指向结构的指针

Question

声明为原子数据类型(如int)时的指针就像这样工作

int a,*b=&a;

printf("\n The address of pointer b = %p",&b); //Here using & operator we get the memory location where the pointer b is stored itself

printf("\n The pointer b points to %p this memory location",b); //The will give me the value of ptr and that value is the memory address of the variable a

printf("\n The value at the memory where pointer b points is %d",*b);//Using * operator we get the value stored at the memory location hold by b

但是当我们使用指向结构的指针时会有一些混乱

#include<stdio.h>
struct A{
    int age;
    int roll_no;
    float marks;
};
int main(void)
{
    struct A obj1;
    struct A *ptr;
    printf("\n The addrees of the obj1 is =%p",&obj1);
    printf("\n The address of the variable age is %p ",&obj1.age);

    ptr=&obj1;
    printf("\n THe pointer ptr points to %p ",ptr); //This will give me the memory location where pointer ptr is pointing to.

    printf("\n The memory address of pointer ptr itself is %p ",&ptr); //This will give the memory location where the pointer ptr is itself store. So far So good

    printf("\n The memory location of variable age is %p",&ptr->age); //Why I have to use this & operator to find the address of the age and we also do not use * opertaor here I guess 

/* Should not ptr->age give me the memory address and *ptr->age give me the value ? */


    return 0;
} 

我对这里运算符的使用感到困惑

Answer 1

语法ptr->age是的缩写(*ptr).age.