访问类中的值类似于boost :: any [英] Accessing Values in a Class Similar to boost::any

问题描述

我为教育目的做一个简单的 boost :: any 类的类，但我不知道如何访问存储的值。我可以完美地设置值，但是当我尝试访问holder类中的任何成员时，编译器只是抱怨该成员在其派生类中找不到。我不能将成员声明为 virtual 因为模板。

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.

这里是相关的代码：

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."
    }
};

推荐答案

将虚拟函数添加到模板函数本身不能是模板。模板类或结构仍然可以有虚拟函数。你需要使用dynamic_cast的魔力。

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."
    }
};

如果更改Get以返回 Datatype * 你可以返回 NULL 而不是throwing。您还没有处理过 StoredValue 的先前值的记忆，但我将其留给您。

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.

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