Finding the type of an object in C++

Question

I have a class A and another class that inherits from it, B. I am overriding a function that accepts an object of type A as a parameter, so I have to accept an A. However, I later call functions that only B has, so I want to return false and not proceed if the object passed is not of type B.

What is the best way to find out which type the object passed to my function is?

Answer 1

dynamic_cast should do the trick

TYPE& dynamic_cast<TYPE&> (object);
TYPE* dynamic_cast<TYPE*> (object);

The dynamic_cast keyword casts a datum from one pointer or reference type to another, performing a runtime check to ensure the validity of the cast.

If you attempt to cast to pointer to a type that is not a type of actual object, the result of the cast will be NULL. If you attempt to cast to reference to a type that is not a type of actual object, the cast will throw a bad_cast exception.

Make sure there is at least one virtual function in Base class to make dynamic_cast work.

Wikipedia topic Run-time type information

RTTI is available only for classes that are polymorphic, which means they have at least one virtual method. In practice, this is not a limitation because base classes must have a virtual destructor to allow objects of derived classes to perform proper cleanup if they are deleted from a base pointer.

Answer 2

Dynamic cast is the best for your description of problem, but I just want to add that you can find the class type with:

#include <typeinfo>

...
string s = typeid(YourClass).name()

Answer 3

This is called RTTI, but you almost surely want to reconsider your design here, because finding the type and doing special things based on it makes your code more brittle.

Answer 4

Just to be complete, I'll build off of Robocide and point out that typeid can be used alone without using name():

#include <typeinfo>
#include <iostream>

using namespace std;

class A {
public:
    virtual ~A() = default; // We're not polymorphic unless we
                            // have a virtual function.
};
class B : public A { } ;
class C : public A { } ;

int
main(int argc, char* argv[])
{
    B b;
    A& a = b;

    cout << "a is B: " << boolalpha << (typeid(a) == typeid(B)) << endl;
    cout << "a is C: " << boolalpha << (typeid(a) == typeid(C)) << endl;
    cout << "b is B: " << boolalpha << (typeid(b) == typeid(B)) << endl;
    cout << "b is A: " << boolalpha << (typeid(b) == typeid(A)) << endl;
    cout << "b is C: " << boolalpha << (typeid(b) == typeid(C)) << endl;
}

Output:

a is B: true
a is C: false
b is B: true
b is A: false
b is C: false

Answer 5

You are looking for dynamic_cast<B*>(pointer)

Answer 6

Probably embed into your objects an ID "tag" and use it to distinguish between objects of class A and objects of class B.

This however shows a flaw in the design. Ideally those methods in B which A doesn't have, should be part of A but left empty, and B overwrites them. This does away with the class-specific code and is more in the spirit of OOP.

Answer 7

Because your class is not polymorphic. Try:

struct BaseClas { int base; virtual ~BaseClas(){} };
class Derived1 : public BaseClas { int derived1; };

Now BaseClas is polymorphic. I changed class to struct because the members of a struct are public by default.

Answer 8

Your description is a little confusing.

Generally speaking, though some C++ implementations have mechanisms for it, you're not supposed to ask about the type. Instead, you are supposed to do a dynamic_cast on the pointer to A. What this will do is that at runtime, the actual contents of the pointer to A will be checked. If you have a B, you'll get your pointer to B. Otherwise, you'll get an exception or null.

Answer 9

As others indicated you can use dynamic_cast. But generally using dynamic_cast for finding out the type of the derived class you are working upon indicates the bad design. If you are overriding a function that takes pointer of A as the parameter then it should be able to work with the methods/data of class A itself and should not depend on the the data of class B. In your case instead of overriding if you are sure that the method you are writing will work with only class B, then you should write a new method in class B.

Answer 10

If you can access boost library, maybe type_id_with_cvr() function is what you need, which can provide data type without removing const, volatile, & and && modifiers. Here is an simple example in C++11:

#include <iostream>
#include <boost/type_index.hpp>

int a;
int& ff() 
{
    return a;
}

int main() {
    ff() = 10;
    using boost::typeindex::type_id_with_cvr;
    std::cout << type_id_with_cvr<int&>().pretty_name() << std::endl;
    std::cout << type_id_with_cvr<decltype(ff())>().pretty_name() << std::endl;
    std::cout << typeid(ff()).name() << std::endl;
}

Hope this is useful.

Answer 11

Use overloaded functions. Does not require dynamic_cast or even RTTI support:

class A {};
class B : public A {};

class Foo {
public:
    void Bar(A& a) {
        // do something
    }
    void Bar(B& b) {
        Bar(static_cast<A&>(b));
        // do B specific stuff
    }
};