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HomeBackend DevelopmentC++How to handle runtime type information in C++ generic programming?

In C++ generic programming, two methods are provided for handling run-time type information (RTTI): The dynamic_cast operator is used to convert a base class pointer or reference to a derived class pointer or reference. The typeid operator returns the type information of an object, and the type name can be obtained through its name() member function. RTTI, while convenient, incurs additional overhead and is therefore only recommended when needed, keeping in mind the binary compatibility issues it may cause.

C++ 泛型编程中如何处理运行时类型信息?

Handling runtime type information (RTTI) in C++ generic programming

In C++ generic programming, we often need Gets type information for an object or reference variable at runtime. C++ provides a runtime type information (RTTI) mechanism for this purpose.

Using dynamic_cast

The dynamic_cast operator is used to convert a base class pointer or reference to a derived class pointer or reference. If the conversion is successful, it returns a pointer or reference to the derived class; otherwise, it returns nullptr.

class Base { };
class Derived : public Base { };

int main() {
  Base* base_ptr = new Derived();

  // 检查 base_ptr 是否指向 Derived 对象
  Derived* derived_ptr = dynamic_cast<Derived*>(base_ptr);
  if (derived_ptr != nullptr) {
    // 转换成功,base_ptr 指向 Derived 对象
  }
}

Use typeid

typeid operator to return the type information of the object, which is a std::type_info Object. You can use the name() member function to obtain the type name, and you can use the before() and after() member functions to compare types.

class Base { };
class Derived : public Base { };

int main() {
  Base obj;
  std::cout << typeid(obj).name() << std::endl; // 输出:Base

  // 检查 obj 是否属于 Derived 类型
  if (typeid(obj).before(typeid(Derived))) {
    std::cout << "obj is not a Derived object" << std::endl;
  }
}

Notes on using RTTI

  • RTTI incurs additional overhead, so it is recommended to only use it when needed.
  • RTTI may cause binary compatibility issues, so be careful when using RTTI in libraries.

Practical case

Scenario:There is a set of shapes (such as circles, rectangles and triangles) that need to be executed differently according to their types operation.

Code:

class Shape {
public:
  virtual void draw() = 0;
};

class Circle : public Shape {
public:
  void draw() override {
    std::cout << "Drawing a circle" << std::endl;
  }
};

class Rectangle : public Shape {
public:
  void draw() override {
    std::cout << "Drawing a rectangle" << std::endl;
  }
};

class Triangle : public Shape {
public:
  void draw() override {
    std::cout << "Drawing a triangle" << std::endl;
  }
};

int main() {
  std::vector<Shape*> shapes{new Circle, new Rectangle, new Triangle};

  for (auto shape : shapes) {
    // 使用 RTTI 获取形状类型
    std::cout << "Drawing a " << typeid(*shape).name() << std::endl;

    // 根据类型调用相应的方法
    shape->draw();
  }
}

Output:

Drawing a Circle
Drawing a Rectangle
Drawing a Triangle

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