结构。但是，当我不使用指针时，它不起作用。虽然我认为它们是相同的。但我还是初学者。所以我需要了解出了什么问题。

不工作代码：

struct Node{
    int data;
    struct Node* next;
};

void insert(struct Node** head_ref,int data){
    //Not Working Here. The Header should change after every insert but it isn't Moving from it's Memory;

    struct Node temp ;

    temp.data = data;
    temp.next = (*head_ref);
    (*head_ref) = &temp;
}

int main(){
    struct Node *head = NULL;

    insert(&head,4);
    insert(&head,2);
    insert(&head,11);
    insert(&head,9);

            while(head->next !=0 ){
                std::cout << head->data <<" ";
                head = head->next;
            }
    return 0;
}

工作代码：

struct Node{
    int data;
    struct Node* next;
};

void insert(struct Node** head_ref,int data){
    //The Problem is in This Line. Pointer to Structure is Working But only Structure isn't
    struct Node* temp = (struct Node*) malloc(sizeof(struct Node)) ;

    temp->data = data;

    temp->next = (*head_ref);
    (*head_ref) = temp;
}

int main(){
    struct Node *head = NULL;

    insert(&head,4);
    insert(&head,2);
    insert(&head,11);
    insert(&head,9);

            while(head->next !=0 ){
                std::cout << head->data <<" ";
                head = head->next;
            }
    return 0;
}

Answer 1

使用代码struct Node temp ; ... (*head_ref) = &temp;，可以存储局部变量的地址。函数insert完成后，存储在此变量中的对象的生命周期结束，并且在此时间之后访问指针是未定义的行为。

这与第二种情况不同，struct Node* temp = (struct Node*) malloc(sizeof(struct Node))动态分配对象;这个对象的生命周期在明确删除时结束，这样你就可以引用它的地址了。

Answer 2

堆和堆栈What and where are the stack and heap?

之间的区别

void insert(struct Node** head_ref,int data){
    //Not Working Here. The Header should change after every insert but it isn't Moving from it's Memory;

    struct Node temp ;

    temp.data = data;
    temp.next = (*head_ref);
    (*head_ref) = &temp;
}

在原始代码中，temp位于stack，因为在范围的末尾自动销毁了

void insert(struct Node** head_ref,int data){
    //The Problem is in This Line. Pointer to Structure is Working But only Structure isn't
    struct Node* temp = (struct Node*) malloc(sizeof(struct Node)) ;

    temp->data = data;

    temp->next = (*head_ref);
    (*head_ref) = temp;
}

这可能有效，因为它位于heap，因此您必须在使用完毕后手动将其删除

指针与结构错误的指针

2 个答案: