Having issues understanding pointers in c++ - c++

Question

I am currently working in QT and recently I noticed something that really confused me.
As far as I can tell normally when we want to create a pointer we have to use the following syntax in C++:
int number = 10;
int* pNumber = &number;
(or something similar to that)
I wanted to create a pointer to a button which was created in QT design. It was for testing purposes only. (I am new to QT and c++ so I wanted to test things out)
But then I noticed something strange that I could not understand. for some reason when I created the pointer of type "OPushButton" with the name of "button" I did not have to use the "&" with the "(*ui).pushButton_5" syntax. (pushButton_5 is the name of my button in my ui)
The code works and the text "5" is added to my "lineEdit" in QT. How does this work? Am I missing something about pointers?
Here is my code:
mainwindow.cpp:
#include "mainwindow.h"
#include "ui_mainwindow.h"
MainWindow::MainWindow(QWidget *parent)
: QMainWindow(parent)
, ui(new Ui::MainWindow)
{
ui->setupUi(this);
QPushButton* button = (*ui).pushButton_5;
ui->lineEdit->setText((*button).text());
}
MainWindow::~MainWindow()
{
delete ui;
}
mainwindow.h:
#define MAINWINDOW_H
#include <QMainWindow>
QT_BEGIN_NAMESPACE
namespace Ui { class MainWindow; }
QT_END_NAMESPACE
class MainWindow : public QMainWindow
{
Q_OBJECT
public:
MainWindow(QWidget *parent = nullptr);
~MainWindow();
private:
Ui::MainWindow *ui;
};
#endif // MAINWINDOW_H

Answer 1

& is not the way to create pointers, it's the way to acquire a pointer to a specific thing that you have access to.
If somebody else tells you where a thing is, you don't need the help of & to find out.
void g(int* q)
{
int* p = q; // 'q' is the location of some unknown 'int', and so is `p`.
}
You need & if you have a thing and want to know where that thing is.
void f()
{
int x = 5;
int* p = &x; // The location of 'x'.
g(&x); // Pass the location of 'x' to 'g'.
}
Also, we usually write x->y rather than (*x).y.
This convention makes a lot of sense if you look at more than one level of indirection – compare x->y->z->w to (*(*(*x).y).z).w.