C++ 链表无法删除头节点

Question

做题时发现单链表删除函数调用后无法删除头结点，其他节点可正常删除。

#include<iostream>
using namespace std;
struct Student {
    int sno;
    char sname[20];
    int ssocre;
    Student* next;
};
int n;
Student* creat() {//创建链表
    Student* head, * p1, * p2;
    head = NULL;
    p1 = p2 = new Student;
    n = 0;
    cin >> p1->sno >> p1->sname >> p1->ssocre;
    while (p1->sno != -1) {
        n = n + 1;
        if (n == 1)head = p1;
        else p2->next = p1;
        p2 = p1;
        p1 = new Student;
        cin >> p1->sno >> p1->sname >> p1->ssocre;
    }
    p2->next = NULL;
    return head;
}
void Stuprint(Student* head) {//输出链表
    Student* p;
    p = head;
    while (p != NULL) {
        cout << p->sno << " " << p->sname << " " << p->ssocre << endl;
        p = p->next;
    }
}
Student* Sinsert(Student* head) {//插入函数（由于不排序，将插入链表放在最后一个结点）
    Student* p0 = NULL, * p1, * p2, * stu;
    p1 = head;
    stu = new Student;
    p0 = stu;
    while (p1->next != NULL) {
        p2 = p1;
        p1 = p1->next;
    }
    p1->next = p0; p0->next = NULL;
    cout << "请输入要插入学生的信息" << endl;
    cin >> p0->sno >> p0->sname >> p0->ssocre;

    n = n + 1;
    return head;

}
struct Student* del(struct Student* head, int num)
{
    struct Student* p1, * p2 = NULL;
    if (head == NULL)
    {
        cout << "list null\n"; return NULL;
    }
    p1 = head;
    while (num != p1->sno && p1->next != NULL)
    {
        p2 = p1; p1 = p1->next;
    }
    if (num == p1->sno)
    {
        if (num == head->sno)head = p1->next;
        else p2->next = p1->next;
        n--;
    }
    else cout << "无此人\n";
    return head;
}

int main() {
    Student* head;
    cout << "请输入需要录入的学生信息(学号，姓名，成绩，全部输入-1时停止录入)" << endl;
    head = creat();
    cout << "***************************************************" << endl;
    Stuprint(head);
    int ans = -1, sno1 = 0;
    while (ans) {
        cout << "***************************************************" << endl;
        cout << "请输入您需要使用的功能（1.删除 2.插入 0.退出程序）" << endl;

        cin >> ans;
        switch (ans) {
        case 1:cout << "请输入需要删除学生的学号" << endl; cin >> sno1; del(head, sno1);
            cout << "***************************************************" << endl;
            Stuprint(head); break;
        case 2:Sinsert(head);   cout << "***************************************************" << endl;
            Stuprint(head); break;
        }
    }
    return 0;

}

运行结果及报错内容 可正常运行

我的解答思路和尝试过的方法

和模板进行了对比，也问了学长和同学

我想要达到的结果

删除函数可以删除所选节点

Answer 1

del函数中删除头结点时修改了del函数中head指针指向的地址 , 你要接收del函数的返回值重新赋值给主函数的head
del(head, sno1) 前面加上 head =

        switch (ans) {
        case 1:cout << "请输入需要删除学生的学号" << endl; cin >> sno1;
            head = del(head, sno1); //前面加上 head =
            cout << "***************************************************" << endl;
            Stuprint(head); break;
        case 2:Sinsert(head);   cout << "***************************************************" << endl;
            Stuprint(head); break;
        }
 

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