What is type erasure in C++?

Question

So I was reading this article about type erasure. But the code in that article seems partially incorrect, for example:

Repaired link

template <typename T>
class AnimalWrapper : public MyAnimal
{
    const T &m_animal;

public:
    AnimalWrapper(const T &animal)
        : m_animal(animal)
    { }

    const char *see() const { return m_animal.see(); }
    const char *say() const { return m_animal.say(); }
};

followed by

void pullTheString()
{
    MyAnimal *animals[] = 
    {
        new AnimalWrapper(Cow()), /* oO , isn't template argument missing? */
        ....
    };
}

These mistakes discouraged me from reading any further in the article.

Anyways; can anyone please teach what type erasure in C++ means, with simple examples?

I wanted to learn about it to understand how std::function works, but couldn't get my head around it.

Answer 1

Here's a very simple example of type erasure in action:

// Type erasure side of things

class TypeErasedHolder
{
  struct TypeKeeperBase
  {
    virtual ~TypeKeeperBase() {}
  };

  template <class ErasedType>
  struct TypeKeeper : TypeKeeperBase
  {
    ErasedType storedObject;

    TypeKeeper(ErasedType&& object) : storedObject(std::move(object)) {}
  };

  std::unique_ptr<TypeKeeperBase> held;

public:
  template <class ErasedType>
  TypeErasedHolder(ErasedType objectToStore) : held(new TypeKeeper<ErasedType>(std::move(objectToStore)))
  {}
};

// Client code side of things

struct A
{
  ~A() { std::cout << "Destroyed an A\n"; }
};

struct B
{
  ~B() { std::cout << "Destroyed a B\n"; }
};

int main()
{
  TypeErasedHolder holders[] = { A(), A(), B(), A() };
}

[Live example]

As you can see, TypeErasedHolder can store objects of an arbitrary type, and destruct them correctly. The important point is that it does not impose any restrictions on the types supported⁽¹⁾: they don't have to derive from a common base, for example.

---

⁽¹⁾ Except for being movable, of course.