c++ - 解决过载问题的原因是什么？

我有一个程序，其中有很多嵌套的if / switch语句，这些语句在多个地方重复出现。我试图将其提取出来，并将开关放入模板方法类中，然后允许客户端重载他们想使用重载专门处理的哪个开关分支：

class TraitsA {};
class TraitsB : public TraitsA {};

class Foo
{
    bool traitsB;
public:
    // Whether or not a Foo has traitsB is determined at runtime. It is a
    // function of the input to the program and therefore cannot be moved to
    // compile time traits (like the Iterators do)
    Foo() : traitsB(false) {}
    virtual ~Foo() {}
    bool HasTraitsB() const { return traitsB; }
    void SetTraitsB() { traitsB = true; }
};

class SpecificFoo : public Foo
{
};

template <typename Client> //CRTP
class MergeFoo
{
protected:
    Foo DoMerge(Foo&, const Foo&, int, TraitsA)
    {
        // Do things to merge generic Foo
    }
public:
    // Merge is a template method that puts all the nasty switch statements
    // in one place.
    // Specific mergers implement overloads of DoMerge to specify their
    // behavior...
    Foo Merge(Foo* lhs, const Foo* rhs, int operation)
    {
        const Client& thisChild = *static_cast<const Client*>(this);

        SpecificFoo* lhsSpecific = dynamic_cast<SpecificFoo*>(lhs);
        const SpecificFoo* rhsSpecific = dynamic_cast<const SpecificFoo*>(rhs);

        // In the real code these if's are significantly worse
        if (lhsSpecific && rhsSpecific)
        {
            if (lhs->HasTraitsB())
            {
                return thisChild.DoMerge(*lhsSpecific,
                               *rhsSpecific,
                               operation,
                               TraitsB());
            }
            else
            {
                return thisChild.DoMerge(*lhsSpecific,
                               *rhsSpecific,
                               operation,
                               TraitsA());
            }
        }
        else
        {
            if (lhs->HasTraitsB())
            {
                return thisChild.DoMerge(*lhs, *rhs, operation, TraitsB());
            }
            else
            {
                return thisChild.DoMerge(*lhs, *rhs, operation, TraitsA());
            }
        }
    }
};

class ClientMergeFoo : public MergeFoo<ClientMergeFoo>
{
    friend class MergeFoo<ClientMergeFoo>;
    Foo DoMerge(SpecificFoo&, const SpecificFoo&, int, TraitsA)
    {
        // Do things for specific foo with traits A or traits B
    }
};

class ClientMergeFooTwo : public MergeFoo<ClientMergeFoo>
{
    friend class MergeFoo<ClientMergeFooTwo>;
    Foo DoMerge(SpecificFoo&, const SpecificFoo&, int, TraitsB)
    {
        // Do things for specific foo with traits B only
    }
    Foo DoMerge(Foo&, const Foo&, int, TraitsA)
    {
        // Do things for specific foo with TraitsA, or for any Foo
    }
};

但是，这无法编译（至少在ClientMergeFooTwo的情况下），表示无法将Foo＆转换为SpecificFoo＆。有什么想法会导致转换失败，而不是在MergeFoo中选择完美的泛型重载吗？

编辑：嗯，鉴于我尝试编写它的速度，这个psuedocode示例显然做得不好。我已经纠正了一些错误...

最佳答案

有什么想法为什么会失败，而不是在MergeFoo中选择完美的泛型重载？


是的，因为有名称隐藏规则。如果派生类中的函数与基类中的函数具有相同的名称，则基类函数为“隐藏”，它甚至不查看所涉及函数的参数。

就是说，解决方案很简单：在派生类中提供基类版本，并在公共部分添加简单的using MergeFoo::DoMerge即可。