Hi,
I am currently evaluating the Qt State Machine Framework, and I am
facing an issue with the QtSignalTransition.
The current implementation does not allow to perform an action on a
QtSignalTransition.
The action can be added to the transition, but the corresponding action
is never performed.
My code is pretty simple:
// Let's create the state machine
m_stateMachine = new QtStateMachine( this );
// Let's create the states
QtState* state1 = new QtState();
QtState* state2 = new QtState();
// Let's add the states to the state machine
m_stateMachine->addState( state1 );
m_stateMachine->addState( state2 );
// Let's add a transition between the 2 states.
// So we create 1 transition
QtSignalTransition* transition12 = new QtSignalTransition(
ui.myButton, SIGNAL(clicked() ) );
// The transition will invoke this->myMethod()
transition12->invokeMethodOnTransition( this, "myMethod" );
// The transition goes from state1 to state2
state1->addTransition( transition12, state2 );
// Let's say we start in state1
m_stateMachine->setInitialState( state1 );
// and start the state machine
m_stateMachine->start();
Am I right in thinking that it is an error, or is it a design point that
I haven't understood?
This behavior is due to the empty implementation of
QtSignalTransition::onTransition()
- If this behaviour is not normal, removing the
QtSignalTransition::onTransition() method, or calling
QtTransition::onTransition() in QtSignalTransition::onTransition() would
solve the problem.
- If this behavior is normal, then I would expect one of the
following two possibilities:
1. QtSignalTransition should implement invokeMethodOnTransition and
setPropertyOnTransition as protected methods. This would return a
compilation error whenever I try to call them.
2. A warning should be displayed in the debug window when calling
one of the functions above.
I understand that I could derive my own signal transition class from
QtSignalTransition and define my own action by overriding onExecute(),
but why should QtSignalTransition not trigger any action?
Best regards
Jean-Yves Reutter
Software Developer
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