How to get this Qt state machine to work?

Question

I have two widgets that can be checked, and a numeric entry field that should contain a value greater than zero. Whenever both widgets have been checked, and the numeric entry field contains a value greater than zero, a button should be enabled. I am struggling with defining a proper state machine for this situation. So far I have the following:

QStateMachine *machine = new QStateMachine(this);

QState *buttonDisabled = new QState(QState::ParallelStates);
buttonDisabled->assignProperty(ui_->button, "enabled", false);

QState *a = new QState(buttonDisabled);
QState *aUnchecked = new QState(a);
QFinalState *aChecked = new QFinalState(a);
aUnchecked->addTransition(wa, SIGNAL(checked()), aChecked);
a->setInitialState(aUnchecked);

QState *b = new QState(buttonDisabled);
QState *bUnchecked = new QState(b);
QFinalState *bChecked = new QFinalState(b);
employeeUnchecked->addTransition(wb, SIGNAL(checked()), bChecked);
b->setInitialState(bUnchecked);

QState *weight = new QState(buttonDisabled);
QState *weightZero = new QState(weight);
QFinalState *weightGreaterThanZero = new QFinalState(weight);
weightZero->addTransition(this, SIGNAL(validWeight()), weightGreaterThanZero);
weight->setInitialState(weightZero);

QState *buttonEnabled = new QState();
buttonEnabled->assignProperty(ui_->registerButton, "enabled", true);

buttonDisabled->addTransition(buttonDisabled, SIGNAL(finished()), buttonEnabled);
buttonEnabled->addTransition(this, SIGNAL(invalidWeight()), weightZero);

machine->addState(registerButtonDisabled);
machine->addState(registerButtonEnabled);
machine->setInitialState(registerButtonDisabled);
machine->start();

The problem here is that the following transition:

buttonEnabled->addTransition(this, SIGNAL(invalidWeight()), weightZero);

causes all the child states in the registerButtonDisabled state to be reverted to their initial state. This is unwanted behaviour, as I want the a and b states to remain in the same state.

How do I ensure that a and b remain in the same state? Is there another / better way this problem can be solved using state machines?

---

Note. There are a countless (arguably better) ways to solve this problem. However, I am only interested in a solution that uses a state machine. I think such a simple use case should be solvable using a simple state machine, right?

Answer 1

The state machine you used above does not correspond to what you described. Using a final state is not correct because after enter a value greater zero I don't see anything that prevents the user from enter zero again. Therefore the valid states can't be final. As far as I can see from your code the user is allowed to change the state of the widgets in any order. Your state machine has to pay attention to this.

I would use a state machine with four child states (no valid input, one valid input, two valid inputs, three valid inputs). You obviously start with no valid input. Each widget can make a transition from no to one an back (same counts for two and three). When three is entered all widgets are valid (button enabled). For all other states the button has to be disabled when the state is entered.

I wrote a sample app. The main window contains two QCheckBoxes a QSpinBox and a QPushButton. There are signals in the main window the ease write down the transitions of the states. There are fired when the state of the widgets are changed.

MainWindow.h

#ifndef MAINWINDOW_H
#define MAINWINDOW_H

#include <QtGui>

namespace Ui
{
    class MainWindow;
}

class MainWindow : public QMainWindow
{
    Q_OBJECT

public:
    MainWindow(QWidget *parent = 0);
    ~MainWindow();

private:
    Ui::MainWindow *ui;
    bool m_editValid;

    bool isEditValid() const;
    void setEditValid(bool value);

private slots:
    void on_checkBox1_stateChanged(int state);
    void on_checkBox2_stateChanged(int state);
    void on_spinBox_valueChanged (int i);
signals:
    void checkBox1Checked();
    void checkBox1Unchecked();
    void checkBox2Checked();
    void checkBox2Unchecked();
    void editValid();
    void editInvalid();
};

#endif // MAINWINDOW_H

MainWindow.cpp

#include "MainWindow.h"
#include "ui_MainWindow.h"

MainWindow::MainWindow(QWidget *parent)
  : QMainWindow(parent), ui(new Ui::MainWindow), m_editValid(false)
{
  ui->setupUi(this);

  QStateMachine* stateMachine = new QStateMachine(this);
  QState* noneValid = new QState(stateMachine);
  QState* oneValid = new QState(stateMachine);
  QState* twoValid = new QState(stateMachine);
  QState* threeValid = new QState(stateMachine);

  noneValid->addTransition(this, SIGNAL(checkBox1Checked()), oneValid);
  oneValid->addTransition(this, SIGNAL(checkBox1Checked()), twoValid);
  twoValid->addTransition(this, SIGNAL(checkBox1Checked()), threeValid);
  threeValid->addTransition(this, SIGNAL(checkBox1Unchecked()), twoValid);
  twoValid->addTransition(this, SIGNAL(checkBox1Unchecked()), oneValid);
  oneValid->addTransition(this, SIGNAL(checkBox1Unchecked()), noneValid);

  noneValid->addTransition(this, SIGNAL(checkBox2Checked()), oneValid);
  oneValid->addTransition(this, SIGNAL(checkBox2Checked()), twoValid);
  twoValid->addTransition(this, SIGNAL(checkBox2Checked()), threeValid);
  threeValid->addTransition(this, SIGNAL(checkBox2Unchecked()), twoValid);
  twoValid->addTransition(this, SIGNAL(checkBox2Unchecked()), oneValid);
  oneValid->addTransition(this, SIGNAL(checkBox2Unchecked()), noneValid);

  noneValid->addTransition(this, SIGNAL(editValid()), oneValid);
  oneValid->addTransition(this, SIGNAL(editValid()), twoValid);
  twoValid->addTransition(this, SIGNAL(editValid()), threeValid);
  threeValid->addTransition(this, SIGNAL(editInvalid()), twoValid);
  twoValid->addTransition(this, SIGNAL(editInvalid()), oneValid);
  oneValid->addTransition(this, SIGNAL(editInvalid()), noneValid);

  threeValid->assignProperty(ui->pushButton, "enabled", true);
  twoValid->assignProperty(ui->pushButton, "enabled", false);
  oneValid->assignProperty(ui->pushButton, "enabled", false);
  noneValid->assignProperty(ui->pushButton, "enabled", false);

  stateMachine->setInitialState(noneValid);

  stateMachine->start();
}

MainWindow::~MainWindow()
{
  delete ui;
}

bool MainWindow::isEditValid() const
{
  return m_editValid;
}

void MainWindow::setEditValid(bool value)
{
  if (value == m_editValid)
  {
    return;
  }
  m_editValid = value;
  if (value)
  {
    emit editValid();
  } else {
    emit editInvalid();
  }
}

void MainWindow::on_checkBox1_stateChanged(int state)
{
  if (state == Qt::Checked)
  {
    emit checkBox1Checked();
  } else {
    emit checkBox1Unchecked();
  }
}

void MainWindow::on_checkBox2_stateChanged(int state)
{
  if (state == Qt::Checked)
  {
    emit checkBox2Checked();
  } else {
    emit checkBox2Unchecked();
  }
}

void MainWindow::on_spinBox_valueChanged (int i)
{
  setEditValid(i > 0);
}

This should do the trick. As you yourself mentioned already there are better ways to achive this behaviour. Especially keep track of all transistions between the state is error prone.

Answer 2

edit

I reopened this test, willing to use it, added to .pro

CONFIG += C++11

and I discovered that lambda syntax has changed... The capture list cannot reference member variables. Here is the corrected code

auto cs = [/*button, check1, check2, edit, */this](QState *s, QState *t, bool on_off) {
    s->assignProperty(button, "enabled", !on_off);
    s->addTransition(new QSignalTransition(check1, SIGNAL(clicked())));
    s->addTransition(new QSignalTransition(check2, SIGNAL(clicked())));
    s->addTransition(new QSignalTransition(edit, SIGNAL(textChanged(QString))));
    Transition *p = new Transition(this, on_off);
    p->setTargetState(t);
    s->addTransition(p);
};

end edit

I used this question as exercise (first time on QStateMachine). The solution is fairly compact, using a guarded transition to move between 'enabled/disabled' state, and lambda to factorize setup:

#include "mainwindow.h"
#include <QLayout>
#include <QFrame>
#include <QSignalTransition>

struct MainWindow::Transition : QAbstractTransition {
    Transition(MainWindow *main_w, bool on_off) :
        main_w(main_w),
        on_off(on_off)
    {}

    virtual bool eventTest(QEvent *) {
        bool ok_int, ok_cond =
            main_w->check1->isChecked() &&
            main_w->check2->isChecked() &&
            main_w->edit->text().toInt(&ok_int) > 0 && ok_int;
        if (on_off)
            return ok_cond;
        else
            return !ok_cond;
    }

    virtual void onTransition(QEvent *) {}

    MainWindow *main_w;
    bool on_off;
};

MainWindow::MainWindow(QWidget *parent)
    : QMainWindow(parent)
{
    QFrame *f = new QFrame(this);
    QVBoxLayout *l = new QVBoxLayout;

    l->addWidget(check1 = new QCheckBox("Ok &1"));
    l->addWidget(check2 = new QCheckBox("Ok &2"));
    l->addWidget(edit = new QLineEdit());

    l->addWidget(button = new QPushButton("Enable &Me"));

    f->setLayout(l);
    setCentralWidget(f);

    QState *s1, *s2;
    sm = new QStateMachine(this);
    sm->addState(s1 = new QState());
    sm->addState(s2 = new QState());
    sm->setInitialState(s1);

    auto cs = [button, check1, check2, edit, this](QState *s, QState *t, bool on_off) {
        s->assignProperty(button, "enabled", !on_off);
        s->addTransition(new QSignalTransition(check1, SIGNAL(clicked())));
        s->addTransition(new QSignalTransition(check2, SIGNAL(clicked())));
        s->addTransition(new QSignalTransition(edit, SIGNAL(textChanged(QString))));
        Transition *tr = new Transition(this, on_off);
        tr->setTargetState(t);
        s->addTransition(tr);
    };
    cs(s1, s2, true);
    cs(s2, s1, false);

    sm->start();
}

Answer 3

When I have to do things like this I usually use signals and slots. Basically each of the widgets and the number box will all emit signals automatically when their states change. If you link each of these to a slot that checks if all 3 objects are in the desired state and enables the button if they are or disables it if they aren't, then that should simplify things.

Sometimes you will also need to change the button state once you've clicked it.

[EDIT]: I'm sure there is some way of doing this using state machines, will you only be reverting in the situation that both boxes are checked and you've added an invalid weight or will you also need to revert with only one checkbox checked? If it's the former then you may be able to set up a RestoreProperties state that allows you to revert to the checked box state. Otherwise is there some way you can save the state before checking the weight is valid, revert all checkboxes then restore the state.

Answer 4

After reading your requirements and the answers and comments here I think merula's solution or something similar is the only pure Statemachine solution.

As has been noted to make the Parallel State fire the finished() signal all the disabled states have to be final states, but this is not really what they should be as someone could uncheck one of the checkboxes and then you would have to move away from the final state. You can't do that as FinalState does not accept any transitions. The using the FinalState to exit the parallel state also causes the parallel state to restart when it is reentered.

One solution could be to code up a transition that only triggers when all three states are in the "good" state, and a second one that triggers when any of those is not. Then you add the disabled and enabled states to the parallel state you already have and connect it with the aforementioned transitions. This will keep the enabled state of the button in sync with all the states of your UI pieces. It will also let you leave the parallel state and come back to a consistent set of property settings.

class AndGateTransition : public QAbstractTransition
{
    Q_OBJECT

public:

    AndGateTransition(QAbstractState* sourceState) : QAbstractTransition(sourceState)
        m_isSet(false), m_triggerOnSet(true), m_triggerOnUnset(false)

    void setTriggerSet(bool val)
    {
        m_triggerSet = val;
    }

    void setTriggerOnUnset(bool val)
    {
        m_triggerOnUnset = val;
    }

    addState(QState* state)
    {
        m_states[state] = false;
        connect(m_state, SIGNAL(entered()), this, SLOT(stateActivated());
        connect(m_state, SIGNAL(exited()), this, SLOT(stateDeactivated());
    }

public slots:
    void stateActivated()
    {
        QObject sender = sender();
        if (sender == 0) return;
        m_states[sender] = true;
        checkTrigger();
    }

    void stateDeactivated()
    {
        QObject sender = sender();
        if (sender == 0) return;
        m_states[sender] = false;
        checkTrigger();
    }

    void checkTrigger()
    {
        bool set = true;
        QHashIterator<QObject*, bool> it(m_states)
        while (it.hasNext())
        {
            it.next();
            set = set&&it.value();
            if (! set) break;
        }

        if (m_triggerOnSet && set && !m_isSet)
        {
            m_isSet = set;
            emit (triggered());

        }
        elseif (m_triggerOnUnset && !set && m_isSet)
        {
            m_isSet = set;
            emit (triggered());
        }
    }

pivate:
    QHash<QObject*, bool> m_states;
    bool m_triggerOnSet;
    bool m_triggerOnUnset;
    bool m_isSet;

}

Did not compile this or even test it, but it should demonstrate the principle

Answer 5

Set up your weight input widget so that there is no way a weight less than zero can be entered. Then you don't need invalidWeight()