Home Backend Development C++ Detailed explanation of C++ function inheritance: How to use RTTI to check type inheritance relationship?

Detailed explanation of C++ function inheritance: How to use RTTI to check type inheritance relationship?

May 03, 2024 pm 02:42 PM
c++ function inheritance

RTTI is a mechanism that allows you to check the inheritance of types at runtime. Using RTTI, you can obtain type information about an object, which is useful for verifying inheritance relationships. To check inheritance, use the typeid operator to obtain an object's type information. To perform dynamic conversions, use the dynamic_cast operator, which converts a base class pointer to a derived class pointer, returning a non-null pointer if the conversion is successful and a null pointer otherwise.

C++ 函数继承详解:如何使用 RTTI 来检查类型继承关系?

C Detailed explanation of function inheritance: How to use RTTI to check type inheritance relationships

Runtime type information (RTTI) is a Mechanism that allows you to obtain type information about an object while a C program is executing. This is useful for checking type inheritance relationships, such as during virtual function overrides or type conversions.

Basics of RTTI

To use RTTI, you need to include the <typeinfo> header file. This will give you two useful classes: typeid and dynamic_cast.

  • typeid: Returns information representing the given object type.
  • dynamic_cast: Allows you to dynamically convert a base class pointer to a derived class pointer.

Checking inheritance

To check whether an object inherits from another class, you can use the typeid operator. This operator returns a typeid object containing details about the object's type.

For example, the following code snippet checks whether obj is an instance of the Foo class:

#include <typeinfo>

class Foo {};

int main() {
  Foo obj;

  if (typeid(obj) == typeid(Foo)) {
    std::cout << "obj is an instance of Foo" << std::endl;
  } else {
    std::cout << "obj is not an instance of Foo" << std::endl;
  }

  return 0;
}
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This code will output the following:

obj is an instance of Foo
Copy after login

Dynamic Conversion

RTTI also allows you to convert a base class pointer to a derived class pointer at runtime. To do this, you can use the dynamic_cast operator.

dynamic_cast The operator accepts a pointer to a base class as its first argument and returns a pointer to a derived class as its result. If the cast is successful, dynamic_cast will return a non-null pointer to an instance of the derived class. Otherwise, it returns a null pointer.

For example, the following code snippet dynamically converts a foo pointer to a Bar class pointer:

#include <typeinfo>

class Foo {};
class Bar : public Foo {};

int main() {
  Foo* foo = new Foo();
  Bar* bar = dynamic_cast<Bar*>(foo);

  if (bar) {
    std::cout << "foo was successfully cast to Bar" << std::endl;
  } else {
    std::cout << "foo could not be cast to Bar" << std::endl;
  }

  delete foo;

  return 0;
}
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This code will output the following:

foo could not be cast to Bar
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Because foo points to an instance of the Foo class, not an instance of the Bar class.

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