Thread: dot operator vs arrow.

Hello everyone. I'm sure this has been asked multiple times and googling I've found plenty of explanations. However, some things just don't come to me unless it's explained with my specific example. I understand that the dot operator is:

Code:

a.b

"a" is the actual object (not a memory location) and "b" is the member. I also understand the arrow to mean:

Code:

a->b

"a" is a pointer to a struct and "b" is it's member so "a" is dereferenced then "b" is given.

Here is where I get a little confused. I have a class:

exampleclass.h

Code:

#ifndef EXAMPLECLASS_H_INCLUDED
#define EXAMPLECLASS_H_INCLUDED

class exampleclass{

    private:
        int number;

    public:
        exampleclass();  //Constructor
        ~exampleclass(); //Destructor
        void changeNumber(int number);
        int getNumber();


};

#endif // EXAMPLECLASS_H_INCLUDED

exampleclass.cpp

Code:

#include "exampleclass.h"

exampleclass::exampleclass(){
        this.number = 0;
}

exampleclass::~exampleclass(){
         
}

void exampleclass::changeNumber(int number){
       this.number = number;
}

int exampleclass::getNumber(){
       return this.number;
}

and last I have my main client code.

Code:

#include <iostream>
#include "exampleclass.h"

int main(){
       exampleclass ec1 = new exampleclass;
       //This is where I'm confused. Is ec1 my object
       //or is ec1 a pointer to the address of my object
      ec1.changeNumber(5); //Is this correct
      ec1->changeNumber(9); //or this?

}

Also if ec1 had a public variable and I wanted to access it would it be referenced the same way as the method I call in ec1?

Thank you very much. One more question. If I were to write my class in the same source file as my client code like many examples would it still be instantiated like so:

Code:

exampleclass* ec1 = new exampleclass;

also is ec1 always a pointer to the object because it is created on the HEAP sorta llike malloc?
Thank you once again.