what is the difference between "." operator and "->" operator, in c++?

#include<iostream>

using namespace std;
class Science {
  int marks;
  public:
    void setMarks(int marks) {
      this -> marks = marks;
    }
  int getMarks() {
    return marks;
  }
};
void change(int * b) {
  * b = 20; //this works fine 
}
int main() {
  int x = 10;
  change( & x);
  cout << x << endl;
  Science z;
  Science * y = & z;
  z.setMarks(70);
  cout << z.getMarks() << endl;
  y -> setMarks(60); //this does not give error
  cout << z.getMarks() << endl;
  * y.setMarks(40); //why writing like this gives error
  cout << z.getMarks() << endl;
  return 0;
}

this code is written just to explain my question
I am also a bit confused in pointers,
as in the change function when I write *b=20 it is working as if writing x=20
so why writing *y.setMarks(40) not same as writing z.setMarks(40)

>Solution :

The . operator works on objects directly.

The -> operator accesses the object using a pointer; i.e. it dereferences the pointer first, and then uses the . operator.

for example, a -> b() is the same as (*a).b().