Question

我做了一个简单的链表实现，当我在Main（）的Insert（）函数调用中运行代码时，我收到此错误，我不明白为什么这样。

#include<stdio.h>
#include<stdlib.h>
  struct Node
  {
     int data;
     struct Node *next;
  };
  struct Node* head;
 int main(){
    head=NULL;
    Insert(1,2);
    Insert(2,1);
    Insert(3,1);
    Insert(4,1);        
 } 
 void Print(){
    struct Node* temp;
    temp=head;
    while(temp!=NULL){
        printf("%d ",tenp->data);
        temp=temp->next;
     }
    printf("\n");
 }
 Node* Insert(Node *head,int data)
 {
    Node *node = new Node();
    node->data = data;              // ASSIGNING VALUES TO NEW NODE
    node->next = NULL;                 // ALSO AS TO ACT AS THE TAIL NODE ADDING THE null VALUE AT THE NEXT POINTER 
    if(head == NULL)               // CASE: WHEN LIST IS EMPTY
        return node;
    Node *temp = head;       // dereferencing value for TEMP,  assignning it the HEAD values, HEAD is the current tail node
    while(temp->next != NULL)         // Traversing till 1 before
        temp = temp->next;             
    temp->next = node;                      // making the last move, assigning the LINK to new node.  
    return head;                                // returinng the funn call with VALUE
}

Answer 1

 Insert(1,2)//this won't work as insert() 
            //expects node* type and not an integer as first argument.

您应该将指针传递给head而不是值。

Answer 2

您正在将1,2,3和4传递给Node *参数。您必须传递实际的头节点并分配结果

head = Insert( head, value );

这就是你如何使用这个功能。

以下代码片段中的错误在哪里，我已经创建了一个链表实现，并且我在LinkedList的尾部添加了元素

2 个答案: