Question

这是我的代码。一个简单的FILO链接列表，但我得到的错误很奇怪。它在Insert函数中我声明了temp，说Node是未声明的。不要理解我是如何完成这样的10次而且没有错误。谢谢你的时间。

//链接列表：在开头插入节点（FILO）

#include <stdio.h>
#include <stdlib.h>

struct Node{
    int data;
    struct Node* next;
};

struct Node* head; //Global


void Insert (int x)
{
    struct Node* temp = (Node*) malloc(sizeof(struct Node));
    temp->data = x;
    temp->next = head;
    head = temp;
}

void Print()
{
    struct Node* temp = head;  //We use a temporary because we dont want to lose the reference to the head
    printf ("List is: ");

    while(temp != NULL)
    {
        printf (" %d", temp->data);
        temp = temp -> next;
    }
    printf ("\n");
}


int main(){

    head = NULL; //empty list
    int n,i,x;

    printf ("How many numbers?\n");
    scanf ("%d",&n);

    for (i=0; i<n; i++)
    {
        printf ("Enter the Number\n");
        scanf ("%d",&x);
        Insert(x);
        Print();
    }
}

Answer 1

DO NOT CAST MALLOC RETURN

您的问题的解决方案只是为了避免转换malloc返回或使用正确的类型：struct node*

#include <stdio.h>
#include <stdlib.h>

struct Node{
    int data;
    struct Node* next;
};

struct Node* head; //Global


void Insert (int x)
{
    struct Node* temp = malloc(sizeof(struct Node));
    temp->data = x;
    temp->next = head;
    head = temp;
}

void Print()
{
    struct Node* temp = head;  //We use a temporary because we dont want to lose the reference to the head
    printf ("List is: ");

    while(temp != NULL)
    {
        printf (" %d", temp->data);
        temp = temp -> next;
    }
    printf ("\n");
}


int main(){

    head = NULL; //empty list
    int n,i,x;

    printf ("How many numbers?\n");
    scanf ("%d",&n);

    for (i=0; i<n; i++)
    {
        printf ("Enter the Number\n");
        scanf ("%d",&x);
        Insert(x);
        Print();
    }

    return 0;
}

简单链接列表（FILO）无法正常工作

1 个答案: