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>

1 files changed, 18 insertions, 0 deletions

diff --git a/tests/auto/corelib/statemachine/qstatemachine/tst_qstatemachine.cpp b/tests/auto/corelib/statemachine/qstatemachine/tst_qstatemachine.cpp
index df6ac88f47..34e9b742b1 100644
--- a/tests/auto/corelib/statemachine/qstatemachine/tst_qstatemachine.cpp
+++ b/tests/auto/corelib/statemachine/qstatemachine/tst_qstatemachine.cpp
@@ -175,6 +175,8 @@ private slots:
     void stopInTransitionToFinalState();
     void stopInEventTest_data();
     void stopInEventTest();
+
+    void initialStateIsEnteredBeforeStartedEmitted();
 };
 
 class TestState : public QState
@@ -3906,5 +3908,21 @@ void tst_QStateMachine::stopInEventTest()
     QVERIFY(machine.configuration().contains(s1));
 }
 
+void tst_QStateMachine::initialStateIsEnteredBeforeStartedEmitted()
+{
+    QStateMachine machine;
+    QState *s1 = new QState(&machine);
+    machine.setInitialState(s1);
+    QFinalState *s2 = new QFinalState(&machine);
+
+    // When started() is emitted, s1 should be the active state, and this
+    // transition should trigger.
+    s1->addTransition(&machine, SIGNAL(started()), s2);
+
+    QSignalSpy finishedSpy(&machine, SIGNAL(finished()));
+    machine.start();
+    QTRY_COMPARE(finishedSpy.count(), 1);
+}
+
 QTEST_MAIN(tst_QStateMachine)
 #include "tst_qstatemachine.moc"