Question

alright, im looking at a code here and the idea is difficult to understand.

#include <iostream>
using namespace std;
class Point
{
public :
    int X,Y;
    Point() : X(0), Y(0) {}
};

void MoveUp (Point * p)
{
    p -> Y += 5;
}

int main()
{
    Point point;
    MoveUp(&point);
    cout << point.X << point.Y;
    return 0;
}

Alright, so i believe that a class is created and X and Y are declared and they are put inside a constructor

a method is created and the argument is Point * p, which means that we are going to stick the constructor’s pointer inside the function;

now we create an object called point then call our method and put the pointers address inside it?

isnt the pointers address just a memory number like 0x255255?

and why wasnt p ever declared?

(int * p = Y)

what is a memory addres exactly? that it can be used as an argument?

Answer 1

p was declared.

void MoveUp (Point * p)
{
    p -> Y += 5;
}

is a function that will take a pointer to a Point and add 5 to its Y value. It’s no different to the following:

void f(int n) {
    printf ("%d\n", n);
}
:
int x = 7;
f(x);

You wouldn’t say n wasn’t defined in that case. It’s the same for p in your case.

Perhaps some comments in the code would help:

#include <iostream>
using namespace std;
class Point
{

public :
    int X,Y;
    Point() : X(0), Y(0) {}       // Constructor sets X and Y to 0.
};

void MoveUp (Point * p)           // Take a Point pointer p.
{
    p -> Y += 5;                  // Add 5 to its Y value.
}
int main()
{
    Point point;                  // Define a Point.
    MoveUp(&point);               // Call MoveUp with its address.
    cout <<point.X << point.Y;    // Print out its values (0,5).
    return 0;
}

Pointers are simply a level of indirection. In the code:

1   int X;
2   int *pX = &X;
3   X = 7;
4   *pX = 7;

the effect of lines 3 and 4 are identical. That’s because pX is a pointer to X so that *pX, the contents of pX, is actually X.

In your case, p->Y is the same as (*p).Y, or the Y member of the class pointed to by p.