【C语言】单向链表反转，运行结果不对

刚刚开始学习数据结构，自己尝试性地写了一个链表反转的代码，编译似乎没有问题，但运行后没有显示任何数字。
想了半天找不到原因。
请问有哪位能帮忙看看问题出在哪里吗？
十分感谢！

#include <stdio.h>
#include <stdlib.h>
struct Node{
    int data;
    struct Node* link;
};
struct Node* head = NULL;
void Print()
{
    struct Node* temp = head;
    while (temp != NULL)
    {
        printf(" %d", temp->data);
        temp= temp->link;
    }
    printf("\n");
}
int Insert(int n)
{
    struct Node* temp = (struct Node*)malloc(sizeof(struct Node*));
    temp = head;
    
    while(temp->link != NULL)
    {
        temp = temp->link;
    }
    struct Node* temp2 = (struct Node*)malloc(sizeof(struct Node*));
    temp2->data = n;
    temp2->link = NULL;
    temp->link = temp2; 
    if(head == NULL)
    {
        head = temp;
    }


}
int Reverse()
{
    struct Node* current, *prev , *next;
    current = head;
    prev = NULL;
    while (current != NULL)
    {
        next = next->link;
        current->link = prev;
        prev = current;
        current = next;
    }
    head = prev;
    
}
int main()
{
    Insert(2);
    Insert(4);
    Insert(6);
    Insert(8);
    Print();
    Reverse();
    Print(); 
}

#include <stdio.h>
#include <stdlib.h>
struct Node
{
    int data;
    struct Node *link;
};
struct Node *head = NULL;

void Print()
{
    struct Node *temp = head->link; //struct Node *temp = head;
    while (temp != NULL)
    {
        printf(" %d", temp->data);
        temp = temp->link;
    }
    printf("\n");
}
void Insert(int n)
{
    //struct Node *temp = (struct Node *)malloc(sizeof(struct Node *));
    //temp = head;
    // while (temp->link != NULL)
    // {
    //     temp = temp->link;
    // }
    struct Node *temp;
    temp = head;
    while (temp->link != NULL)
    {
        temp = temp->link;
    }
    struct Node *temp2 = (struct Node *)malloc(sizeof(struct Node *));
    temp2->data = n;
    temp2->link = NULL;
    temp->link = temp2;
    // if (head == NULL)
    // {
    //     head = temp;
    // }
}

void Reverse()
{
    struct Node *current, *prev, *next;
    next = current = head->link; //current = head;
    prev = NULL;
    while (current != NULL)
    {
        next = next->link;
        current->link = prev;
        prev = current;
        current = next;
    }
    head->link = prev; //head = prev;
}
int main()
{
    head = (struct Node *)calloc(1, sizeof(struct Node *));
    Insert(2);
    Insert(4);
    Insert(6);
    Insert(8);
    Print();
    Reverse();
    Print();
    return 0;
}

next都没有赋值，怎么能next = next->link呢

int Reverse()
{
    struct Node* current,  *next;
    current = head;
    if(current == NULL)
        return 0;
    next = current->link;
    current->link = NULL;
    while (next != NULL)
    {
        head = next;
        head->link = current;
        current = next;
        next = next->link;
    }
}

c语言单向链表反转6,单链表反转的问题怎么解决（C语言）_weixin_39602005的博客-CSDN博客 匿名用户1级2015-02-07 回答单链表反转：比如原链表为 head->1->2->3->NULL;反转后：head->3->2->1->NULL;实现代码：#include#includetypedefstructNode{intdata;structNode*next;}*List;#definenodestructNod... https://blog.csdn.net/weixin_39602005/article/details/117137563
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C语言 单链表的反转_Genven_Liang的博客-CSDN博客_c语言反转链表 C语言 单链表的反转一、简述 记--简单的将单链表的数据顺序进行反转。如将原来的顺序1 2 3 4 5 6 7 反转为：7 6 5 4 3 2 1二、方式1：头插法 2.1 头插法1--类似新建链表 2.1.1 思路：断开链表头，然后以头插法的方式将原链表的数据添加链表。 2.1.2 测... https://blog.csdn.net/nanfeibuyi/article/details/90116419
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