这种重载分辨率头痛的原因是什么？ [英] What is the cause of this overload resolution headache?

问题描述

我有一个程序，我有很多嵌套的if / switch语句，在几个地方重复。我试图提取出来，并把开关放在一个模板方法类，然后允许客户端重载他们想要使用重载特别处理的开关分支：

I've got a program where I've got a lot of nested if/switch statements which were repeated in several places. I tried to extract that out and put the switches in a template method class, and then allow clients to overload which switch branches they wanted to specifically handle using overloading:

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 'case），说它不能转换Foo&转换为SpecificFoo&任何想法为什么它不能转换，而不是在 MergeFoo ？

However, this fails to compile (At least in ClientMergeFooTwo's case), saying it cannot convert a Foo& into a SpecificFoo&. Any ideas why it's failing that conversion instead of choosing the perfectly good generic overload in MergeFoo?

编辑中选择完美的通用重载：好吧，这个伪代码示例显然没有这么好，给定我有多快我试图写它。我已更正了一些错误...

EDIT: Well, this psuedocode example apparently didn't do so well given how fast I tried to write it. I have corrected some of the mistakes...

推荐答案

任何想法，在MergeFoo中选择完美的通用重载？

Any ideas why it's failing that conversion instead of choosing the perfectly good generic overload in MergeFoo?

是的，因为名称隐藏规则。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。

Yes, because of name hiding rules. If a function in the derived class has the same name as a function in the base class, the base class function is "hidden", it doesn't even look at the parameters of the involved function.

也就是说，解决方案很简单：使用一个简单的在公共部分使用MergeFoo :: DoMerge 在派生类中提供基类版本。

That said, the solution is easy: Make the base class version available in the derived class with a simple using MergeFoo::DoMerge in the public part.

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