C + 11战略模式与国家

Question

本书的头部优先设计模式的战略模式的一个例子是在[ 这里 ]用C++编写的.我正在练习将其转换为C++ 11风格,根据有效的GoF模式,使用C++ 11和Boost,如下所示.

该嘎嘎行为:

struct Quack {
    static void quack()
    {
        std::cout << __FUNCTION__ << std::endl;
    }
};

struct MuteQuack {
    static void quack()
    {
        std::cout << __FUNCTION__ << std::endl;
    }
};

该飞的行为:

struct FlyWithWings {
public:
    static void fly()
    {
        std::cout << __FUNCTION__ << std::endl;
    }
};

struct FlyNoWay {
public:
    static void fly()
    {
        std::cout << __FUNCTION__ << std::endl;
    }
};

该鸭层次:

class Duck
{
public:
    typedef std::function<void(void)> QUACK;
    typedef std::function<void(void)> FLY;

public:
    Duck(const QUACK &q, const FLY &f)
        : m_Quack(q), m_Fly(f) {}

    virtual ~Duck()
    {
    }

    void perform_quack()
    {
        m_Quack();
    }
    void perform_fly()
    {
        m_Fly();
    }

protected:
    QUACK   m_Quack;
    FLY     m_Fly;

private:
    Duck(const Duck&) = delete;
    Duck& operator=(const Duck&) = delete;
};

class MallardDuck
    : public Duck
{
public:
    MallardDuck()
        : Duck(&Quack::quack, &FlyWithWings::fly)
    {
    }
};

class PaintedDuck
    : public Duck
{
public:
    PaintedDuck()
        : Duck(&MuteQuack::quack, &FlyNoWay::fly)
    {
    }
};

到目前为止,客户运作良好.

int main()
{
    MallardDuck x1;
    x1.perform_quack();
    x1.perform_fly();

    PaintedDuck x2;
    x2.perform_quack();
    x2.perform_fly();

    return 0;
}

现在我想扩展到一个新的类RubberDuck,以鸭层次,并RubberDuck采用了新飞的行为FlyWithRocket具有对象的状态.如下:

一个新的Fly行为:

class FlyWithRocket {
public:
    FlyWithRocket() : m_Energy(3) {}
    void fly()
    {
        if(m_Energy > 0)
        {
            fly_with_rocket();
            --m_Energy;
        }
        else
        {
            fly_out_of_energy();
        }
    }

private:
    void fly_with_rocket()
    {
        std::cout << __FUNCTION__ << std::endl;
    }
    void fly_out_of_energy()
    {
        std::cout << __FUNCTION__ << std::endl;
    }

    unsigned int m_Energy;
};

一种新的鸭子类型:

class RubberDuck
    : public Duck
{
public:
    RubberDuck()
        : Duck(&MuteQuack::quack, std::bind(&FlyWithRocket::fly, std::ref(m_flyrocket)))
        , m_flyrocket()
    {
    }
private:
    FlyWithRocket m_flyrocket;
};

从现在起我想知道成员初始化顺序的规则.基数Duck在成员之前初始化m_flyrocket,但请注意,使用尚未Duck初始化的绑定初始化基数m_flyrocket.因为我在VS2013中运行它,这在运行时没有错.

但是代码实际上不安全吗？如果没有,我怎么能修改为更好的设计？

Answer 1

It's not safe, but it's unlikely to break unless you call m_Fly() from the base class constructor.

You can easily avoid this though, by either:

1. giving the base class constructor a dummy or default-constructed std::function, and re-assigning m_Fly to your bind functor in the derived class constructor

   RubberDuck()
       : Duck(&MuteQuack::quack, std::function<void()>())
   {
       m_Fly = std::bind(&FlyWithRocket::fly, std::ref(m_flyrocket));
   }
   

2. making FlyWithRocket a functor itself (just rename void fly to void operator()) and passing it by value instead of keeping a private member (it will be owned by the m_Fly function object, and you can access it via std::function::target<FlyWithRocket>() if you need)

   class FlyWithRocket {
   public:
       FlyWithRocket() : m_Energy(3) {}
       void operator() () {
   // ...
   
   RubberDuck()
       : Duck(&MuteQuack::quack, FlyWithRocket()) {}