The difference between *s.pd and s->pd

I was reading about copy constructors for structs and i found this example:

#include <iostream>
#include <string>
using namespace std;

struct SomeData {
    int * pd;
    string id;
    SomeData(SomeData & ref) {
        cout << "Copy Constructor called" << endl;
        pd = new int (*ref.pd);
        id = "Copy Constructed";
    }
    SomeData(string name) {
        pd = new int(0);
        id = name;
        cout << "Constructor for " << id << endl;
    };
    ~SomeData() {
        cout << "Destructor for " << id << endl;
        delete pd;
    }
};

int main() {
    SomeData s("First");
    *s.pd = 9;
    SomeData s2=s;
    cout << *s2.pd << endl;
    return 0;
}

in the main, the member pd of SomeData is accessed using the dereference, but why is that, isn't the correct way is

s->pd=9;

why was it written like that in the example?