Question

所以我有一个任务涉及在C ++中创建一个虚构的内存管理器。以下是说明

编写一个模拟操作系统分配职责的C ++程序记忆到某些程序。这将是一个非常简单的基于页面的内存视图管理。在启动时，您的程序将有大约32页连续，未使用记忆。每页的长度为4 KB

我的问题是删除特定的＆＃34;程序＆＃34;来自链表。这是我目前的DestroyProgram方法

void DestroyProgram(string proName) { //Deletes
    Node* iterator = head;
    while (iterator != NULL) {
        if (iterator->programName == "FREE") {
            int count = 0;
            while (iterator != NULL && iterator->programName == "FREE") {
                iterator->programName = "FREE";
                iterator = iterator->nextProgram;

            }
            return;
            count++;
        }
        else {
            iterator = iterator->nextProgram;
        }
    }
    cout << iterator->programName << " is not there.";
}

如果未使用内存，则标记为FREE。因此，我试图说出程序名是否等于用户输入的程序名称，然后将其删除。我很困惑为什么它不起作用

Answer 1

我在这里猜测，但做出这些假设：
*您的链接列表包含由内存管理器管理的所有内存块 *如果表示存储块的节点是空闲的，则programName设置为“FREE”
*如果节点由程序占用，则programName设置为程序名称
*您实际上并没有删除任何节点，只是将它们重新标记为“FREE”

void DestroyProgram(string proName) { //Deletes
    Node* iterator = head;
    while (iterator != NULL) {
        if (iterator->programName == proName) {
            int count = 0;
            while (iterator != NULL && iterator->programName == proName) {
                iterator->programName = "FREE";
                iterator = iterator->nextProgram;
            }

            // If you return here, you're assuming that programs always have memory allocated in consecutive blocks
            // If they request memory more than once, the memory blocks may not be together
            // In that case, you have to keep searching through to the end of the list
            return;

            count++;
        }
        else {
            iterator = iterator->nextProgram;
        }
    }
    cout << proName << " is not there.";
}

Answer 2

好的，我只是给你一个通用版本来从链表中删除一个节点。从列表中删除节点时必须提出三种情况。

如果列表为空

如果您要删除的节点是

如果节点在其他任何地方

void DestroProgram(string proName) //Deletes
  {
    Node *iterator = head;
    if(head == NULL)
    {
    cout << "List is empty" << endl;
    }
    else if(iterator->programName == proName)
    {
       head = head->nextProgram   //point to your new head
       delete iterator;    //delete your old head
    }
    else
    {
       Node *previousNode = head;
       while(head != NULL)
       {
         if(iterator->programName == proName)
         {
          previousNode->next = iterator->next  //make the new connection between nodes
          delete iterator
         }
          previousNode = iterator;  //sets previousNode one node behind iterator
          iterator = iterator->nextProgram;
        }
     }
  }

从C ++中的链接列表中删除特定项目

2 个答案: