这完全是使用 dynamic_cast 的错误位置.您应该使用多态.每个 Animal 类都应该具有一个 virtual 函数，例如 process ，在这里您应该只调用 animal-> process().

This is EXACTLY the wrong place to use dynamic_cast. You should be using polymorphism. Each of the Animal classes should have a virtual function, say, process and here you should just call animal->process().

class Animal {
    virtual void Process() = 0;
}

class Cat : public Animal {
    void Process() { std::cout << " I am a tiny cat"; }
}

class Bear : public Animal {
    void Process() { std::cout << "I am a big bear"; }
}

void func(Animal * animal) {
    if (animal != nullptr) { animal->Process(); }
}

其他问题.

如果 animal 是 Dog ，但由于错误导致 animal_type 说它是 Cat ，该怎么办?

What if animal is a Dog, but due to a bug animal_type says its a Cat?

有时需要 static_cast ，并且如果可能的话，请使用它代替 dynamic_cast .动态转换具有静态转换所没有的额外性能成本.为此，您必须确保知道传入的类型，因为 static_cast 更不安全.

There are times when static_cast is necessary, and if possible use it instead of dynamic_cast. Dynamic cast has the additional performance cost that static cast does not. For this, you need to be sure you know the type that is coming in, since static_cast is more unsafe.

至少， animal_type 应该是 Animal 的成员.