这里 我们可以读到没有复制构造和复制赋值运算符可评估.但是这里我们可以读到qRegisterMetaType
和 Q_DECLARE_METATYPE
必须有公共默认构造函数、公共复制构造函数和公共析构函数.问题是:谁在说谎?还是我没有正确理解?
Here we can read that no copy construct and copy assignment operator evaluable. But here we can read that qRegisterMetaType
and Q_DECLARE_METATYPE
have to have public default constructor, public copy constructor and public destructor. The question is: who is telling a lie? Or I did not understand it correctly?
一切都是真的:
1.QObject
不能被复制,它的所有后代也不能被复制.
2. Q_DECLARE_METATYPE
接受带有公共构造函数、复制构造函数和析构函数的对象.
Everything is true:
1. QObject
can't be copied and all its descendants can't be copied also.
2. Q_DECLARE_METATYPE
accepts objects with public constructor, copy constructor and destructor.
没有矛盾,因为你不能用Q_DECLARE_METATYPE
注册QObject
后代.
There is no contradiction, because you can't register QObject
descendants with Q_DECLARE_METATYPE
.
当您将类转换为 QVariant
时,它使用复制构造函数来复制您的对象:
When you convert your class to QVariant
it uses a copy constructor to make a copy of your object:
void *ptr = QMetaType::construct(x->type, copy);
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