c++ - 为什么装饰模式适用于指针而不适用于引用？

Question

我刚刚了解了装饰器模式，并尝试编写一个使用该代码的示例。这个例子是关于饮料和一些调味品的。在装饰器内部，我有一个 引用变量 到饮料。提供的饮料是Decaf和 Espresso .可用的调味品是Soy和 Caramel .如果我定义一个 Decaf不止一个 Caramel例如，我得到的结果只是一个带有一个装饰器的 Decaf。所以定义 Caramel -> Caramel -> Decaf给我 Caramel -> Decaf .定义 Caramel -> Soy -> Caramel -> Decaf工作正常。定义 Caramel -> Soy -> Caramel -> Caramel -> Decaf给我 Caramel -> Soy -> Caramel -> Decaf .长话短说，我不能一个接一个地吃两种或多种相同类型的调味品。它们只成为一种调味品。如果我使用指针它工作正常。

编码:

#include <iostream>
//#include "Decaf.h"
//#include "Espresso.h"
//#include "SoyDecorator.h"
//#include "CaramelDecorator.h"

class Beverage
{
public:

    virtual std::string GetDescription() const = 0;
    virtual int GetCost() const = 0;
};

class CondimentDecorator : public Beverage
{
public:

    Beverage& beverage;
    CondimentDecorator(Beverage& beverage) : beverage(beverage) {}
};

class Espresso : public Beverage
{
    virtual std::string GetDescription() const override
    {
        return "Espresso";
    }

    virtual int GetCost() const override
    {
        return 5;
    }
};

class Decaf : public Beverage
{
    virtual std::string GetDescription() const override
    {
        return "Decaf";
    }

    virtual int GetCost() const override
    {
        return 4;
    }
};

class CaramelDecorator : public CondimentDecorator
{
public:

    CaramelDecorator(Beverage& beverage) : CondimentDecorator(beverage) {}

    virtual std::string GetDescription() const override
    {
        return this->beverage.GetDescription() + " with Caramel";
    }

    virtual int GetCost() const override
    {
        return this->beverage.GetCost() + 2;
    }
};

class SoyDecorator : public CondimentDecorator
{
public:

    SoyDecorator(Beverage& beverage) : CondimentDecorator(beverage) {}

    virtual std::string GetDescription() const override
    {
        return this->beverage.GetDescription() + " with Soy";
    }

    virtual int GetCost() const override
    {
        return this->beverage.GetCost() + 1;
    }
};

int main()
{
    Decaf d;
    SoyDecorator s(d);
    CaramelDecorator c(s);
    CaramelDecorator cc(c);

    std::cout << cc.GetDescription() << std::endl;
    std::cout << cc.GetCost() << std::endl;
}

输出:

Decaf with Soy with Caramel
7

// Expected:
// Decaf with Soy with Caramel with Caramel
// 9

这是相同的代码，但使用指针并且工作正常:
https://ideone.com/7fpGSp

Answer 1

随着从指针到引用的切换，OP 构造函数签名变得与(默认)复制构造函数非常相似。

    CondimentDecorator(Beverage &beverage) : beverage(beverage) {}

对比

    CondimentDecorator(const Beverage&); // generated by compiler

首先，我认为删除复制构造函数就足够了，但编译器仍然尝试使用已删除的构造函数，因为它不能再使用相应的投诉。

最后，我能够通过提供 resp 来解决 OP 的问题。阻止使用复制构造函数的候选对象。

(实际上不再需要删除复制构造函数，但我把它留在了。)

class CondimentDecorator : public Beverage
{
public:

    Beverage& beverage;
    CondimentDecorator(Beverage &beverage) : beverage(beverage) {}
    CondimentDecorator(CondimentDecorator &beverage) : beverage(beverage) {}
    CondimentDecorator(const CondimentDecorator&) = delete;
};

对于派生类也必须这样做:

class CaramelDecorator : public CondimentDecorator
{
public:

    CaramelDecorator(Beverage &beverage) : CondimentDecorator(beverage) {}
    CaramelDecorator(CaramelDecorator &beverage) : CondimentDecorator(beverage) {}
    //CaramelDecorator(const CaramelDecorator&) = delete;

    virtual std::string GetDescription() const override
    {
        return this->beverage.GetDescription() + " with Caramel";
    }

    virtual int GetCost() const override
    {
        return this->beverage.GetCost() + 2;
    }
};

我只修复了 CaramelDecorator用于演示，但实际上，这必须对 class CondimentDecorator 的所有派生类进行。 .

OP的固定MCVE:

#include <iostream>
//#include "Decaf.h"
//#include "Espresso.h"
//#include "SoyDecorator.h"
//#include "CaramelDecorator.h"

class Beverage
{
public:
    virtual std::string GetDescription() const = 0;
    virtual int GetCost() const = 0;
};

class CondimentDecorator : public Beverage
{
public:

    Beverage& beverage;
    CondimentDecorator(Beverage &beverage) : beverage(beverage) {}
    CondimentDecorator(CondimentDecorator &beverage) : beverage(beverage) {}
    CondimentDecorator(const CondimentDecorator&) = delete;
};

class Espresso : public Beverage
{
    virtual std::string GetDescription() const override
    {
        return "Espresso";
    }

    virtual int GetCost() const override
    {
        return 5;
    }
};

class Decaf : public Beverage
{
    virtual std::string GetDescription() const override
    {
        return "Decaf";
    }

    virtual int GetCost() const override
    {
        return 4;
    }
};

class CaramelDecorator : public CondimentDecorator
{
public:

    CaramelDecorator(Beverage &beverage) : CondimentDecorator(beverage) {}
    CaramelDecorator(CaramelDecorator &beverage) : CondimentDecorator(beverage) {}
    //CaramelDecorator(const CaramelDecorator&) = delete;

    virtual std::string GetDescription() const override
    {
        return this->beverage.GetDescription() + " with Caramel";
    }

    virtual int GetCost() const override
    {
        return this->beverage.GetCost() + 2;
    }
};

class SoyDecorator : public CondimentDecorator
{
public:

    SoyDecorator(Beverage &beverage) : CondimentDecorator(beverage) {}

    virtual std::string GetDescription() const override
    {
        return this->beverage.GetDescription() + " with Soy";
    }

    virtual int GetCost() const override
    {
        return this->beverage.GetCost() + 1;
    }
};

int main()
{
    Decaf d;
    SoyDecorator s(d);
    CaramelDecorator c(s);
    CaramelDecorator cc(c);

    std::cout << cc.GetDescription() << std::endl;
    std::cout << cc.GetCost() << std::endl;
}

输出:

Decaf with Soy with Caramel with Caramel
9

Live Demo on coliru

为什么需要额外的候选人？
CondimentDecorator源自 Beverage .

因此对于:

CondimentDecorator d;
CondimentDecorator d2(d);

编译器有两种选择来构造 d2 :

自定义构造函数 CondimentDecorator::CondimentDecorator(Beverage &beverage)

(默认)复制构造函数 CondimentDecorator::CondimentDecorator(const CondimentDecorator&) .

对于第一个，必须应用隐式强制转换，但对于复制构造函数，不需要强制转换(或至多是 const 强制转换)。

因此，编译器更喜欢复制构造函数(不幸的是，即使它被删除了)。

因此，必须提供另一个候选人，它需要像复制构造函数这样的隐式强制转换:

另一个自定义构造函数 CondimentDecorator::CondimentDecorator(CondimentDecorator&) .

进一步阅读:Overload Resolution