C编程语言中的指针混淆

Question

Hello guys and a happy new year!

I have difficulty understanding pointers in C language. As far as I learned a pointer is a special variable that stores addresses of regular variables.

Ι am posting two samples of code which are equivalent. In the first I typed into scanf the &d1.am .

In the second sample if I change the &d1.am to ptd1.am it pops up a compile error and I cannot understand why.

struct student{
    int am;
    char stname[20];
    char stsurname[20];
};

int main(){

struct student d1;
printf("1st student\n");
printf("Enter am\n");
scanf("%d", &d1.am)

Second equivalent sample:

struct student{
    int am;
    char stname[20];
    char stsurname[20];
};

int main(){

struct student d1;
struct student *ptd1;
ptd1=&d1;
printf("1st student\n");
printf("Enter am\n");
scanf("%d", &(*ptd1).am);

I know the correct is to type &(*ptd1).am instead but I can't figure why. How &(*ptd1).am is equal to &d1.am and ptd1.am is not? I typed clearly that ptd1=&d1 !

Thanks in advance for your help!

Answer 1

. operator has higher precedence than unary & . &d1.am is equivalent to &(d1.am) while ptd1.am is equivalent to (&d1).am , that said &d1.am != (&d1).am .

Answer 2

The structure access operator ( . ) has higher precedence than the address-of operator ( & ). So &d1.am is the address of the am member of d1 , which is an int* , which differs from the type of ptd1 ( struct student * ).

Answer 3

Has other have said the . operator has higher precedence than & . So that ptd1.am is equivalent to (&d1).am (see for example the answer from @haccks).

I want to add that ptd1 is a pointer to struct and thus to get a member you should use -> . The call to scanf should be:

scanf("%d", &(ptd1->am));

Answer 4

As u have written ptd1=&d1; that means that *ptd1 will point to d1 .

So if you want to access d1.am you need to write (*ptd1).am .

But here you are passing to scanf and it needs memory address so need to pass with & and so you need to put &(*ptd1).am.

As others have mentioned (.) has higher priority so (.) will choose it operands first and expression will be equal to
&((*ptd1).am)

Better to use &(ptd1->am)