(->) arrow operator and (.) dot operator , class pointer

Question

In C++ we know that for a pointer of class we use (->) arrow operator to access the members of that class like here:

#include <iostream>
using namespace std;

class myclass{
    private:
        int a,b;
    public:
        void setdata(int i,int j){
            a=i;
            b=j;
        }
};

int main() {
    myclass *p;
    p = new myclass;
    p->setdata(5,6);
    return 0;
}

Then I create an array of myclass.

p=new myclass[10];

when I go to access myclass members through (->) arrow operator, I get the following error:

base operand of '->' has non-pointer type 'myclass'

but while I access class members through (.) operator, it works. These things make me confused. Why do I have to use the (.) operator for an array of class?

Answer 1

you should read about difference between pointers and reference that might help you understand your problem.

In short, the difference is:
when you declare myclass *p it's a pointer and you can access it's members with ->, because p points to memory location.

But as soon as you call p=new myclass[10]; p starts to point to array and when you call p[n] you get a reference, which members must be accessed using ..
But if you use p->member = smth that would be the same as if you called p[0].member = smth, because number in [] is an offset from p to where search for the next array member, for example (p + 5)->member = smth would be same as p[5].member = smth