c++ - 为什么我不能分配这个 Qt 属性？

Question

Qt 在运行时抛出以下错误。

Unable to assign LIMITS_T to LIMITS_T

我假设 Qt 需要更多的元数据信息，但我不知道我错过了什么。我已经做了一切来声明元类型:

limits.h

class LIMITS_T : public QObject{
    Q_OBJECT
    Q_PROPERTY(float min READ readMin WRITE writeMin NOTIFY minChanged)

public:
    LIMITS_T() : QObject() {}
    LIMITS_T(const LIMITS_T& limit) : QObject()
    {
        this->min = limit.min;
    }

    float min = 0;

    float readMin() { return min; }
    void writeMin(float min) { this->min = min; }

    bool operator = (const LIMITS_T &limit)
    {
        this->min = limit.min;
    }

signals:
    void minChanged();
};

Q_DECLARE_METATYPE(LIMITS_T)

这是 splitBarGauge 类的简化版本

splitDialGauge.h

class SplitDialGauge : public QQuickPaintedItem
{
    Q_OBJECT
    Q_PROPERTY(LIMITS_T limits READ getLimits WRITE setLimits NOTIFY limitsChanged)

public:
    SplitDialGauge(QQuickItem *parent = 0);

protected:
    LIMITS_T limits;
    virtual LIMITS_T getLimits();
    virtual void setLimits(LIMITS_T value);
}

splitDialGauge.cpp

#include "splitBarGauge.h"

SplitDialGauge::SplitDialGauge(QQuickItem *parent = 0);
    : QQuickPaintedItem(parent)
{
}

LIMITS_T SplitDialGauge::getLimits()
{
    return this->limits;
}

void SplitDialGauge::setLimits(LIMITS_T limits)
{
    this->limits = limits;
    update();
}

然后我用 Qt 元数据系统注册这个类

#include "limits.h"
#include "splitDialGauge.h"
int main(int argc, char *argv[])
{
    QCoreApplication::setAttribute(Qt::AA_EnableHighDpiScaling);
    QGuiApplication app(argc, argv);

    qmlRegisterType<SplitDialGauge>("ExampleModule", 1, 0, "SplitDialGauge");
    qmlRegisterType<LIMITS_T>("ExampleModule", 1, 0, "Limits");
    qRegisterMetaType<LIMITS_T>();

    QQmlApplicationEngine engine;
    engine.load(QUrl(QLatin1String("qrc:/main.qml")));
    if (engine.rootObjects().isEmpty())
        return -1;

    return app.exec();
}

这是 QML 文件的片段

import ExampleModule 1.0

ApplicationWindow {
    visible: true
    width: 800
    height: 480

    SplitDialGauge {
        Limits {
            id: qtyLimits
            min: 4
        }
        height: 200
        width: 50
        limits: qtyLimits
    }
}

Answer 1

您必须将该属性声明为指针。根据 docs，所有 QObject 都应该作为指针进行操作。 :

No Copy Constructor or Assignment Operator QObject

has neither a copyconstructor nor an assignment operator. This is by design. Actually,they are declared, but in a private section with the macroQ_DISABLE_COPY(). In fact, all Qt classes derived from QObject (director indirect) use this macro to declare their copy constructor andassignment operator to be private. The reasoning is found in thediscussion on Identity vs Value on the Qt Object Model page.

The main consequence is that you should use pointers to QObject (or toyour QObject subclass) where you might otherwise be tempted to useyour QObject subclass as a value. For example, without a copyconstructor, you can't use a subclass of QObject as the value to bestored in one of the container classes. You must store pointers.

在你的情况下:

#ifndef SPLITDIALGAUGE_H
#define SPLITDIALGAUGE_H

#include "limits_t.h"

#include <QPainter>
#include <QQuickPaintedItem>    

class SplitDialGauge : public QQuickPaintedItem {
    Q_OBJECT
    Q_PROPERTY(LIMITS_T *limits READ getLimits WRITE setLimits NOTIFY limitsChanged)
    LIMITS_T *limits;    
public:
    SplitDialGauge(QQuickItem *parent = 0) : QQuickPaintedItem(parent), limits(nullptr) { }
    void paint(QPainter *painter) {
        [...]
    }    
    LIMITS_T *getLimits() const { return limits; }
    void setLimits(LIMITS_T *value) {
        if (limits == value) return;
        limits = value;
        [...]
        emit limitsChanged();
    }     
signals:
    void limitsChanged();
};
#endif // SPLITDIALGAUGE_H

可以在以下link 中找到一个功能示例.