Question

我正在编写一个程序，该程序接受8个用户输入的整数，并从中得出一个链表。我让程序打印链接列表，然后删除最后一个节点，然后反向打印列表。在测试过程中，我一直在进行编程以确保每个部分都工作正常，并且可以完成打印原始链接列表的工作。

当我完成修改的代码编写并打印部分时，我遇到了一个问题-在打印出原始列表后，该程序将不会输出任何内容。因此，例如，如果我输入1,2,3,4,5,6,7,8，它将输出： 1个 2 3 4 5 6 7 8 就是这样。我已经尝试过将cout <<“ testing”;在不同的位置查看我的代码停止输出的位置，以及成功输出的最新位置就在while循环之前。

我不确定为什么while循环会导致程序停止输出任何内容，甚至是与while循环本身无关的任意cout语句，因此我想在这里问一下。如果有帮助，我正在使用Visual Studio 2017。感谢您提供的所有帮助！

#include <iostream>
using namespace std;

void getdata(int & info); //function that assigns a user inputted value to each node
const int nil = 0;
class node_type         // declaration of class
{
public:
    int info;
    node_type *next;
};

int main()
{
    node_type *first, *p, *q, *r, *newnode;
    first = new node_type;
    newnode = new node_type;
    int info;
    getdata(info); //first node
    (*first).info = info;
    (*first).next = nil;
    getdata(info); //second node
    (*newnode).info = info;
    (*first).next = newnode;
    (*newnode).next = nil;
    p = newnode;
    for (int i = 2; i < 8; i++) //nodes 3-8
    {
        newnode = new node_type;
        getdata(info);
        (*newnode).info = info;
        (*p).next = newnode;
        p = newnode;
        (*newnode).next = nil;
    }
    q = first;
    while (q != nil) // printing linked list
    {
        cout << (*q).info << "\n";
        q = (*q).next;
    }

    //deletes last node then reverses list
    p = first;
    q = (*p).next;
    r = (*q).next;

    if (first == nil) //if list is empty
        cout << "Empty list";
    else if ((*first).next == nil) //if list has one node
        first = nil;
    else if (r == nil) //if list has two nodes
        q = nil;
    else //general case
    {
        (*first).next = nil; //last line where when i put a cout << ""; it prints in the output window
        while ((*r).next != nil)
        {
            (*q).next = p;
            (*r).next = q;
            p = q;
            q = r;
            r = (*r).next;
        }
        (*q).next = p;
        first = q;
    }
    q = first;
    while (q != nil) // printing newly modified list.
    {
        cout << (*q).info << "\n";
        q = (*q).next;
    }
    return 0;
}
void getdata(int & info)
{
    cout << "Enter number: \n";
    cin >> info;
}

Answer 1

class HTMLBaseElement extends HTMLElement {
  constructor(...args) {
    const self = super(...args)
    self.parsed = false // guard to make it easy to do certain stuff only once
    self.parentNodes = []
    return self
  }

  setup() {
    // collect the parentNodes
    let el = this;
    while (el.parentNode) {
      el = el.parentNode
      this.parentNodes.push(el)
    }
    // check if the parser has already passed the end tag of the component
    // in which case this element, or one of its parents, should have a nextSibling
    // if not (no whitespace at all between tags and no nextElementSiblings either)
    // resort to DOMContentLoaded or load having triggered
    if ([this, ...this.parentNodes].some(el=> el.nextSibling) || document.readyState !== 'loading') {
      this.childrenAvailableCallback();
    } else {
      this.mutationObserver = new MutationObserver(() => {
        if ([this, ...this.parentNodes].some(el=> el.nextSibling) || document.readyState !== 'loading') {
          this.childrenAvailableCallback()
          this.mutationObserver.disconnect()
        }
      });

      this.mutationObserver.observe(this, {childList: true});
    }
  }
}

在反转列表时会陷入无限循环。这就是为什么它什么都不输出的原因。将元素2链接到元素3后，下一步将元素3链接到元素2。
在遍历列表的同时反转它会导致无限循环。

在一般情况下将其反转的正确方法如下：

class MyComponent extends HTMLBaseElement {
  constructor(...args) {
    const self = super(...args)
    return self
  }

  connectedCallback() {
    // when connectedCallback has fired, call super.setup()
    // which will determine when it is safe to call childrenAvailableCallback()
    super.setup()
  }

  childrenAvailableCallback() {
    // this is where you do your setup that relies on child access
    console.log(this.innerHTML)

    // when setup is done, make this information accessible to the element
    this.parsed = true
    // this is useful e.g. to only ever attach event listeners once
    // to child element nodes using this as a guard
  }
}

在旁注中，应使用else //general case { (*first).next = nil; //last line where when i put a cout << ""; it prints in the output window while ((*r).next != nil) { (*q).next = p; (*r).next = q; p = q; q = r; r = (*r).next; } (*q).next = p; first = q; }而不是else //general case { node_type* current = first; node_type* next = first->next; (*first).next = nil; //last line where when i put a cout << ""; it prints in the output window while (next) { node_type* temp = next->next; next->next = current; current = next; next = temp; } first = current; }取消引用。

特定行后未输出Visual Studio 2017代码

1 个答案: