使用指针构造读取成员变量

Question

#include<stdio.h>

struct data{
    int i;
    struct data *p;
};

int main() {
    struct data *p=malloc(sizeof(struct data));

    //How do i use pointer to structure to read a integer in member variable i?

    scanf("%d",&p->i);    // I am advised to use this,Can you interpret this??
    scanf("%d",&(*p).i);  // Is this valid?
    scanf("%d",p->i);     // Why is this not valid since p is nothing but a pointer 
}

1. Interpret this &p->i . Why this represents the address of the member variable i?

2. Is this scanf("%d",&(*p).i); valid? Why?

Answer 1

In your case

• &p->i is the same as &(p->i) because of the operator precedence .
• &(*p).i is the same as &(p->i) .

and they both produce a pointer to an integer, as required by the scanf() function argument based on the supplied conversion specifier.

However,

 scanf("%d",p->i);

is not valid, as p->i gives you an int , whereas, you need a pointer to integer.

Answer 2

scanf expects a pointer to something, in order to store data according to the format you provide to the function.

scanf("%d",&p->i); // I am advised to use this,Can you interpret this??

p->i gives you the integer i from the structure pointed to by p .
&p->i gives the address of i , required by scanf.

scanf("%d",&(*p).i);  //Is this valid?

Yes, that's the same as above. (*p).i is p->i

scanf("%d",p->i);  //Why is this not valid since p is nothing but a pointer 

scanf needs a pointer to store a "%d", meaning an integer ; though, here you give the value of i , not a pointer to i .