Why do dynamically allocated pointer arrays not need dereferencing to get to their actual members

Question

So I recently came across the following piece of code:

struct Student
{
    int *number;
    char *name;
    double *marks;
};

int main(){
    int n;
    Student *s;
    s = new Student;
    cout << "Enter the number of subjects the student learns: ";
    cin >> n;
    s->number= new int;
    s->name=new char[20];
    s->marks=new double[n];
    cout << "Enter the name of the student: ";
    cin >> s->name;
    cout << "Enter the number in class of " << s->name << ": ";
    cin >> *(s->number);
    for (int i = 0 ; i < n ; i++){
        cout << "Enter mark No" << i+1 << " of the student: ";
        cin >> s->marks[i];
    }
}

When I worked with one dimensional arrays of pointers there was the need to use dereference twice. Once to reach the n-th pointer of the array ("s->marks[i]" in this case) and a second time to get the actual value that is pointed which I thought meant to write it like this:

*(s->marks[i])

This is apparently not needed, although I thought this would return the memory serial number contained by the pointer "marks[i]". On the other hand, there is the need to dereference the "number" pointer that is a single variable like that:

*(s->number)

This I perfectly understand.

Can someone please explain to me (or point me to a good article) why there is no need to use the dereference operator when working with dynamically allocated pointer arrays like, in this case, "marks". I am also confused with the use of the "name" char array pointer which is used rather like a normal char variable.

Thanks in advance for all the help.

Edit to sum up:
I just now realised that

/*the following two statements declare an array of pointers hence allow for the
use of double dereference*/
int* marks[n]; //static memory allocation
int** marks=new int*[n]; //dynamic memory allocation

/*the following two statements declare an array of variables hence allow for the
use of only one dereference (just a simple 1 dimensional array)*/
int marks[n]; //static memory allocation
int* marks=new int[n]; //dynamic memory allocation

I thought that I was dealing with something similar to the first pair of statements. In reality, I had to deal with the second pair.

Answer 1

s->marks is a pointer to the first element of an array of double.

s->marks + i is a pointer to the ith element.

*(s->marks + i) dereferences that to give the double element itself.

s->marks[i] is a convenient way to write *(s->marks + i). It includes the dereference operation, so there's no need for another one.

Answer 2

Let assume that you have

int x = 10;
int *p = &x;

then expression

*p = 20;

is equivalent to expression

p[0] = 20;

Relative to your code you could write

cin >> s->number[0];

instead of

cin >> *(s->number);

The same way you could write

cin >> *( s->marks + i );

instead of

cin >> s->marks[i];

According to the C++ Standard

The expression E1[E2] is identical (by definition) to *((E1)+(E2))

Again returning to your code expression

cin >> *(s->number);

can be written as

cin >> *(s->number + 0);

that is

cin >> s->number[0];