CODEWITHSUNDEEP
c++pointerssmart-pointers
Alexander
Alexander

Reputation: 779

How can I determine type of object pointer points to?

Im playing with my smart pointer class and I want to implement ++ and -- operators with the following behavior: If pointer points to single variable or if it points to array and ++(--) moves pointer out of array bounds an exception should be thrown when trying to ++(--).

Something like that:

class A;

SmartPtr<A> s(new A[3]);
SmartPtr<A> s1(new A());
++s;//ok
--s;//ok
--s;//exception OutOfBounds thrown
++s1;//exception OutOfBounds thrown
--s1;//exception OutOfBounds thrown

I tried to use typeid. But it returns A type anyway.

A* arr=new A[3];
typeid(arr).name();//type is P1A
typeid(--arr).name();//type is P1A
typeid(arr+7).name();//type is P1A

So is there any way to determine does pointer point to "my" type of object after ++(--)?

Upvotes: 1

Views: 184

Answers (4)

gbjbaanb
gbjbaanb

Reputation: 52679

stop using raw pointers/raw arrays and start using a vector. Then you can get the size of the vector and simply store the position of the element in your smart pointer class.

Upvotes: 1

James Hopkin
James Hopkin

Reputation: 13973

You can only do this kind of bounds checking if you use a utility function, rather than array new directly, e.g.:

template <class Type, std::size_t size>
SmartPtr<Type> MakeArrayPtr()
{
  return SmartPtr<Type>(new Type[size], size);
}

If the smart pointer deletes the object, remember to have it use delete[] in the case of arrays.

Upvotes: 1

Juho
Juho

Reputation: 1006

No, if you want to implement this kind of behaviour, you must store somewhere the size of the allocated array.

Upvotes: 1

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272497

new A[3] returns an A*, just like new A does. So you can't distinguish them.

If you want your class to do bounds checking, then you will need to explicitly tell it how many items are in the array.

Upvotes: 3

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