访问类中的值类似于 boost::any

发布于 2024-10-17 20:49:55 字数 861 浏览 7 评论 0原文

我正在制作一个简单的类似 boost::any 的类用于教育目的,但我不知道如何访问存储的值。我可以完美地设置该值,但是当我尝试访问“holder”类中的任何成员时,编译器只是抱怨在它派生的类中找不到该成员。由于模板的原因,我无法将成员声明为virtual

这是相关代码:

class Element
{
    struct ValueStorageBase
    {
    };

    template <typename Datatype>
    struct ValueStorage: public ValueStorageBase
    {
        Datatype Value;

        ValueStorage(Datatype InitialValue)
        {
            Value = InitialValue;
        }
    };

    ValueStorageBase* StoredValue;

public:

    template <typename Datatype>
    Element(Datatype InitialValue)
    {
        StoredValue = new ValueStorage<Datatype>(InitialValue);
    }

    template <typename Datatype>
    Datatype Get()
    {
        return StoredValue->Value; // Error: "struct Element::ValueStorageBase" has no member named "Value."
    }
};

I'm making a simple boost::any-like class for educational purposes, but I can't figure out how to access the stored value. I can set the value perfectly, but when I try to access any member in the "holder" class the compiler just complains that the member wasn't found in the class it was derived from. I can't declare the members as virtual because of the templates.

Here's the relevant code:

class Element
{
    struct ValueStorageBase
    {
    };

    template <typename Datatype>
    struct ValueStorage: public ValueStorageBase
    {
        Datatype Value;

        ValueStorage(Datatype InitialValue)
        {
            Value = InitialValue;
        }
    };

    ValueStorageBase* StoredValue;

public:

    template <typename Datatype>
    Element(Datatype InitialValue)
    {
        StoredValue = new ValueStorage<Datatype>(InitialValue);
    }

    template <typename Datatype>
    Datatype Get()
    {
        return StoredValue->Value; // Error: "struct Element::ValueStorageBase" has no member named "Value."
    }
};

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评论(2

半葬歌 2024-10-24 20:49:55

将虚拟函数添加到模板中很好 - 只是函数本身不能是模板。模板化的类或结构仍然可以有虚函数。你需要使用dynamic_cast的魔力。

class Element
{
    struct ValueStorageBase
    {
        virtual ~ValueStorageBase() {}
    };

    template <typename Datatype>
    struct ValueStorage: public ValueStorageBase
    {
        Datatype Value;

        ValueStorage(Datatype InitialValue)
        {
            Value = InitialValue;
        }
    };

    ValueStorageBase* StoredValue;

public:

    template <typename Datatype>
    Element(Datatype InitialValue)
    {
        StoredValue = new ValueStorage<Datatype>(InitialValue);
    }

    template <typename Datatype>
    Datatype Get()
    {
        if(ValueStorage<DataType>* ptr = dynamic_cast<ValueStorage<DataType>*>(StoredValue)) {
            return ptr->Value;
        else
            throw std::runtime_error("Incorrect type!"); // Error: "struct Element::ValueStorageBase" has no member named "Value."
    }
};

如果您将 Get 更改为返回 Datatype*,则可以返回 NULL 而不是抛出异常。您还没有处理 StoredValue 先前值的内存,但我将其留给您。

It's fine to add virtual functions to templates- just the functions themselves cannot be templates. A templated class or struct can still have virtual functions just fine. You need to use the magic of dynamic_cast.

class Element
{
    struct ValueStorageBase
    {
        virtual ~ValueStorageBase() {}
    };

    template <typename Datatype>
    struct ValueStorage: public ValueStorageBase
    {
        Datatype Value;

        ValueStorage(Datatype InitialValue)
        {
            Value = InitialValue;
        }
    };

    ValueStorageBase* StoredValue;

public:

    template <typename Datatype>
    Element(Datatype InitialValue)
    {
        StoredValue = new ValueStorage<Datatype>(InitialValue);
    }

    template <typename Datatype>
    Datatype Get()
    {
        if(ValueStorage<DataType>* ptr = dynamic_cast<ValueStorage<DataType>*>(StoredValue)) {
            return ptr->Value;
        else
            throw std::runtime_error("Incorrect type!"); // Error: "struct Element::ValueStorageBase" has no member named "Value."
    }
};

If you change Get to return a Datatype* you can return NULL instead of throwing. You also haven't handled the memory of the previous value of StoredValue, but I'm leaving that up to you.

相对绾红妆 2024-10-24 20:49:55

您需要首先将其转换为ValueStorage
还将虚拟析构函数添加到 ValueStorageBase 类,以具有多态类。没有它,您将无法运行时检查您的转换是否正常:)。

之后你可以写:

template <typename Datatype>
    Datatype Element_cast()
    {
        //throw exception for a wrong cast
        if(typeid(*StoredValue)!=typeid(ValueStorage<Datatype>) )
               throw exception;

        //we already checked, that our type casting is OK,
        //So no need for dynamic_cast anymore
        return static_cast<ValueStorage*> (StoredValue)->value;
    }

You need to cast it to ValueStorage first.
Also add virtual destructur to ValueStorageBase class, to have polymorphic class. Without it you can't runtime check if your casting is OK :).

After it you can write:

template <typename Datatype>
    Datatype Element_cast()
    {
        //throw exception for a wrong cast
        if(typeid(*StoredValue)!=typeid(ValueStorage<Datatype>) )
               throw exception;

        //we already checked, that our type casting is OK,
        //So no need for dynamic_cast anymore
        return static_cast<ValueStorage*> (StoredValue)->value;
    }
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