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Alex Rivera
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If the classes below were not templates I could simply have x in the derived class. However, with the code below, I have to use this->x . Why? template <typename T> class base { protected: int x; }; template <typename T> class derived : public base<T> { public: int f() { return this->x; } }; int main() { derived<int> d; d.f(); return 0; }
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