Question

我有一个可变参数模板类，用于保留自己的网络类的答案

enum class TaskType {readReg, writeReg, readRnd, writeRnd, readBlock, writeBlock, pause};

template<typename... Args> class Reply {
public:
    Reply(TaskType t, Args...params): t(t), tuple(std::make_tuple(params...)) {}
    Reply();
    TaskType t;
    auto getData() const {return std::get<4>(tuple);}   //simplified getter of safe function that deals with oversizing
private:
    std::tuple<Args ...> tuple;
};

我注册了模板签名以将其保留在Qvariant中

using ReadRegReply = Reply<TaskType, uint, uint, uint, uint, uint> ;
using WriteReply = Reply<TaskType, uint, uint, uint> ;
using ReadRndReply = Reply<TaskType, uint, uint, uint, QVector<uint>, QVector<uint>> ;
using ReadBlockReply = Reply<TaskType, uint, uint, uint, uint, QVector<uint>> ;

Q_DECLARE_METATYPE(QVector<uint>)
Q_DECLARE_METATYPE(ReadRegReply)
Q_DECLARE_METATYPE(WriteReply)
Q_DECLARE_METATYPE(ReadRndReply)
Q_DECLARE_METATYPE(ReadBlockReply)

然后我像这样使用它：

class Task: public QObject{ 
public:
//c-tor and some functions, virtual functions etc
template <class TReply> bool applyReply(TReply reply){
    varReply = QVariant::fromValue(reply);
}
auto getData(){  //here I should return data from tuple inside reply.
   QVariant::Type t = varReply.type();
   auto l = varReply.value<t>();// t is not a constexp // t is not a reply type but enum QVariant::type, as I understand.
   return l.getData(); // l has no method getData
}
QVariant varReply;      //this is the qvariant that contains templated reply;
}

我在QVariant中怀念某些东西。我认为注册类型应该以某种方式存储在Qvariant中，但事实并非如此。另一个问题：我不能使用c ++ 17;它会在许多具有不同回复签名的项目中使用。有什么方法可以保留这些类型并在将来进行添加而无需完全重构？我曾考虑过某种经理阶层，但我可能会想得太多

Answer 1

QVariant仅保存Qt中使用的内置类型的有用类型信息。如果您有其他类型，则需要将其记录在其他位置

class Task: public QObject{ 
public:
    template <class TReply> bool applyReply(TReply reply){
        varReply = QVariant::fromValue(reply);
    }
    template <class TReply> TReply getData(){reply.   
       return varReply.value<TReply>();
    }
    QVariant varReply;
}

void usesWriteTask() {
    Task task;
    task.applyReply(WriteReply());
    // ...
    WriteReply back = tast.getData<WriteReply>();
}

或者，如果您有权使用增强功能，则可以使用boost::variant，它启发了std::variant

class Task: public QObject{ 
public:
     using VarReply = boost::variant<ReadRegReply, WriteReply, ReadRndReply, ReadBlockReply>;
     bool applyReply(VarReply reply) {
         varReply = reply;
     }
     template <class Result> Result getData(boost::static_visitor<Result> & visitor) {
        return varReply.apply_visitor(visitor);
     }
     VarReply varReply;
}

从QVariant接收未知的模板对象

1 个答案: