搭建链表，然后输出出来，出错了

这个是什么问题啊，可以来帮帮我嘛，我搭建链表，尾插法搭建的，每一scanf输入，然后for循环输出出来，但是出错了

创建函数修改为：loand* Createlist2(loand *l,int n)
{
...........
return l;
}
主函数中调用：
int main()
{
loand* l;
l = Createlist2(l, c);
.........
}
修改如下，供参考：

#include <stdio.h>
#include <malloc.h>
typedef struct loand {
    int data;
    struct loand* next;
}Loand;

Loand * Createlist2(Loand * l, int n)
{
    l = (Loand*)malloc(sizeof(loand));
    l->next = NULL;
    Loand* r = l;
    for (int i = 0; i < n; i++)
    {
        Loand* s = (Loand*)malloc(sizeof(Loand));
        s->next = NULL;
        printf("请输入一个数:");
        scanf("%d", &s->data);
        r->next = s;
        r = s;
    }
    return l;
}
int main()
{
    loand* l;
    int c = 5, d;
    l = Createlist2(l, c);
    Loand* p = l->next;
    while (p)
    {
        printf("%d ", p->data);
        p = p->next;
    }
    return 0;
}

这里的知识点是： 函数传参，拷贝传参。（你可以百度了解）
如果你要在函数内部申请内存或者改变变量，通过参数的方式使用，要么使用楼上说的返回值赋值的方式，要么参考我的传地址，而不是传值（拷贝了）

struct Data {
    int val;
    struct Data* next;
};

void createdata(struct Data* list, int n)
{
    list = (struct Data*)malloc(sizeof(struct Data*));
    if (list == NULL)
    {
        printf("error malloc \n");
        return;
    }
    list->next = NULL;
    list->val = 1;
}

void createdata1(struct Data** list, int n)
{
    *list = (struct Data*)malloc(sizeof(struct Data*));
    if (*list == NULL)
    {
        printf("error malloc \n");
        return;
    }
    (*list)->next = NULL;
    (*list)->val = 1;
}
int main(void)
{  
    struct Data* list = NULL;
    createdata(list, 1);

    if (list != NULL)
    {
        printf("%d \n", list->val);
    }
    else
    {
        printf("list is null \n");
    }

    createdata1(&list, 1);
    if (list != NULL)
    {
        printf("%d \n", list->val);
    }
    else
    {
        printf("list is null \n");
    }
    return 0;
}