Enter initial state before QStateMachine::started() is emitted
The documentation says that started() "is emitted when the state machine has entered its initial state", but the implementation didn't adhere to that. The consequence is that if you e.g. emitted a signal from a slot connected to started(), and that signal was used by a transition from the initial state, the signal would effectively get ignored and the state machine would remain in the initial state. Task-number: QTBUG-24307 Change-Id: Ibbeb627d517eaff821d88e256a949eacf6aae350 Reviewed-by: Eskil Abrahamsen Blomfeldt <eskil.abrahamsen-blomfeldt@nokia.com>
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@ -1203,7 +1203,6 @@ void QStateMachinePrivate::_q_start()
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#endif
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state = Running;
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processingScheduled = true; // we call _q_process() below
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emit q->started();
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QState *start = startState();
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Q_ASSERT(start != 0);
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@ -1230,6 +1229,9 @@ void QStateMachinePrivate::_q_start()
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#ifdef QSTATEMACHINE_DEBUG
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qDebug() << q << ": initial configuration:" << configuration;
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#endif
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emit q->started();
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_q_process();
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}
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@ -175,6 +175,8 @@ private slots:
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void stopInTransitionToFinalState();
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void stopInEventTest_data();
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void stopInEventTest();
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void initialStateIsEnteredBeforeStartedEmitted();
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};
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class TestState : public QState
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@ -3906,5 +3908,21 @@ void tst_QStateMachine::stopInEventTest()
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QVERIFY(machine.configuration().contains(s1));
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}
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void tst_QStateMachine::initialStateIsEnteredBeforeStartedEmitted()
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{
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QStateMachine machine;
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QState *s1 = new QState(&machine);
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machine.setInitialState(s1);
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QFinalState *s2 = new QFinalState(&machine);
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// When started() is emitted, s1 should be the active state, and this
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// transition should trigger.
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s1->addTransition(&machine, SIGNAL(started()), s2);
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QSignalSpy finishedSpy(&machine, SIGNAL(finished()));
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machine.start();
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QTRY_COMPARE(finishedSpy.count(), 1);
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}
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QTEST_MAIN(tst_QStateMachine)
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#include "tst_qstatemachine.moc"
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