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C++ Cookbook
book

C++ Cookbook

by D. Ryan Stephens, Christopher Diggins, Jonathan Turkanis, Jeff Cogswell
November 2005
Beginner to intermediate content levelBeginner to intermediate
594 pages
16h 23m
English
O'Reilly Media, Inc.
Content preview from C++ Cookbook

8.6. Determining an Object’s Type at Runtime

Problem

At runtime, you need to interrogate dynamically the type of particular class.

Solution

Use runtime type identification (commonly referred to as RTTI) to query the address of the object for the type of object it points to. Example 8-6 shows how.

Example 8-6. Using runtime type identification

#include <iostream>
#include <typeinfo>

using namespace std;

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

int main() {

   Base b, bb;
   Derived d;

   // Use typeid to test type equality
   if (typeid(b) == typeid(d)) { // No
      cout << "b and d are of the same type.\n";
   }
   if (typeid(b) == typeid(bb)) { // Yes
      cout << "b and bb are of the same type.\n";
   }
   if (typeid(d) == typeid(Derived)) { // Yes
      cout << "d is of type Derived.\n";
   }
}

Discussion

Example 8-6 shows you how to use the operator typeid to determine and compare the type of an object. typeid takes an expression or a type and returns a reference to an object of type_info or a subclass of it (which is implementation defined). You can use what is returned to test for equality or retrieve a string representation of the type’s name. For example, you can compare the types of two objects like this:

if (typeid(b) == typeid(d)) {

This will return true if the type_info objects returned by both of these are equal. This is because typeid returns a reference to a static object, so if you call it on two objects that are the same type, you will get two references to the same thing, which is why the equality test ...

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Publisher Resources

ISBN: 0596007612Supplemental ContentErrata Page