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Buckeye

Deleting pointers to inherited objects

3 posts in this topic

EDIT: topic subject should be "...inheriting objects." (How do I change the subject line?)

 

Question: Why did I not get memory leaks when deleting a base class pointer? Please note that the code below is poorly formed and I know that. My question is "why did it work anyway?"

 

Background:

 

Using VS2013 Express for Desktop, Windows 7, I am programming a graphics editor in C++ and DirectX 9. I had a situation similar to:

class UndoRedo
{
   UndoRedo();
   ~UndoRedo(); // deletes pointers in actions

   std::vector<BaseClass*> actions;

   // other functions, e.g., rule of three, etc.
};

class BaseClass
{
   virtual void Undo() { }
   virtual void Redo() { }
};

class Derived : BaseClass
{
   Derived( int inX, int inY, int inZ ) : x(inX), y(inY), z(inZ) { }
   void Undo();
   void Redo();
   // e.g.,
   int x, y, z;
};


Implementation of actions was initiated as:

Derived* derived = new Derived( ... data ... );
UndoRedoPtr->actions.push_back( (BaseClass*)derived );


The destructor for UndoRedo:

for( size_t i = 0; i < actions.size(); i++ ) delete actions[i];


Why are there no memory leaks, considering that the allocation for Derived should be larger than for BaseClass?

 

I eventually DID create another derived class, the deletion of which DID result in a memory leak. That was resolved by changing BaseClass to:

class BaseClass
{
   BaseClass() { }
   virtual ~BaseClass() { }
   virtual void Undo() { }
   virtual void Redo() { }
private:
   EditAction(const EditAction&) { }
   EditAction& operator= (const EditAction&) { }
};


In spite of my original best efforts to incorrectly implement BaseClass, the process did not result in memory leaks for any but the last derived class I added.

 

So, again, why did I not get memory leaks when deleting BaseClass* without a BaseClass virtual destructor when that pointer was, in fact, Derived* ?

 

EDIT: Is there, e.g., a minimum ("new") allocation block which gets deleted whether the entire allocation is used or not?

Edited by Buckeye
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You are aware of the fact that you should have virtual destructors in order to avoid memory leaks, but I think you are misunderstanding the reason why.

 

The presence of a virtual destructor helps make sure that the destructor of your derived class is called. This is very important if your derived class has allocated some memory or other resources. If the virtual destructor is missing, the derived class's destructor isn't called and resources that the derived class has allocated are not released.

 

It sounds like you think that the virtual destructor is involved in the process of telling delete the size of the class it's going to delete. It's not involved in that process at all. Instead, typically, when you ask for a block of memory with new, a memory manager will also allocate a small header (say 16 bytes) which is located directly before the pointer that it returns to you. The size of the object that was allocated will be encoded within that header. When you call delete on your pointer, the memory manager can subtract 16 bytes and read the size from the header block.

 

So, this question "Is there, e.g., a minimum ("new") allocation block which gets deleted whether the entire allocation is used or not?" is kind of irrelevant to explaining the behaviour you're seeing, but probably yes, there is often a minimum allocation size or alignment. So allocating 100 bytes with 100 calls to new, will likely use a couple of KB at least due to the header overhead and alignment/minimum sizes.

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Deleting a derived object through a pointer to the base when there is no virtual destructor in the base class is undefined behavior. This means anything can happen including memory being freed properly.

Indeed. Asking why something works if it's behavior is undefined is moot. Thanks for the response, albeit highlights my ignorance. As explained below, none of the classes involve explicit allocations.
 
 
 

The presence of a virtual destructor helps make sure that the destructor of your derived class is called. This is very important if your derived class has allocated some memory or other resources. If the virtual destructor is missing, the derived class's destructor isn't called and resources that the derived class has allocated are not released.



Actually, I do understand that. By design, none of the derived classes allocate memory, and none have destructors. Some of the construction parameters are std::vector, but none of  those vectors are for objects that perform explicit allocations themselves. I.e., they're all of type <int>, <unsigned int>, or structures comprised only of same.
 
Appreciate both responses. Edited by Buckeye
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