在Q_PROPERTY中使用自定义类型

如何解决在Q_PROPERTY中使用自定义类型

我无法在Q_PROPERTY MEMBER中使用自定义类型。

header.h

class Custom {
  Q_GADGET
  Q_PROPERTY(int mode MEMBER mode STORED true)

 public:
  Custom();
  int mode;
};

class Test {
  Q_GADGET
  Q_PROPERTY(QString ip MEMBER ip STORED true)
  Q_PROPERTY(Custom discovery MEMBER discovery STORED true)

 public:
  test();
  QString ip;
  Custom discovery;
};

Q_DECLARE_MetaTYPE(Custom)
Q_DECLARE_MetaTYPE(Test)

main.cpp

#include "header.h"

    Test::test() { qRegisterMetaType<Test>("Test"); }
    
    Custom::Custom() { qRegisterMetaType<Custom>("Custom"); }
    
    int main(int argc,char *argv[]) {
      QCoreApplication a(argc,argv);
      Test reg;
      return a.exec();
    }

当我编译代码时，我得到下面的错误。

/testing/build-test-Desktop_Qt_5_12_4_GCC_64bit-Debug/moc_CustomeTypes.cpp:175: error: no match for ‘operator!=’ (operand types are ‘Custom’ and ‘Custom’)
moc_CustomeTypes.cpp:175:31: error: no match for ‘operator!=’ (operand types are ‘Custom’ and ‘Custom’)
             if (_t->discovery != *reinterpret_cast< Custom*>(_v)) {
                 ~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

是否必须为自定义类提供 operator！= ？

仅提及一下，如果我将下面的Test类更改为可编译的。

class Test {
  Q_GADGET
  Q_PROPERTY(QString ip MEMBER ip STORED true)
  Q_PROPERTY(Custom discovery READ discovery WRITE setdiscovery STORED true)

 public:
  test();
  Custom discovery() const { return m_discovery; }
  void setdiscovery(const Custom& c) { m_discovery = c; }
  QString ip;
  Custom m_discovery;
};

为什么不能使用成员作为自定义类型？