打印链栈元素之后，程序异常退出。

Question

下面是我做的链栈的基本内容，测试发现打印栈的功能出现了问题，循环体结束之后，程序会异常退出，但是没有找到原因，打印链表的函数我之前一直用在其他文件里面，没出现过这样的问题，但是复制到这里来就用不了了。vs没有报warning或者error。

```c

#include "stdio.h"
#include "stdlib.h"
typedef struct node
{
    int data;
    struct node* next;
}node;

int push(node* s,int x)
{
    if (s) 
    {
        node* p;
        p = (node*)malloc(sizeof(node));
        p->data = x;
        if (s->next == NULL)
            s->next = p;
        else
        {
            p->next = s->next;
            s->next = p;
        }
        return 1;
    }
    return 0;
}

int pop(node* s)
{
    if (s->next == NULL)
    {
        printf("空栈\n");
        return 0;
    }
    node* p;
    p = s->next;
    s->next = s->next->next;
    free(p);
    return 1;
}

int  displist(node* l)
{
    node* p;
    if (l)
    {
        p = l;
        while (p->next != NULL)
        {
            p = p->next;
            printf("%d\t", p->data);
        }
        printf("\n");
        return 1;
    }
    return 0;
}

int init(node* s)
{
    s->next = NULL;
    return 1;
}

int gettop(node* s)
{
    return s->next->data;
}

int main()
{
    node* s;
    int y, e, status = 0;
    exe:
    printf("菜单：\n1.栈初始化\t2.入栈\n3.出栈\t4.取栈顶元素\n5.打印栈\n");
    scanf("%d", &y);
    switch (y)
    {
    case 1:status = init(&s); break;
    case 2:printf("请输入入栈元素:"); scanf("%d", &e); status = push(&s, e); break;
    case 3:status = pop(&s); break;
    case 4:e = gettop(&s); printf("%d\n", e); break;
    case 5:status = displist(&s); break;
    default:return 0;
        break;
    }
    if (status == 1)
        printf("exe succeeded\n");
    else printf("exe failed\n");
    goto exe;
    return 0;
}

![img](https://img-mid.csdnimg.cn/release/static/image/mid/ask/225170679846173.png "#left")

Answer 1

push函数有问题

int push(node* s,int x)
{
    if (s)
    {
        node* p;
        p = (node*)malloc(sizeof(node));
        p->data = x;
        if (s->next == NULL)
        {
            s->next = p;
            s->next->next=NULL;//加了这句 
        }
        else
        {
            p->next = s->next;
            s->next = p;
        }
        return 1;
    }
    return 0;
}

修改后可以运行