QTimer won't start when passing a variable - c++

Question

I have a QTimer object that kicks off a function every second. The timer runs correctly if I do the following
MainWindow::MainWindow(QWidget *parent) :
QMainWindow(parent)
{
// Set a Qtimer to update the OSD display every 1 second
QTimer *timer = new QTimer(this);
connect(timer, SIGNAL(timeout()), this, SLOT(print()));
timer->start(1000);
}
void MainWindow::print()
{
printf("hello world\n");
}
But I want to pass a variable to print(). But when I do this, I never see my print statement.
MainWindow::MainWindow(QWidget *parent) :
QMainWindow(parent)
{
// Set a Qtimer to update the OSD display every 1 second
QTimer *timer = new QTimer(this);
int val = 42;
// Now pass val to print()
connect(timer, SIGNAL(timeout()), this, SLOT(print(val)));
timer->start(1000);
}
void MainWindow::print(int val)
{
// I never see my print statement
printf("hello world, val=%d\n", val);
}
header.h
class MainWindow : public QMainWindow
{
Q_OBJECT
public:
explicit MainWindow(QWidget *parent = 0);
~MainWindow();
public slots:
void print(int val);
Why does this not work? What can I do to pass a variable into print() using QTimer?

Answer 1

connect(timer, SIGNAL(timeout()), this, SLOT(print(val)));
Qt signal/slot connections don't work that way. The text within the SIGNAL() and SLOT() macros has to be the signature of the signal/slot method, verbatim; you can't put variables or other non-method-signature text there.
If you look at your program's stdout while it runs, you'll see an error message printed by connect() telling you that it can't find any slot-method named print(val).
If you want to provide a separate value for your slot, you could either make val a member-variable of your class, and have print() look at the member variable instead of an argument, or you could use an intermediary slot, like this:
public slots:
void print(int val);
void print42() {print(val);}
... and then connect your signal to SLOT(print42()) instead.