填充链表的分段错误

Question

我已经在这工作了 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

// 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

ma​​in.cpp

// This program demonstrates the SimpleVector template.
#include <iostream>
#include "SimpleVector.h"
using namespace std;


int main(){
int SIZE = 10; // Number of elements

// Create a SimpleVector of ints.
SimpleVector intTable;

intTable.linkList(SIZE);

intTable.displayList();

return 0;
}

Answer 1

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

假设第一个节点包含 10，然后结束包含 9（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 表示10 -> 8 前 9 个丢失。 . . .

最后会变成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变量指向前一个节点，然后在previous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。它会删除整个列表