Question

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Is it safe to delete a void pointer?

Say I have a new allocation to a class called MyClass and allocation is as simple as:

MyClass *myClassPtr = new MyClass();

And I store reference to the list of void* where I simply say

myListOfPointers.add(static_cast<void*>(myClassPtr)); // this has to be void*

And later I release memory so instead of doing:

delete myClassPtr

I use:

delete MyListOfPointer.get(0)

(Say myClassPtr reference is at zero-index.) Also, please note that it has to be void* since this list can store different types of pointers so I wouldn’t know the type of pointer that I am deleting:

So I can’t do any thing like:

delete static_cast<MyClass*>(MyListOfPointer.get(0))

Is this way going to release the correct memory size? (sizeof(MyClass))?

Note:
I am not looking for any answer pointing to smart pointers.

Answer 1

Deleting through a void* results in undefined behavior, so you are guaranteed nothing.

5.3.5 Delete [expr.delete]

1 The delete-expression operator destroys a most derived object (1.8) or array created by a new-expression.
[…]
The operand shall have a pointer to object type, or a class type having a single non-explicit conversion function (12.3.2) to a pointer to object type. The result has type void.⁷⁸

78) This implies that an object cannot be deleted using a pointer of type void* because void is not an object type.

Emphasis mine.

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So even though you said not to say it, the answer is to create some form of smart pointer. It would need to use type-erasure to hide the type externally (allowing the heterogeneous list), but internally keep track of the type it was given and how to delete it. Something much like boost::any.