显示链表c++后停止

Stop after display linked list c++

本文关键字：c++ 链表 显示      更新时间：2023-10-16

我运行我的程序，当它运行完display(head)时，它将停止，不会执行最后一行cout << "Done";

任何人都可以帮我解决这个问题：D

这是我的代码：

#include <iostream>
#include <cstring>
#include <stdlib.h>
using namespace std;
struct Node{
    Node* next;
    char* data;
};
void initNode(struct Node *head, char *n){
    head->data = n;
    head->next = NULL;
}
void addNode(struct Node *head, char *n){
    Node *newNode = new Node;
    newNode->data = n;
    newNode->next = NULL;
    Node *cur = head;
    while(cur){
        if(cur->next == NULL){
            cur->next = newNode;
            return;
        }
        cur = cur->next;
    }
}
void display(struct Node *head){
    Node *list = head;
    while(list != NULL){
        cout << list->data <<endl;
        list = list->next;
    }
}
int main(){
    struct Node *head = new Node;
    char str[] = "-  This is a sample string";
    char * pch;
    pch = strtok (str," ");
    initNode(head, pch);
    while (pch != NULL){
        pch = strtok(NULL, " ");
        addNode(head, pch);
  }
  display(head);
  cout << "Done";
}

TonyD指出，对strtok（）的最后一次调用会给你空pch，你试图将它作为最后一个元素添加到链表中。这是一个简单的修复将使您的代码运行：

while (pch != NULL)
{
    pch = strtok(NULL, " ");
    if(pch!=NULL) // do not add empty data to your linked list
    {
        addNode(head, pch);
    }
}
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