c - 结构和指针运算

Question

#include <stdio.h>

struct student
{
  int roll_no;
  int total_marks;
} s1;

int main()
{
  struct student *stu;
  stu = &s1;
  printf("Address of member roll_no accessed by s1 = %u", &s1.roll_no);
  printf("Address of member total_marks accessed by s1 = %u", &s1.total_marks);
  printf("Value of stu = %u", stu); // which is address of s1.roll_no

同样，如何使用 stu 获取 s1.total_marks 的地址？

完全想通了。假设我们有这样的结构:这些成员中的每一个都可以通过以下方式访问，

   #include <stddef.h>   

   struct student    
   {   
      int roll_no;   
      char name[50];   
      double mark;   
      double percetage;   
   };   

   int main()   
   {   
   struct student s = {1, "Jug", 98.99, 97.5678};   
    
       printf("%d %s\n", *(int*)((char*)&s + offsetof(struct student,    
   roll_no)), (char*)((char*)&s + offsetof(struct student, name)));      
       printf("%lf %lf", *(double*)((char*)&s + offsetof(struct student,  
  mark)), *(double*)((char*)&s + offsetof(struct student, percentage)));        
   }```

Answer 1

成员访问 运算符(.、->)具有更高的 precedence比 address-of 运算符 (&)。所以 &p->m 不等同于 (&p)->m，而是 &(p->m) 。

因此，您可以使用

&stu->total_marks

以及你已经写过的

&s1.total_marks