Template parameter interpreted as base, not derived

[User 2601640]

I have a template function to check the type passed to it. I am templatizing it on a derived type but when I break in the function, the template parameter is the base class. A sample app is below. Can someone please explain where I am going wrong? Thanks

class Base
{
public:
virtual ~Base() {}
};

class Derived1 : public Base {};

class Derived2 : public Base {};

template<class T> bool CheckType(boost::shared_ptr<Base> pBase)
{
T* pTemplate = NULL; //Why is pTemplate Base* and not Derived1* when passing a Derived1 to template parameter?
Base* pBaseRaw = pBase.get();
if(typeid(pBaseRaw) == typeid(pTemplate))
{
return true;
}

return false;

}

int _tmain(int argc, _TCHAR* argv[])
{
boost::shared_ptr<Base> pBase1(new Derived1);
boost::shared_ptr<Base> pBase2(new Derived2);
bool bIsDerived = CheckType<Derived1>(pBase1);
bIsDerived = CheckType<Derived2>(pBase2);

return 0;

}

[Stuart Dootson]

A few questions:

1. If I understand you correctly, CheckType is returning true, but you expected false?
2. What compiler (& version) are you using?

I've just tried your code with g++ 4.0.1 and it gives the results I'd expect. I added some code to print out type names:

template<class T> bool CheckType(boost::shared_ptr<Base> pBase)
{
std::cout << typeid(T).name() << std::endl;
T* pTemplate = NULL; //Why is pTemplate Base* and not Derived1* when passing a Derived1 to template parameter?
Base* pBaseRaw = pBase.get();
std::cout << typeid(pBaseRaw).name() << std::endl;
if(typeid(pBaseRaw) == typeid(pTemplate))
{
return true;
}

return false;

}

. I also printed out the result of CheckType and got these results:

8Derived1
P4Base
0
8Derived2
P4Base
0

So - the template parameter is Derived, the parameter is a pointer to Base, CheckType is returning false. I'm suspecting that the compiler you're using isn't fully standard compliant if it does something different?

[Rajkumar R]

Member 2603772 wrote:

T* pTemplate = NULL; //Why is pTemplate Base* and not Derived1* when passing a Derived1 to template parameter?

nope, you confused with the result, check by printing std::cout << typeid(pTemplate).name() I think you expected the result as true and you got false (bIsDerived = true you expected, as the pointer you passed is derived). to get what inside the pointer you can dereference it say typeid(*pBaseRaw) so your code can be,

template<class T> bool CheckType(boost::shared_ptr<Base> pBase)
{
Base* pBaseRaw = pBase.get();
if(typeid(*pBaseRaw) == typeid(T)) {
return true;
}
return false;
}

modified on Friday, January 23, 2009 1:13 PM

[User 2601640]

Thanks for the response. Your output is what I expected. I was looking at the pointer types in the debugger, not printing out the type info to the console. The print statements give me what you got. I am using VS2008 v9 SP1. I was performing the wrong check to get a type match. I changed the function to the following:

template<class T> bool CheckType(boost::shared_ptr<Base> pBase)
{
Base* pBaseRaw = pBase.get();
if(pBaseRaw == null)
{
return false;
}

if(typeid(\*pBaseRaw) == typeid(T)) //Now dereferencing pBaseRaw. Bad idea?
{
	return true;
}

return false;

}

I am now dereferencing the base class pointer returned from shared_ptr which casts it to the right type. I believe this is a sketchy implicit cast to the derived type. Should I just dynamic_cast or does the typeid comparison take care of type issues? Thanks:thumbsup:

[User 2601640]

Thanks. I just posted a similar result based on the previous response. I appreciate the help.:thumbsup:

[Stuart Dootson]

IIRC, typeid uses the same mechanism as dynamic_cast, so typeid on its own is fine.