C ++ std容器-没有指针的多态性.是否有可能? [英] C++ std containers - polymorphism without pointers. Is it possible?

问题描述

是否可以在不使用指针的情况下保持任何 std c ++容器中派生类的知识，从而动态地从容器中转换返回值?我知道我可以创建向量或某些基类类型的指针，并保留它们的子类.但是问题是我必须使用指针吗?

Is it possible to maintain knowledge of the derived class in any std c++ container without using pointers, dynamically casting return values from the container? I know I can create a vector or what ever of pointers of some base class type, and have them retain their sub classes. But the question is do I have to use pointers?

示例:

struct A {
  int x = 0, y = 0, z = 0;
  virtual void foo() { cout << "A" << endl; };
};

struct B : public A {
  int a = 1, b = 1, c = 1;
  virtual void foo() { cout << "B" << endl; };
};

int main() {  
  <SOMECONTAINER><A> a(2);
  a[0] = A();
  a[1] = B();
  B * p;
  B& n = dynamic_cast<B&>(a[1]); // Always throws?
  p = dynamic_cast<B*>(&a[1]); // Always zero?
  cout << p << endl;
}

推荐答案

是的，您必须使用指针.否则，尝试将B放入A的容器中会导致切片:B被切成A(这不仅限于容器，如果执行A a = B()或将B传递给期望A).

Yes, you do have to use pointers. Otherwise, attempting to put a B into a container of A results in slicing: the B gets cut down into an A (this is not limited to containers, the exact same thing happens if you do A a = B() or if you pass a B to a function expecting an A).

当您稍后将其取出时，它是一个A，它完全不知道其谱系包括类型为B的杰出祖先-无论您以何种方式看待A，您都可以使其成为B.

When you later take it back out, it's an A that has absolutely no knowledge its lineage includes an illustrious forefather of type B -- and no matter what way you look at an A, you can't make it a B.

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