c - 类型转换，结构与转换

Question

我该如何转换这些Student1和Student2变量并访问其成员
可变的？我在student1和student 2变量中有转换错误。

typedef struct {
    int id;
    int hw_score[3];
    int final;
} Student;

int  intComparator(void* p, void* q) {
     return ((int)p - (int)q);
}

int   idComparator(void *student1, void *student2){

      int result;
      Student* s1, s2;

      s1 = (Student*)student1;
      s2 = (Student*)student2;

      result = intComparator((void*)(s1->id), (void*)(s2->id));

      if (result > 0) {return 1;}
      else if (result < 0) {return -1;}
      else {return 0;}

}


int main()
{
    int result;
    Student s1, s2;
    s1.id = 8;
    s2.id = 10;

    result = idComparator((void*)&s1, (void*)&s2);
    printf("result: %d\n", result);

    return 0;
}

我很难理解void *。

Answer 1

除了该策略过于复杂以至于无法执行简单的减法运算外，您的代码中还存在一些错误：

#include <stdio.h>       // a

typedef struct {
    int id;
    int hw_score[3];
    int final;
} Student;

int  intComparator(void* p, void* q) {
     return (*(int*)p - *(int*)q);   // b
}

int   idComparator(void *student1, void *student2){

      int result;
      Student* s1, *s2;      // c

      s1 = (Student*)student1;
      s2 = (Student*)student2;

      result = intComparator(&(s1->id), &(s2->id)); // d

      if (result > 0) {return 1;}
      else if (result < 0) {return -1;}
      else {return 0;}

}


int main()
{
    int result;
    Student s1, s2;
    s1.id = 8;
    s2.id = 10;

    result = idComparator(&s1, &s2);
    printf("result: %d\n", result);

    return 0;
}

a）printf需要stdio.h

b）* void需要取消引用的类型： Dereference void pointer

c）s2不是指针： Declaring pointers; asterisk on the left or right of the space between the type and name?

d）您不必强制转换为void，但是该函数需要一个地址： Why type cast a void pointer?

这是 live demo，但再次考虑改进您的设计，使其更清晰，更简单。