Question

我有ModeSessions类和2个派生类：ReadModeSessions和WriteModeSessions; ModeSessions具有唯一的Mode类型。在Logic中，我使用自己的唯一类型在容器中收集每个会话。但是现在我坚持使用模板方法，以通过给定的模式返回正确的会话。

#include <QVector>
#include <QSharedPointer>
namespace states {
    enum class Mode{ read1, read2, write1, write2};
    constexpr bool modeIsReadType(Mode m){ return (m == Mode::read1 || m == Mode::read2)? true: false;}
    constexpr bool modeIsWriteType(Mode m){ return (m == Mode::write1 || m == Mode::write2)? true: false;}
}

class ModeSession{
public:
    states::Mode type;
};
using namespace states;
class ReadModeSession: public ModeSession{
public:
    ReadModeSession(Mode m){ModeSession::type = m;}
};
class WriteModeSession : public ModeSession{
public:
    WriteModeSession(Mode m){ModeSession::type = m;}
};
class Logic
{
public:
    Logic(){
        allModeSessions << QSharedPointer<ReadModeSession>(new ReadModeSession(Mode::read1));
        allModeSessions << QSharedPointer<ReadModeSession>(new ReadModeSession(Mode::read2));
        allModeSessions << QSharedPointer<WriteModeSession>(new WriteModeSession(Mode::write1));
        allModeSessions << QSharedPointer<WriteModeSession>(new WriteModeSession(Mode::write2));
    }
    QVector<QSharedPointer<ModeSession>> allModeSessions;
    template<class Tm> QSharedPointer<ReadModeSession>
        getMs(Tm m, typename std::enable_if<states::modeIsReadType(m), Tm>::type){
        for (QSharedPointer<ModeSession> ms: allModeSessions){
            if (ms->type == m) return ms.dynamicCast<ReadModeSession>();
        }
        return nullptr;
    }
    template<class Tm> QSharedPointer<WriteModeSession>
        getMs(Tm m, typename std::enable_if<states::modeIsWriteType(m), Tm>::type){ //to stack overflow....
        for (QSharedPointer<ModeSession> ms: allModeSessions){
            if (ms->type == m) return ms.dynamicCast<WriteModeSession>();
        }
        return nullptr;
    }
};

两种模板方法都具有

In file included from ..\enableIf\logic.cpp:1:0:
..\enableIf\logic.h:35:74: error: template argument 1 is invalid
         getMs(Tm m, typename std::enable_if<states::modeIsReadType(m), Tm>::type){
                                                                          ^
..\enableIf\logic.h:35:74: error: template argument 1 is invalid
..\enableIf\logic.h:35:74: error: template argument 1 is invalid
..\enableIf\logic.h:35:74: error: template argument 1 is invalid
..\enableIf\logic.h:35:35: error: invalid use of template-name 'std::enable_if' without an argument list
         getMs(Tm m, typename std::enable_if<states::modeIsReadType(m), Tm>::type){
                                   ^
..\enableIf\logic.h:35:44: error: expected ',' or '...' before '<' token
         getMs(Tm m, typename std::enable_if<states::modeIsReadType(m), Tm>::type){
                                            ^
..\enableIf\logic.h:42:75: error: template argument 1 is invalid
         getMs(Tm m, typename std::enable_if<states::modeIsWriteType(m), Tm>::type){ //to stack overflow....
                                                                           ^
..\enableIf\logic.h:42:75: error: template argument 1 is invalid
..\enableIf\logic.h:42:75: error: template argument 1 is invalid
..\enableIf\logic.h:42:75: error: template argument 1 is invalid
..\enableIf\logic.h:42:35: error: invalid use of template-name 'std::enable_if' without an argument list
         getMs(Tm m, typename std::enable_if<states::modeIsWriteType(m), Tm>::type){ //to stack overflow....
                                   ^
..\enableIf\logic.h:42:44: error: expected ',' or '...' before '<' token
         getMs(Tm m, typename std::enable_if<states::modeIsWriteType(m), Tm>::type){ //to stack overflow....

我不确定是否做对了，尝试将constexpr函数放进去，但是据我了解，应该有bool参数，下面的示例包含使用bool作为返回值的std :: base_of或std :: is_integer。，有什么区别？

我可能会考虑另一种方法，但是重点是，我需要关心具有一些通用方法并且对于每种类型都是唯一的会话。（也许要去掉那些方法才能在另一个新的sessionManager类中实现它）

std :: enable_if-如何使用自己的函数作为模板参数

0 个答案: