- arrow operator and dot operator class pointer

Question

In C++, we know that given a class pointer, we use the (->) arrow operator to access the members of that class, as seen 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'

However, when I access class members using the (.) operator, it works. 

These are the things that perplex me. Why do I have to use the (.) operator for a class array?

Answer 1

You should learn about the distinction between pointers and references since it can clarify your issue.

The key distinction is that, when you declare myclass *p, it is a pointer and you may use > to access any of its members since p

points to a memory address.

However, after you use p=new myclass[10]; p begins to point to an array, and when you call p[n], you obtain a reference, whose members must be accessed using.

However, using p->member = smth would have the same effect as calling p[0].

For example, (p + 5)->member = smth would be the same as p[5] since the number in [] represents an offset from p to where search for the next array member is performed. smth = member