我对如何测试QStateMachine感到有点困惑。我有一个项目组织得很好,一边是源代码,另一边是测试代码。
标题
class Foo
{
signals:
void sigGoToStateOne();
void sigGoToStateTwo();
void sigGoToStateThree();
private:
QStateMachine *stateMachine;
QState *state1;
QState *state2;
void initStateMachine();
}在源文件中
Foo::initStateMachine()
{
// constructors
state1->addTransition(this,SIGNAL(sigGoToStateTwo()),this->state2);
state2->addTransition(this,SIGNAL(sigGoToStateOne()),this->state1);
}我想知道是否有一种很好的方法来测试我的stateMachine是否正确。换句话说,如果我在状态机上发出sigGoToStateThree(),那么状态机是如何反应的,等等。
我看到的解决方案:1-获取stateMachine的地址(并最终获得所有其他状态)并测试它(但我不知道如何)2-从测试文件中模拟信号(sigGoToStateX()) (同样,不知道是否可以在另一个类中发出我的类Foo的信号)
我唯一的要求是我不想修改源文件的核心。
谢谢已经提前了。
发布于 2014-06-10 21:33:43
在Qt 5中,信号总是公共方法。要使代码与Qt 4兼容,可以将信号显式地公开如下:
class Foo {
public:
Q_SIGNAL void sigGoToStateOne();
...
}或者,您可以保持任意信号的可见性,并声明一个朋友测试类:
class Foo {
friend class FooTest;
...
}最后,您可以创建一个测试项目,其中使用Qt的测试框架测试Foo类的行为。下面的代码在Qt4和Qt5中都有效。
// main.cpp
#include <QCoreApplication>
#include <QStateMachine>
#include <QEventLoop>
#include <QtTest>
#include <QTimer>
class Waiter {
QTimer m_timer;
public:
Waiter() {}
Waiter(QObject * obj, const char * signal) {
m_timer.connect(obj, signal, SIGNAL(timeout()));
}
void stop() {
m_timer.stop();
QMetaObject::invokeMethod(&m_timer, "timeout");
}
void wait(int timeout = 5000) {
QEventLoop loop;
m_timer.start(timeout);
loop.connect(&m_timer, SIGNAL(timeout()), SLOT(quit()));
loop.exec();
}
};
class SignalWaiter : public QObject, public Waiter {
Q_OBJECT
int m_count;
Q_SLOT void triggered() {
++ m_count;
stop();
}
public:
SignalWaiter(QObject * obj, const char * signal) : m_count(0) {
connect(obj, signal, SLOT(triggered()), Qt::QueuedConnection);
}
int count() const { return m_count; }
};
#if QT_VERSION >= QT_VERSION_CHECK(5,0,0)
typedef QSignalSpy SignalSpy;
#else
class SignalSpy : public QSignalSpy, public Waiter {
public:
SignalSpy(QObject * obj, const char * signal) :
QSignalSpy(obj, signal), Waiter(obj, signal) {}
};
#endif
class Foo : public QObject {
Q_OBJECT
friend class FooTest;
QStateMachine m_stateMachine;
QState m_state1;
QState m_state2;
Q_SIGNAL void sigGoToStateOne();
Q_SIGNAL void sigGoToStateTwo();
public:
explicit Foo(QObject * parent = 0) :
QObject(parent),
m_state1(&m_stateMachine),
m_state2(&m_stateMachine)
{
m_stateMachine.setInitialState(&m_state1);
m_state1.addTransition(this, SIGNAL(sigGoToStateTwo()), &m_state2);
m_state2.addTransition(this, SIGNAL(sigGoToStateOne()), &m_state1);
}
Q_SLOT void start() {
m_stateMachine.start();
}
};
class FooTest : public QObject {
Q_OBJECT
void call(QObject * obj, const char * method) {
QMetaObject::invokeMethod(obj, method, Qt::QueuedConnection);
}
Q_SLOT void test1() {
// Uses QSignalSpy
Foo foo;
SignalSpy state1(&foo.m_state1, SIGNAL(entered()));
SignalSpy state2(&foo.m_state2, SIGNAL(entered()));
call(&foo, "start");
state1.wait();
QCOMPARE(state1.count(), 1);
call(&foo, "sigGoToStateTwo");
state2.wait();
QCOMPARE(state2.count(), 1);
call(&foo, "sigGoToStateOne");
state1.wait();
QCOMPARE(state1.count(), 2);
}
Q_SLOT void test2() {
// Uses SignalWaiter
Foo foo;
SignalWaiter state1(&foo.m_state1, SIGNAL(entered()));
SignalWaiter state2(&foo.m_state2, SIGNAL(entered()));
foo.start();
state1.wait();
QCOMPARE(state1.count(), 1);
emit foo.sigGoToStateTwo();
state2.wait();
QCOMPARE(state2.count(), 1);
emit foo.sigGoToStateOne();
state1.wait();
QCOMPARE(state1.count(), 2);
}
};
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
FooTest test;
QTest::qExec(&test, a.arguments());
QMetaObject::invokeMethod(&a, "quit", Qt::QueuedConnection);
return a.exec();
}
#include "main.moc"我强制从事件循环执行所有的信号调用,这样事件转换只能在事件循环运行时发生。这使得测试代码在每次转换后一致等待。否则,第二个wait将超时:
Q_SLOT void test1() {
SignalSpy state1(&m_foo.m_state1, SIGNAL(entered()));
SignalSpy state2(&m_foo.m_state2, SIGNAL(entered()));
m_foo.start();
state1.wait();
QCOMPARE(state1.count(), 1);
emit m_foo.sigGoToStateTwo(); // The state2.entered() signal is emitted here.
state2.wait(); // But we wait for it here, and this wait will time out.
QCOMPARE(state2.count(), 1); // But of course the count will match.
emit m_foo.sigGoToStateOne();
state1.wait(); // This would timeout as well.
QCOMPARE(state1.count(), 2);
}这可以通过使用内部使用排队连接的信号间谍类来实现,而无需使用显式排队调用。
发布于 2014-06-11 01:01:55
Kuba很好地分析了如何使用测试框架& SignalSpy对您的状态机进行深入测试。
如果您所要做的只是从一个测试文件生成一个sigGoToStateX(),那么不要忘记,您可以将信号链在一起。
例如,给定一个类"Tester":
class Tester : public QObject {
Q_OBJECT
public:
Tester(Foo *fooClass) {
//Connecting signals gives you the kind of behaviour you were asking about
connect(this, SIGNAL(testTransitionToState1()), fooClass, SIGNAL(sigGoToState1()));
connect(this, SIGNAL(testTransitionToState2()), fooClass, SIGNAL(sigGoToState2()));
connect(this, SIGNAL(testTransitionToState3()), fooClass, SIGNAL(sigGoToState3()));
}
void SwitchState(int newState) {
//Now any time we emit the test signals, the foo class's signals will be emitted too!
if (newState == 1) emit testTransitionToState1();
else if (newState == 2) emit testTransitionToState1();
else if (newState == 3) emit testTransitionToState1();
}
signals:
void testTransitionToState1();
void testTransitionToState2();
void testTransitionToState3();
}例如,调用SwitchState( 1 )将调用正确的信号以切换到状态1,如果这个简单的情况是测试所需的全部,那么这就是您真正需要的。
如果您需要更复杂的内容,请使用完整的SignalSpy示例。
https://stackoverflow.com/questions/24148670
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