抽象类选择性地基于派生类型揭示方法

Question

我正在编写一个实现工厂timeManager生成两种类型对象的程序； 计时器和stopwatch。这两个对象均来自抽象类timepiece，并共享常见方法。我希望工厂返回类型timepiece，但是由于计时器实现了额外的方法，所以我不能。如何使timepiece公开timer的额外方法，而无需将它们暴露于sepwatch创建这些相应类型时？我希望这有意义，因为很难解释。

timemanager.h

enum TimePieceType{TIMER, STOP_WATCH};

class TimeManager {
public:
    TimeManager();
    ~TimeManager();

    void Tick();
    TimePiece* AddTimePiece(TimePieceType timeType);
};

timepiece.h

enum RunningState { STOPPED, RUNNING, PAUSED };

class TimePiece {
private:
    unsigned long startTime = 0;
    unsigned long currentTime = 0;
    unsigned long pauseOffset = 0;

protected:
    Time time;
    RunningState state = STOPPED;

public:
    TimePiece() = default;
    virtual ~TimePiece() = default;

    void Stop();
    void Start();
    void Pause();
    void Resume();
    void Reset();
    virtual void Tick();

    Time GetTime();
    RunningState GetState();
    virtual uint8_t GetId() = 0;
};

stepwatch.h

class StopWatch : public TimePiece {
private:
    uint8_t watchId = 0;
    static int8_t currentId;
    
public:
    StopWatch();
    virtual ~StopWatch();

    uint8_t GetId();
};

Timer.h

class Timer : public TimePiece {
private:
    uint8_t timerId = 0;
    static int8_t currentId;
    unsigned long timeout;

public:
    Timer();
    ~Timer();

    void Tick() override;

    void SetTimeoutMs(unsigned long timeout);
    void SetTimeoutSec(unsigned long timeout);

    uint8_t GetId();
    Time GetRemainingTime();
    unsigned long GetTimeout();
};

Answer 1

您不应返回timepiece*，因为两个子类共享timepiece其实现而不是其接口。

一种简单的方法是给出timeManager两个单独的方法：

Timer* AddTimer();
StopWatch* AddStopWatch();

如果需要remove> remove方法：您可以利用多态性行为：

void RemoveTimePiece(TimePiece*);