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making an object from an unknown type?

 
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making an object from an unknown type?
The Architect 2.0
post 15 Jul, 2008 - 08:54 PM
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how would you go about making an object of an unspecified type?

basically, i want to 'take' the type of a STL container and make an object that is of the same type as its elements, WITHOUT knowing what the type is.

I need to add another element into the container, without knowing the actual type of the object.

ex.

if i was given a list<Dog> conainer, the function would create a Dog object. if i was given a list<String> container, i would be given a String. all in a generic manner


also, provided the above is possible, how would i initiate its default constructor(provided it's not 'covered' in the declaration).
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captainhampton
post 16 Jul, 2008 - 05:07 AM
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I think you may be hinting towards the use of templates:
http://www.cplusplus.com/doc/tutorial/templates.html
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polymath
post 16 Jul, 2008 - 07:22 AM
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Definatly look at that link.
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The Architect 2.0
post 16 Jul, 2008 - 03:35 PM
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cpp

template<class T>
T& makeObject(list<T> &container) 
{
	T newobject = new T;
	return newobject;	
}


basically, what i want to do is that. i think its pretty obvious what its doing, but just incase; it is taking a container and making a new object on the free store that is the same type as the container's element type.

i THINK the way i written it should result in ONE object bring created that still exists after the function returns and that the reference being returned is still valid.
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skaoth
post 16 Jul, 2008 - 05:51 PM
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Almost.

You don't need to pass in a list of type T to create a new object of type T. The template declaration should take care of that. Here is a fixed version
CODE

template<class T>  
T* makeObject()  
{  
     T* newobject = new T();  
     return newobject;    
}
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The Architect 2.0
post 16 Jul, 2008 - 07:01 PM
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um....i want to make an object from w/e type the list is using as its value-type
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skaoth
post 16 Jul, 2008 - 07:18 PM
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what does w/e mean?

*Edit

I think I understand what your looking for.
You are correct in saying that you need to pass in the list of type T if you want the compiler to implicitly determine the type.

CODE

template<class T>  
T* makeObject(list<T> list)  
{  
     T* newobject = new T();  
     return newobject;    
}

// ...  in main
void main(void)
{
        std::list<double> mylist;
    double *d = makeObject(mylist);
}


Then again I don't know the context (I would have to see more) from which you are making the call to makeObject().
If you are doing it like the example above then there is no reason to pass in the list to
the function as it could be called like makeObject<double>()

This post has been edited by skaoth: 16 Jul, 2008 - 08:10 PM
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The Architect 2.0
post 16 Jul, 2008 - 08:40 PM
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thanks skaoth. the pointer solution works fine.

but what is the issue with using a reference?
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