C ：如何覆寫具有相同介面的特定類的方法-有解無憂

我有一種情況，我需要從兩個具有相同介面的類繼承，但要分別覆寫它們，我絕對不能調整介面。請參閱下面的代碼示例

template<typename T>
struct Foo
{
    virtual ~Foo() = default;
    virtual void foo() = 0;
};

struct Derived : public Foo<int>, public Foo<double>
{
#if 0 // having something like this would be great, but unfortunately it doesn't work
    void Foo<int>::foo() override
    {
        std::cout << "Foo<int>::foo()" << std::endl;
    }

    void Foo<double>::foo() override
    {
        std::cout << "Foo<double>::foo()" << std::endl;
    }
#endif
};

uj5u.com熱心網友回復：

您始終可以定義宣告自己的介面的中間類：

template<typename T>
struct Foo
{
    virtual ~Foo() = default;
    virtual void foo() = 0;
};

struct ProxyFooInt : public Foo<int>
{
    virtual void fooInt() = 0;
    
    void foo() override
    {
        return fooInt();
    }
};

struct ProxyFooDouble : public Foo<double>
{
    virtual void fooDouble() = 0;
    
    void foo() override
    {
        return fooDouble();
    }
};

struct Derived : public ProxyFooInt, public ProxyFooDouble
{
    void fooInt() override
    {
        std::cout << "Foo<int>::foo()" << std::endl;
    }

    void fooDouble() override
    {
        std::cout << "Foo<double>::foo()" << std::endl;
    }
};

更高級的解決方案是使用 CRTP：

template<typename D>
struct CrtpFooInt : public Foo<int>
{
    void foo() override
    {
        return static_cast<D*>(this)->fooInt();
    }
};

template<typename D>
struct CrtpFooDouble : public Foo<double>
{
    void foo() override
    {
        return static_cast<D*>(this)->fooDouble();
    }
};

struct Derived : public CrtpFooInt<Derived>, public CrtpFooDouble<Derived>
{
    void fooInt()
    {
        std::cout << "Foo<int>::foo()" << std::endl;
    }

    void fooDouble()
    {
        std::cout << "Foo<double>::foo()" << std::endl;
    }
};

uj5u.com熱心網友回復：

使用 CRTP 解決此問題的另一種創造性方法是使用添加一些多載的能力來區分的不同呼叫foo()：


template<typename T>
struct Foo
{
    virtual ~Foo() = default;
    virtual void foo() = 0;
};

template<typename T>
struct Tag{};

template<typename T>
struct Crtp : public Foo<T>
{
    virtual void foo(Tag<T>) = 0;

    void foo() override
    {
        foo(Tag<T>{});
    }
};

struct Derived : public Crtp<int>, public Crtp<double>
{
    void foo(Tag<int>) override
    {
        std::cout << "Foo<int>::foo()" << std::endl;
    }

    void foo(Tag<double>) override
    {
        std::cout << "Foo<double>::foo()" << std::endl;
    }
};

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標籤：C 遗产界面

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