CODEWITHSUNDEEP
c++crtp
dani
dani

Reputation: 3897

If I cast the this pointer of a base class to the derived type, is it safe and well defind?

If I have a template base class that takes the type of the derived class, is safe to cast the this pointer of this base class to the type of the derived class?

Consider this code, where the base class A casts the this pointer to a template parameter (Derived). It also checks if the provided type is actually derived from this class. It obviously works (here), but is it well defined?

#include <iostream>

class D;

template<typename Derived, typename T>
class A
{
public:
    Derived *thisToDerived() {
        static_assert(std::is_base_of< A<Derived, T>, Derived>::value, "not");
        return static_cast<Derived*>(this);
    }
private:
    T m;
};


class D : public A<D, int>
{
private:
    double d;
    float f;
};


int main() {
    D d;
    std::cout<<"this:    "<<&d<<"\nderived: "<<d.thisToDerived();
}

Upvotes: 0

Views: 303

Answers (1)

user0042
user0042

Reputation: 8018

If I have a template base class that takes the type of the derived class, is safe to cast the this pointer of this base class to the type of the derived class?
...
It obviously works (here), but is it well defined?

Yes, it's safe and is well defined. It's actually a well known and commonly used pattern (see CRTP) and aka Static Polymorphism.

Examples of usage:

Derived *thisToDerived() {
    // Your static_assert is superfluos
    // static_assert(std::is_base_of< A<Derived, T>, Derived>::value, "not");
    return static_cast<Derived*>(this);
        // ^^^^^^^^^^^^^^^^^^^^^
        // Already guarantees that Derived is inherited from A
}

Upvotes: 2

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