CODEWITHSUNDEEP
c++genericstemplatesdynamic
genesys
genesys

Reputation: 3647

c++ getting dynamic generic type of pointer?

the title probably is misleading, but i didn't really know how to name it.

let's say I have the following structs

template <typename T>
struct SillyBase{
   void doFunnyStuff(vector<T> vec){
        dummyField = T();
        for(int i=0; i<10; i++)
            vec.push_back(dummyField++);
    }
    T dummyField;
 };

struct A : public SillyBase<char>{};

struct B : public SillyBase<float>{};

now let's further assume i have a pointer

ISillyBase* ptr;

which is pointing to an object of a DECENDANT class (A or B) of SillyBase - however, i DON'T KNOW which one (i just know it's either A or B);

Is there ANY way for me to call doFunnyStuff() ?

maybe something like:

vector<dynamic_generic_type_of(ptr)> vec;
ptr->doFunnyStuff(vec);

thanks!

Upvotes: 3

Views: 911

Answers (6)

Jeff
Jeff

Reputation: 1

I would probably go with the dynamic_cast keyword. Although you would run into a problem when you declare the vector. It would have to be of type SillyBase. The problem with this is that this requires the use of template arguments, which would be unkown and impossible to get. To remedy this problem in the past, I've just put the vector in the Base class of type base, and then providing a functions to modify it. The inheireted class would provide a variable of the same name of the vector's in the base class. The inheireted class, inheiriting the base class's functions would hopefully fill the correct vector. I don't know if this is a good idea or a bad idea, but it's the way I attempted to overcome this problem within my projects.

Upvotes: 0

speedplane
speedplane

Reputation: 16141

I think what you're looking for is the built in dynamic_cast template function. It dynamically determines the type of a data structure and returns null if it is not an instantiation. The syntax is 

dynamic_cast<the_template_definition>(the_template_instantiation)

The one important catch is that the template classes must derive from a common virtual class. So to do what you want, you can use the following code:

#include <stdio.h>
class Top /* The parent class */ {
public: 
virtual ~Top() {}; // You need one base virtual function in the base class
};

template <typename T> class SillyBase : public Top {
   void doFunnyStuff(T dummyVar){
        dummyField  = dummyField + dummyVar;
    }
    T dummyField;
 };

class A : public SillyBase<char>{};
class B : public SillyBase<float>{};

int main(){
   A a;
   Top *ptr = (Top*)(&a);
   if(dynamic_cast<A*>(ptr) != NULL)
      printf("ptr is of type A*\n");
   if(dynamic_cast<B*>(ptr) != NULL)
      printf("ptr is of type B*\n");
   return 0;
}

The output of the program will be:

ptr is of type A*

Upvotes: 0

stefanB
stefanB

Reputation: 79840

In your example you can't have SillyBase* because SillyBase is defined as 

template <typename T> struct SillyBase {...}

So you need to provide type ...

Another problem is that you pass a copy of vector<T> into doFunnyStuff() which you then populate ... that does not seems right because when the method returns you lose your vec, was it supposed to be reference vector<T>& ?

Upvotes: 1

Gregory Pakosz
Gregory Pakosz

Reputation: 70234

EDIT: even with ISillyBase, there is no such thing as virtual template members so you won't be able to define one that takes vector<T> parameter.

Well in your example, there is no such non template type as SillyBase hence you cannot declare SillyBase* ptr;

And in the end I don't really see what you're trying to achieve :/

Upvotes: 0

alemjerus
alemjerus

Reputation: 8268

These are actually two absolutely different classes. There is no way to call function by name in classes that have met 100 years ago on their aunt wedding. You have to redesign the solution to include some base class for your template classes like

class SillyBaseBase
{
    public:
        void function doFunnyFunnyStuff(SillyBase<char> *)
        {
            SillyBase<char>::doFunnyStuff();
        }

        void function doFunnyFunnyStuff(SillyBase<float> *)
        {
            SillyBase<float>::doFunnyStuff();
        }

}

Upvotes: 0

Jerry Coffin
Jerry Coffin

Reputation: 490398

The SillyBase *ptr; shouldn't even compile. Without a template parameter list, SillyBase isn't a type, so attempting to use it in a context that requires a type should fail.

Since you can't create the pointer in the first place, there's no real answer to how to dereference it.

Upvotes: 0

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