Question

我无法实现以下功能：

class Based {
  public:
    template <typename T> void foo(T t) { boo(t); }
    template <typename T> virtual void boo(T t) = 0; // Error!
 };

class Derived : public Based {
  public:
    template <typename T> void boo(T t) {
      std::cout<<"Derived::boo called\n";
    }
};

int main() {

  Derived derived;
  derived.foo(0);
  derived.foo('c');
  //etc.
}

我知道C ++不允许使用虚拟模板化方法，但看不到任何其他方法来执行此操作。任何帮助将不胜感激！

Answer 1

出于我的目的，我找到了这个解决方案。

template <typename X>
class Based {
  public:
    template <typename T> void foo(T t) { static_cast<X *>(this)->boo(t); }
};

class Derived : public Based<Derived> {
  public:
    template <typename T> void boo(T t) {
      std::cout<<"Derived::boo called\n";
    }
};

int main() {

  Derived derived;
  derived.foo(0);
  derived.foo('c');
  //etc.
}

感谢您的建议！

如何避免模板化虚拟方法？

1 个答案: