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 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
.
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; }
This code will output the following:
obj is an instance of Foo
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; }
This code will output the following:
foo could not be cast to Bar
Because foo
points to an instance of the Foo
class, not an instance of the Bar
class.
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