Question

I know this is already a pointer, but I want to understand what this is exactly. Is it an address or is it a * type pointer?

If I were to store this into a variable how would I have to define that variable?

Lets take this as an example

class Adult
{
private:
 Child child;

public:
 Adult(){
  child = new Child(this);
  //to have something to hold onto and get back to the upper level of the hierarchy.
 }
};


class Child
{
private:
 Adult* my_adult;

public:
 Child();
 Child(Adult &hand){
  my_adult = hand;
 }
}

Okay, so where I run into the trouble is with the line of code that goes my_adult = hand;
It outputs this when I try to build the project, I will be horribly shocked if there are more behind this one.

sys/chin.cpp:19:14: error: cannot convert ‘Adult’ to ‘Adult*’ in assignment

So how does this work when being dealt with as a data type opposed to accessing members?

Answer 1

my_adult is a pointer, hand is not, pure and simple.

You either need:

class Child
{
private:
 Adult my_adult; //no pointer

public:
 Child();
 Child(Adult &hand){
  my_adult = hand;
 }
};

or

class Child
{
private:
 Adult* my_adult;

public:
 Child();
 Child(Adult* hand){ //pointer parameter
  my_adult = hand;
 }
};

Your first example is also illegal:

child = new Child(this);

shouldn’t compile, since child is an object, not a pointer.

child = Child(this);

would be the correct version.