c++ - Implementing a "variant" class

Question

Note: I'm aware of boost::variant, but I am curious about the design principles. This question mostly for self-education.

Original post

At my current job I found an old variant class implementation. It's implemented with a union and can only support a handful of datatypes. I've been thinking about how one should go about designing an improved version. After some tinkering I ended up with something that seems to work. However I'd like to know your opinion about it. Here it is:

#include <iostream>
#include <map>
#include <stdexcept>
#include <string>
#include <typeinfo>
#include <boost/shared_ptr.hpp>

class Variant
{
public:
    Variant() { }

    template<class T>
    Variant(T inValue) :
        mImpl(new VariantImpl<T>(inValue)),
        mClassName(typeid(T).name())
    {
    }

    template<class T>
    T getValue() const
    {
        if (typeid(T).name() != mClassName)
        {
            throw std::logic_error("Non-matching types!");
        }

        return dynamic_cast<VariantImpl<T>*>(mImpl.get())->getValue();
    }

    template<class T>
    void setValue(T inValue)
    {
        mImpl.reset(new VariantImpl<T>(inValue));
        mClassName = typeid(T).name();
    }

private:
    struct AbstractVariantImpl
    {
        virtual ~AbstractVariantImpl() {}
    };

    template<class T>
    struct VariantImpl : public AbstractVariantImpl
    {
        VariantImpl(T inValue) : mValue(inValue) { }

        ~VariantImpl() {}

        T getValue() const { return mValue; }

        T mValue;
    };

    boost::shared_ptr<AbstractVariantImpl> mImpl;
    std::string mClassName;
};

int main()
{
    // Store int
    Variant v(10);
    int a = 0;
    a = v.getValue<int>();
    std::cout << "a = " << a << std::endl;

    // Store float
    v.setValue<float>(12.34);
    float d = v.getValue<float>();
    std::cout << "d = " << d << std::endl;

    // Store map<string, string>
    typedef std::map<std::string, std::string> Mapping;
    Mapping m;
    m["one"] = "uno";
    m["two"] = "due";
    m["three"] = "tre";
    v.setValue<Mapping>(m);
    Mapping m2 = v.getValue<Mapping>();
    std::cout << "m2["one"] = " << m2["one"] << std::endl;
    return 0;
}

Output is correct:

a = 10
d = 12.34
m2["one"] = uno

My SO questions are:

• Is this implementation correct?
• Will the dynamic cast in getValue() work as expected (I'm not certain)
• Should I return T as a const reference instead? Or can I count on return-value-optimization to kick in?
• Any other problems or suggestions?

Update

Thanks to @templatetypedef for his suggestions. This updated version only uses dynamic_cast to check if the types match. Type mismatches caused by differences in constness are now avoided thanks to the TypeWrapper classes (which I have shamelessly stolen from the Poco C++ project).

So this is the current version. It's likely to contain a few errors though, as I'm not familiar with the idea of modifying const/ref on template templates. I'll have a fresh look tomorrow.

template <typename T>
struct TypeWrapper
{
    typedef T TYPE;
    typedef const T CONSTTYPE;
    typedef T& REFTYPE;
    typedef const T& CONSTREFTYPE;
};

template <typename T>
struct TypeWrapper<const T>
{
    typedef T TYPE;
    typedef const T CONSTTYPE;
    typedef T& REFTYPE;
    typedef const T& CONSTREFTYPE;
};

template <typename T>
struct TypeWrapper<const T&>
{
    typedef T TYPE;
    typedef const T CONSTTYPE;
    typedef T& REFTYPE;
    typedef const T& CONSTREFTYPE;
};

template <typename T>
struct TypeWrapper<T&>
{
    typedef T TYPE;
    typedef const T CONSTTYPE;
    typedef T& REFTYPE;
    typedef const T& CONSTREFTYPE;
};

class Variant
{
public:
    Variant() { }

    template<class T>
    Variant(T inValue) :
        mImpl(new VariantImpl<typename TypeWrapper<T>::TYPE>(inValue))
    {
    }

    template<class T>
    typename TypeWrapper<T>::REFTYPE getValue()
    {
        return dynamic_cast<VariantImpl<typename TypeWrapper<T>::TYPE>&>(*mImpl.get()).mValue;
    }

    template<class T>
    typename TypeWrapper<T>::CONSTREFTYPE getValue() const
    {
        return dynamic_cast<VariantImpl<typename TypeWrapper<T>::TYPE>&>(*mImpl.get()).mValue;
    }

    template<class T>
    void setValue(typename TypeWrapper<T>::CONSTREFTYPE inValue)
    {
        mImpl.reset(new VariantImpl<typename TypeWrapper<T>::TYPE>(inValue));
    }

private:
    struct AbstractVariantImpl
    {
        virtual ~AbstractVariantImpl() {}
    };

    template<class T>
    struct VariantImpl : public AbstractVariantImpl
    {
        VariantImpl(T inValue) : mValue(inValue) { }

        ~VariantImpl() {}

        T mValue;
    };

    boost::shared_ptr<AbstractVariantImpl> mImpl;
};

Answer 1

This implementation is close to correct, but it looks like it has a few bugs. For example, this code:

if (typeid(T).name() != mClassName)

is not guaranteed to work correctly because the .name() function in type_info is not guaranteed to return a unique value for each type. If you want to check if the types match, you should probably use something like this:

if (typeid(*mImpl) == typeid(VariantImpl<T>))

Which more accurately checks if the type matches. Of course, you need to watch out for const issues, since storing a const T and storing a T will yield different types.

As for your question about dynamic_cast, in the case you've described you don't need to use the dynamic_cast because you already have a check to confirm that the type will match. Instead, you can just use a static_cast, since you've already caught the case where you have the wrong type.

More importantly, though, what you've defined here is an "unrestricted variant" that can hold absolutely anything, not just a small set of restricted types (which is what you'd normally find in a variant). While I really like this code, I'd suggest instead using something like Boost.Any or Boost.Variant, which has been extensively debugged and tested. That said, congrats on figuring out the key trick that makes this work!