通过将基类强制转换为派生并设置数据来扩展基类

Question

Assume I have this:

class Base {
public:
    int a;
    Base() : a(5) {}
};

template<class T>
class Derived : public Base {
public:
    T value;
};

The code below works but I want to know what can be the challenges of using such approach:

Base * base = new Base;
Derived<int> * derived = static_cast<Derived<int>*>(base);
derived->value = 5;
Derived<String> * derived1 = static_cast<Derived<String>*>(base);
derived1->value = "test";

Derived<String> * newderived = static_cast<Derived<String>*>(base);
std::cout << newderived->value;
Derived<int> * newderived1 = static_cast<Derived<int>*>(base);
std::cout << newderived1->value;
//Output: test1

Or how can I achieve such thing in a different, safer way. I want to pass a class through 5 functions that will manipulate it.

Answer 1

What you're doing here will fail horribly at some point because the size of the derived class is larger than the base class and you write after the end of the base class. The above write operation will overwrite memory that belongs to another object. You can have a SetValue() method in the base class and implement it in the derived class.

Answer 2

The code does not work. All you objects even after casting are still a Base because you constructed them as a base. The casts just say: Hey, I know it's a Derived<xxx> , so please just interpret that this way. You don't know this here, in fact you know it is NOT a Derived .

To properly use the objects, you need to create a Derived<xxx> and afterwards cast. If you use a dynamic_cast here all cases should come back as null as they are Base .

Given that you wanted to "pass a class through 5 functions" you'd probably want the inverted setup. Create Derived<xxx> objects and hold them as a pointer to Base . This works without casting as it should. Then pass the Base* through your functions. Polymorphism will take care that everything works fine.