超载操作员=在模板类型之间切换

Question

我有一个模板类，其中多个类正在继承，从根本上制造了一些用户友好的构建器，从而隐藏了在干燥时在不同构建器上不需要的功能。

但是，我在构建器类型之间切换时遇到了麻烦。我宁愿使用 operator = 切换而不麻烦，并具有以下代码。

template<class T> class BuilderFunctions
{
protected:
    std::string text;

public:
    template<class Other>
    BuilderFunctions<T>& operator=(const Other& other);
};

template <class T>
template <class Other>
BuilderFunctions<T>& BuilderFunctions<T>::operator=(const Other& other)
{
    // yes, I need same object protection
    text = other.text;
    return *this;
}

我可能会做我不应该在模板类

class BuilderGenericList : public BuilderFunctions<BuilderGenericList> 
{
public:
    BuilderGenericList() = default;
};

// Build configuration details
class BuilderRootList : public BuilderFunctions<BuilderRootList>
{
private:
    // I'll delete the functions I don't want accessed here

public:
    BuilderRootList() = default;
};

除了

// Real world you'd build the root first and switch out to another builder or be using a prototype to seed another builder
BuilderRootList cmakeRoot;
BuilderGenericList list;

// Separate to make sure it's not trying to use cop constructor
list = cmakeRoot;

中做的事情，尽管其他人似乎在操作员中取得了成功，所以我可能会假设某种错误。

---

更多信息：我遇到的错误是：

Error   C2679   binary '=': no operator found which takes a right-hand operand 
of type 'BuilderRootList' (or there is no acceptable conversion)

因此，它肯定是在寻找我的 operator = ，它只是没有从模板中生成一个

Answer 1

当您执行模板继承时，必须在基本类成员的情况下明确说明。 More read:

• Why我是否必须通过此指针访问模板基类成员？
• 派生的模板级访问基本成员data

在您的情况下，成员 text 和 operator = ，可以是使用声明通过带来。

class BuilderGenericList : public BuilderFunctions<BuilderGenericList> 
{
public:
    BuilderGenericList() = default;
    using BuilderFunctions<BuilderGenericList>::text;
    using BuilderFunctions<BuilderGenericList>::operator=;
    // ... so on, other members from base if needed!
};

// Build configuration details
class BuilderRootList : public BuilderFunctions<BuilderRootList>
{
public:
    BuilderRootList() = default;
    using BuilderFunctions<BuilderRootList>::text;
    using BuilderFunctions<BuilderRootList>::operator=;
    // ... so on, other members from base if needed!
};

live demo

请注意，这将使成员使用builderFunctions＆lt;文本公开，如果不是想要的，请在其他答案中考虑@jarod42 的建议。

Answer 2

builderFunctions＆lt; t1＆gt; 无法访问 private /受保护 builderFunctions＆lt; t2＆gt; （带有 t1 t1 ） /code>！= t2 ）：添加访问者，或使它们 friend 。

此外，您还必须使用 operator = ：

template<class T>
class BuilderFunctions
{
    template <typename U> friend class BuilderFunctions;
protected:
    std::string text;

public:
    template<class Other>
    BuilderFunctions<T>& operator=(const Other& other);
};

template <class T>
template <class Other>
BuilderFunctions<T>& BuilderFunctions<T>::operator=(const Other& other)
{
    //yes, I need same object protection
    text= other.text;
    return *this;
}


class BuilderGenericList : public BuilderFunctions<int>{
public:
    using BuilderFunctions<int>::operator=;

    BuilderGenericList() = default;
};

//Build configuration details
class BuilderRootList : public BuilderFunctions<float>
{
private:
//I'll delete the functions I don't want accessed here

public:
    using BuilderFunctions<float>::operator=;

    BuilderRootList() = default;
};

demo