函数返回值类型为指针，出现段错误。

Question

struct student{
int age;
int sex;
};
int main(){
struct student *p;
p->sex=1;
p->age=20;
struct student *q;
q=p;
/printf("age:%d\n",p->age);
printf("sex:%d\n",p->sex);/
printf("age:%d\n",q->age);
printf("sex:%d\n",q->sex);
return 0;
}

为何我这样输出printf("age:%d\n",p->age);
printf("sex:%d\n",p->sex);提示段错误，而我再申明一个指针来指向他，这样输出才是对的呢？

Answer 1

修改如下，供参考：

#include<stdio.h>
struct student{
       int age;
       int sex;
};
int main()
{
     struct student *p;//p 这里只是一个指针
     p = (struct student *)malloc(sizeof(struct student));
     p->sex=1;
     p->age=20;
     struct student *q;
     q=p;
     printf("age:%d\n",p->age);
     printf("sex:%d\n",p->sex);
     printf("age:%d\n",q->age);
     printf("sex:%d\n",q->sex);
     
     return 0;
}