// Does this leak `lambdaObject`?

// An interface for any type with an implemented operator().
template<typename ReturnType, typename... Args>
struct Function {
    virtual ReturnType operator() (Args... args) const = 0;
};

template<typename LambdaType, typename ReturnType, typename... Args>
struct Lambda: public Function<ReturnType, Args...> {
    LambdaType lambdaObject;
    ReturnType (LambdaType::*rawFnPtr)(Args...)const;

    Lambda(const LambdaType &lambdaObject):
    lambdaObject(lambdaObject),
    rawFnPtr(&LambdaType::operator()) {}

    ReturnType operator() (Args... args) const {
        return (lambdaObject.*rawFnPtr)(args...);
    }
};