Unique class type Id that is safe and holds across library boundaries

Question

I would appreciate any help as C++ is not my primary language.

I have a template class that is derived in multiple libraries. I am trying to figure out a way to uniq

Answer 1

In the modern C++ (03 - assuming you're using a recent compiler like gcc) you can use the typeid keyword to get a type_info object that provides basic type informations at least at runtime - that's a standard (and then cross-platform) feature.

I took the example from wikipedia and added a template/inheritance check, it seems to works well but i'm not certain for the int version (that is a hack exploiting the assumption that the compiler will have the types names somewhere in a read only memory space...that might be a wrong assumption).

The string identifier seems far better for cross-platform identification, if you can use it in you case. It's not cross-compiler compatible as the name it gives you is "implementation defined" by the standard - as suggested in comments.

The full test application code:

#include 
#include   //for 'typeid' to work

class Person 
{
public:
   // ... Person members ...
   virtual ~Person() {}
};

class Employee : public Person 
{
   // ... Employee members ...
};

template< typename DERIVED >
class Test
{
public:
    static int s_id()
    {
        // return id unique for DERIVED
        // NOT SURE IT WILL BE REALLY UNIQUE FOR EACH CLASS!!
        static const int id = reinterpret_cast(typeid( DERIVED ).name());
        return id;
    }

    static const char* s_name()
    {
        // return id unique for DERIVED
        // ALWAYS VALID BUT STRING, NOT INT - BUT VALID AND CROSS-PLATFORM/CROSS-VERSION COMPATBLE
        // AS FAR AS YOU KEEP THE CLASS NAME
        return typeid( DERIVED ).name();
    }
};

int wmain () 
{
    Person person;
    Employee employee;
    Person *ptr = &employee;



    std::cout << typeid(person).name() << std::endl;   // Person (statically known at compile-time)
    std::cout << typeid(employee).name() << std::endl; // Employee (statically known at compile-time)
    std::cout << typeid(ptr).name() << std::endl;      // Person * (statically known at compile-time)
    std::cout << typeid(*ptr).name() << std::endl;     // Employee (looked up dynamically at run-time
                                                    // because it is the dereference of a pointer to a polymorphic class)

    Test test;
    std::cout << typeid(test).name() << std::endl;    
    std::cout << test.s_id() << std::endl;    
    std::cout << test.s_id() << std::endl;    
    std::cout << test.s_id() << std::endl;    
    std::cout << test.s_name() << std::endl;    

    Test< Person > test_person;
    std::cout << test_person.s_name() << std::endl;    
    std::cout << test_person.s_id() << std::endl;    

    Test< Employee > test_employee;
    std::cout << test_employee.s_name() << std::endl;    
    std::cout << test_employee.s_id() << std::endl;    

    Test< float > test_float;
    std::cout << test_float.s_name() << std::endl;    
    std::cout << test_float.s_id() << std::endl;    


    std::cin.ignore();
    return 0;
}

Outputs :

class Person
class Employee
class Person *
class Employee
class Test
3462688
3462688
3462688
int
class Person
3421584
class Employee
3462504
float
3462872

This works at least on VC10Beta1 and VC9, should work on GCC. By the way, to use typeid (and dynamic_cast) you have to allow runtime type infos on your compiler. It should be on by default. On some plateform/compiler (I'm thinking about some embedded hardwares) RTTI is not turned on because it have a cost, so in some extreme cases you'll have to find a better solution.