Question

我试图使用双向链表并从文本文件中读入以创建一个非常基本的缓存。目前缓存容量为5，我从txt文件读取值为1-20的20个值。当我尝试删除头部然后打印链表时，它会打印0,2,3-20，因此将0替换为0。据我所知，我需要创建一个临时节点来设置头值，然后将头指向下一个节点，最后删除临时节点，这就是我所做的，但我显然缺少一些重要的东西

int main(int argc, char** argv) {
node* head;
node* tail;
node* n;

int capacity = 5; 
int size = 0;

std::string fileName;   
std::ifstream inFile;

int start_block, num_blocks, ignore, req_num;
std::cout << "Project 3 milestone: LRU" << std::endl;
std::cout << "Enter filename: " << std::endl;
std::cin >> fileName;


inFile.open(fileName.c_str(), std::ifstream::in);
if (!inFile)    {
    std::cerr << "Cannot open file!  " << fileName << std::endl;
}

while(inFile.is_open()) {
    inFile >> start_block >> num_blocks >> ignore >> req_num;
    n = new node;
    n->data = start_block;
    n->prev = NULL; // 1st node
    head = n;
    tail = n;
    size++;

    while(!inFile.eof())    {
        inFile >> start_block >> num_blocks >> ignore >> req_num;
        n = new node;
        n->data = start_block;
        n->prev = tail;
        tail->next = n;
        tail = n;
        size++;
        //std::cout << start_block << " " << num_blocks << " " << ignore << " " << req_num << std::endl;    
        if  (size == capacity)  {
            cout << "Reached capacity:" << capacity << endl;
            // this is where I would delete the head node
        }   
    }           
    inFile.close();
}

PrintForward(head);
//PrintReverse(tail);
SearchRecursive(head,18);
DeleteHead(head, tail);
PrintForward(head);
//DeleteHead(head, tail);
//PrintForward(head);
return 0;
}

void SearchRecursive(node* ptr, int searchValue)    {
if(ptr == NULL) {   // if we pssed through list and didnt find value
    cout << searchValue << " was NOT found in the list\n";
}       
else if (ptr->data == searchValue)  {   // if we DID find it
    cout << searchValue << " IS in the list!\n";
}
else    {
    SearchRecursive(ptr->next, searchValue);    // else search recursively
}
}

void DeleteHead(node* head, node* tail) {
if (head == tail)   {   // if only 1 element
    cout << "Only 1 element here" << endl;
    delete head;
    head = NULL;
    tail = NULL;
}
else    {
    cout << "More than 1 element here" << endl;
    node *temp = head;
    head = head->next;
    delete temp;
}
}

编辑：我改进了SearchRecursive函数，现在可以删除除头部和尾部之外的节点。这是我正在使用的：

void SearchRecursive(node* ptr, int searchValue)    {
if(ptr == NULL) {   // if we pssed through list and didnt find value
    cout << searchValue << " was NOT found in the list\n";
}       
else if (ptr->data == searchValue)  {   // if we DID find it
    cout << searchValue << " IS in the list!\n";
    ptr->prev->next = ptr->next;
    ptr->next->prev = ptr->prev;
    delete ptr;
}
else    {
    SearchRecursive(ptr->next, searchValue);    // else search recursively
}
}

Answer 1

因为你删除了头指针，在你调用DeleteHead之后头指针现在无效，因为指针没有改变，函数的变化只反映了{{1}点的值。函数里面的参数并没有反映在这个函数之外，你想要的是改变传递指针的值，这是通过引用来完成的。所以你的函数现在正在获取对指针和函数签名的引用，现在是

head

并在函数内部执行void DeleteHead(node** head, node* tail)

或者你可以从函数

返回新的头部值

*head = (*head)->next;

Answer 2

我最终做的是上述答案和评论的结合。这是我现在的DeleteHead代码

void DeleteHead(node*& head, node* tail)    {
cout << "Deleting head..." << endl;
node* temp = head;
head = head->next;
size--;
delete temp;
}

尝试删除所选节点的双链表

2 个答案: