Question

我正在尝试开发一个通用函数，该函数确定两个QObject是否相等。为了使之成为可能，被比较的函数必须具有一个“等于”方法，该方法比较每个函数中的各个函数值，如果它们都相等，则返回true。此外，必须使用Q_INVOKABLE声明此“等于”方法。

但是，当我尝试为'equals'方法调用invokeMethod时，它失败并显示错误“ QMetaObject :: invokeMethod：没有这样的方法F1 :: equals（QObject *）（QObject *）”。

这是我的测试项目和文件：

项目文件：


CONFIG += c++11 console
CONFIG -= app_bundle

# The following define makes your compiler emit warnings if you use
# any feature of Qt which as been marked deprecated (the exact warnings
# depend on your compiler). Please consult the documentation of the
# deprecated API in order to know how to port your code away from it.
DEFINES += QT_DEPRECATED_WARNINGS

# You can also make your code fail to compile if you use deprecated APIs.
# In order to do so, uncomment the following line.
# You can also select to disable deprecated APIs only up to a certain version of Qt.
#DEFINES += QT_DISABLE_DEPRECATED_BEFORE=0x060000    # disables all the APIs deprecated before Qt 6.0.0

SOURCES += \
        main.cpp

HEADERS += \
    f1.h \
    assert1.h

assert1.h

#ifndef ASSERT1_H
#define ASSERT1_H

#include <QObject>
#include <QDebug>

class Assert1 : public QObject
{
 Q_OBJECT
public:
 explicit Assert1(QObject *parent = nullptr) {}
 static bool isEqual(QString msg, QObject* o1, QObject* o2)
 {
  if(o1 != nullptr && o2 != nullptr)
  {
   if(o1->metaObject()->className() != o2->metaObject()->className())
   {
    qDebug() << msg << " not same class type!";
    return false;
   }
   const QMetaObject* metaObject =o1->metaObject();
   int ix = metaObject->indexOfMethod(QMetaObject::normalizedSignature("equals(QObject *)"));
   qDebug() << QMetaObject::normalizedSignature("equals(QObject *)");
   if(ix == -1)
   {
    qDebug() << msg << tr("indexOfMethod(\"equals\") returns %1").arg(ix);
    return false;
   }
   else
   {
     bool rslt = false;
     if(!QMetaObject::invokeMethod(o1, QMetaObject::normalizedSignature("equals(QObject *)"),
                                   Qt::DirectConnection,
                                   Q_RETURN_ARG(bool, rslt),
                                   Q_ARG(QObject*, o2)))
         qDebug() << msg << tr("invoke method 'equals' failed for %1").arg(o1->metaObject()->className());
     if(!rslt)
         qDebug() << msg  << tr(" objects not equal");
     return false;
    }
  }
  qDebug() << msg << "not equal";
 }

signals:

public slots:
};

#endif // ASSERT1_H

f1.h

#ifndef F1_H
#define F1_H

#include <QObject>
#include <QDebug>

class F1 : public QObject
{
    Q_OBJECT
public:
    explicit F1(int p1, QString p2, QObject *parent = nullptr) : QObject(parent)
    {
      this->p1 = p1;
      this->p2 = p2;
    }

    void setP1(int p) {this->p1 = p;}
    void setP2(QString p) {this->p2 = p;}
    Q_INVOKABLE bool equals(QObject* other)
    {
     if(qobject_cast<F1*>(other) != nullptr)
     {
       if(this->p1 != ((F1*)other)->p1)
        return false;
       if(this->p2 != ((F1*)other)->p2)
        return false;
     }
     return true;
    }
    Q_INVOKABLE QString toString()
    {
     qDebug() << "p1 '" << p1 << " p2 = '" << p2 << "'";
    }


signals:

public slots:
private:
    int p1;
    QString p2;
};

#endif // F1_H

main.cpp

#include <QCoreApplication>
#include "f1.h"
#include "assert1.h"
#include <QDebug>

int main(int argc, char *argv[])
{
    QCoreApplication a(argc, argv);

    F1* tf1 = new F1(1, "1");
    F1* tf2 = new F1(1, "a");
    F1* tf3 = new F1(1, "a");
    F1* tf4 = new F1(4, "abc");

    qDebug() << "tf1->equals(tf4) returns: " << (tf1->equals(tf4)?"true":"false");
    qDebug() << "tf2->equals(tf3) returns: " << (tf2->equals(tf3)?"true":"false");
    Assert1::isEqual("should be equal", (QObject*)tf2, (QObject*)tf3);

    //return a.exec();
}

运行测试会产生以下输出：

Debugging starts
tf1->equals(tf4) returns:  false
tf2->equals(tf3) returns:  false
"equals(QObject*)"
QMetaObject::invokeMethod: No such method F1::equals(QObject*)(QObject*)
"should be equal" "invoke method 'equals' failed for F1"

如何使invokeMethod起作用？

Answer 1

您遇到的错误是invokeMethod仅等待Q_SLOT或Q_INVOKABLE的名称，不需要或不需要签名，但是您传递了QMetaObject::normalizedSignature("equals(QObject*)")并返回“ equals（QObject *）”，所以解决方案就是传递等于：

if(!QMetaObject::invokeMethod(o1, "equals",
    Qt::DirectConnection,
    Q_RETURN_ARG(bool, rslt),
    Q_ARG(QObject*, o2)))

无法解决Qt的QMetaObject :: invokeMethod：没有此类方法错误

1 个答案: