Question

我已经在这里工作了4.5个小时试图弄清楚为什么这不起作用。仍然没有运气。我不断收到分段错误，或者尽管构建成功，但列表永远不会显示。

SimpleVector.h

stop-and-wait

的main.cpp

// SimpleVector class template
#ifndef SIMPLEVECTOR_H
#define SIMPLEVECTOR_H
#include <iostream>
#include <iomanip>
using namespace std;

class SimpleVector
{
private:
    struct Link{
        int data;
        Link *next;
    };
   Link *head;
public:
   // Default constructor
   SimpleVector()
      {head = NULL;}

   // Destructor declaration
   ~SimpleVector();

   void linkList(int);

   //void insertLink(T);

   void displayList();
};

//Destructor for SimpleVector

SimpleVector::~SimpleVector(){
    Link *linkPtr;
    Link *nextPtr;

    nextPtr = head;

    while(linkPtr !=  NULL){
        nextPtr = linkPtr->next;
    }

    delete linkPtr;

    linkPtr = nextPtr;
}

//Creation of List

void SimpleVector::linkList(int size){    
    Link *newLink = new Link; //create first link
    head = newLink; //

    head->data = size--;         //Fill the front with data
    head->next = NULL;     //Point the front to no where

    do{
        Link *end = new Link;   //Create a new link
        end->data = size--;     //Fill with data
        end->next = NULL;       //Point to no where
        head->next = end;       //Previous link will point to the end
       // head = end;             //Move to the end
    }while(size > 0);          //Repeat until filled
}

//Creation of Link and insertion
/*
template <class T>
void SimpleVector<T>::insertLink(T){

}
*/

//Function to print the entire list

void SimpleVector::displayList(){
    Link *linkPtr;

    linkPtr = head;

    while(linkPtr != NULL){
        cout<<setprecision(3)<<linkPtr->data;
        linkPtr = linkPtr->next;
    }
}

#endif

Answer 1

你在linkList()函数中做错了。这一行head->next = end;

让我们说第一个节点包含10然后结束包含9个（新节点）
现在head->next = end

表示10 -> 9

现在新节点end变为8

再次head->next = end 意味着10 -> 8 之前的9人输了。。。

最后它将成为10 -> 1

试试这个。

void SimpleVector::linkList(int size){    
    Link *newLink = new Link; //create first link
    head = newLink; //

    head->data = size--;         //Fill the front with data
    head->next = NULL;     //Point the front to no where
    Link *temp = head;
    do{
        Link *end = new Link;   //Create a new link
        end->data = size--;     //Fill with data
        end->next = NULL;       //Point to no where
        temp->next=end;//Previous link will point to the end
        temp=end;  //Now this has become previous link     
       // head = end;             //Move to the end
    }while(size > 0);          //Repeat until filled
}

它使用temp变量指向上一个节点，然后在之前和end（新创建的节点）之间创建链接，然后temp变为end。

编辑：

Link *linkPtr=head;
    Link *nextPtr;// = linkPtr->next;
    do
    {
        nextPtr = linkPtr->next;
        cout<<linkPtr->data<<"\t";
        delete linkPtr;
        linkPtr = nextPtr;
    }while(linkPtr!=NULL);

试试这个。它删除整个列表

填充链表中的分段错误

1 个答案: