How can we call "delete this; " in a const-member function?

Question

I saw the code snippet as follows:

class UPNumber {
public:
  UPNumber();
  UPNumber(int initValue);
  ...

  // pseudo-destructor (a const member function, because
  // even const objects may be destroyed)
  void destroy() const { delete this; } // why this line is correct???

  ...

private:
  ~UPNumber();
};

First, I am sure that above class definition is correct. Here is my question, why we can define the function 'destroy' as above? The reason being asking is that why we can modify 'this' in a const-member function?

Telpher

Answer 1

The const qualifier applied to a method have the effect of making the this passed to it a const pointer; in particular, in your case it will be a const UPNumber *.

Still, this is not a problem for delete: actually you can use delete on a const pointer without having to cast anything, as specified at §5.3.5 ¶2:

[Note: a pointer to a const type can be the operand of a delete-expression; it is not necessary to cast away the constness (5.2.11) of the pointer expression before it is used as the operand of the delete-expression. ]

Notice that, before the standard was completed, there have been many discussion about whether this was or wasn't a good idea, so some pre-standard compilers will issue an error when trying to delete const pointers.

The idea behind allowing this behavior is that otherwise you would have no way to delete const objects without using a const_cast; see this question for more info.

Sackett

Answer 2

That works for the same reason that this work:

const int* upn = new int();
delete upn;

You can delete a pointer to a const-qualified object. The const-qualification on the member function just means that this has a type const UPNumber*.

Bobwhite

@James, here, we cannot call "delete upn" b/c the destructor is private. -- thx – Telpher

@q0987: Class members have access to private members. The function destroy requires access to the destructor, because of the delete this expression, and has access because it's a member of the class. – Embellishment

@GMan: I think the OP was complaining about my example, which at first used his UPNumber class instead of int. – Bobwhite

@James: When reading your code I thought you were trying to make a reference to the old UPN network. :) – Imes

Answer 3

Where did you get the idea that we are modifying this? In fact, this is not an lvalue. It It cannot ever be modified, regardless of whether the member function is const or not.

Applying delete-expression to a pointer (any pointer, not just this) is not in any way considered a modification of that pointer. Moreover, the argument of delete-expression is treated as rvalue, meaning that it cannot possibly be modified by delete.

So, there are no problems with that application of delete in your code.

Lieu

if I understood your point right, the const-member function ONLY promises NOT change the values pointed by this. --thx – Telpher

I think this is more than needs to be said about the fact that the questioner has mistakenly written "why we can modify 'this' in a const-member function?", despite meaning something more like, "why can we modify the object through this". The unspoken assumption being that deleting an object is a form of modification (and really it is, since the destructor is free to modify members on its way out). – Usher

Answer 4

As it is technically possible and defined to call delete this provided that you are careful, the constness of the "suicide" method is technically pointless, since the object must not be touched after the call.

That's why it is semantically incorrect to have a const method calling delete this.

An interesting point brought Steve Jessep's comment. Running the destructor doesn't keep an object unchanged, so applying the concept of constness is questionable.

Decrepitude