c++ - std::enable_if - 如何使用自己的函数作为模板参数

Question

我有类 ModeSessions 和 2 个派生类：ReadModeSessions 和 WriteModeSessions；ModeSessions 有一个独特的 Mode 类型；在 Logic 中，我在容器中收集每个具有自己独特类型的会话。但是现在我坚持使用模板方法以通过给定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....

我不确定我是否做得对，试图将我的 constexpr 函数放在里面，但据我所知，应该有 bool 参数，我遵循的示例有 std::base_of 或 std::is_integer 使用 bool 作为返回，所以是什么区别？

我可能会考虑另一种方法，但重点是，我需要关注具有一些通用方法且每种类型都独一无二的会话；（也许取出这些方法在不同的新 sessionManager 类中实现它）

标签： c++qttemplatesenable-if