修复算术中的循环依赖 class

Question

我有一组 classes 实现奇怪的重复模板模式。然而，诀窍在于基础 class 需要 return 个子 class 实例。这是一个例子：

template <typename SubType>
class ArithmeticBase
{
public:
    template <typename OtherType>
    const Addition operator+(const OtherType &other)
        {return Addition(get_subclass(), other);}

    // ...
    // Operators for subtraction, multiplication, division, ...

private:
    const SubType &get_subclass() const
        {return *static_cast<const SubType*>(this);}
};

template <typename OperatorType1, typename OperatorType2>
class Addition : ArithmeticBase<Addition<OperatorType1, OperatorType2>>
{
public:
    Addition(const OperatorType1 &op1, const OperatorType2 &op2)
        : op1(op1)
        , op2(op2)
    {}

private:
    const OperatorType1 &op1;
    const OperatorType2 &op2;
};

// ...
// Additional classes for subtraction, multiplication, division, ...

编译失败，因为 Addition class 在 ArithmeticBase class:

中使用之前未定义

arithmetic.cpp:6:8: error: unknown type name 'Addition'
        const Addition operator+(const OtherType &other)
              ^

我该如何解决这个问题？

Answer 1

您可以在基础 class 之前转发声明 Addition。

template <typename OperatorType1, typename OperatorType2>
class Addition;
template <typename SubType>
class ArithmeticBase
{
...
};

这允许编译器知道有一个类型 Addition 在它被定义之前就已经存在。

Answer 2

除了转发声明 Addition class（如 bhzag 的回答所示），您还需要将 operator+ 的定义移动到定义之后 [=16] =].否则你会在下一行得到一个错误。

确保定义在头文件中。如果不是，您将收到链接器错误。

Answer 3

或使用在Addition之后声明的非成员形式:

template <typename OperatorType1, typename OperatorType2>
class Addition;

template <typename SubType>
class ArithmeticBase
{
public:

     template <typename OneType, typename OtherType>
     friend const Addition<OneType, OtherType> operator+(const ArithmeticBase<OneType>& one, const OtherType &other);

private:
    const SubType &get_subclass() const
    {
        return *static_cast<const SubType*>(this);
    }
};

class ArithmeticType : public ArithmeticBase < ArithmeticType > {};

template <typename OperatorType1, typename OperatorType2>
class Addition : ArithmeticBase<Addition<OperatorType1, OperatorType2>>
{
public:
    Addition(const OperatorType1 &op1, const OperatorType2 &op2)
        : op1(op1)
        , op2(op2)
    {}

private:
    const OperatorType1 &op1;
    const OperatorType2 &op2;
};


template <typename OneType, typename OtherType>
const Addition<OneType, OtherType> operator+(const ArithmeticBase<OneType>& one, const OtherType &other)
{
    return Addition<OneType, OtherType>(one.get_subclass(), other);
}

int main()
{
    ArithmeticType a, b;
    a + b;
}

修复算术中的循环依赖 class

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