Deleting void pointer

**Bargi** · 07-18-2008

Hi,

I have a following program:

Code:

#include <iostream>
using namespace std;
class Object {
   void* data; // Some storage
   const int size;
   const char id;
public:
   Object(int sz, char c) : size(sz), id(c) {
     data = new char[size];
     cout << "Constructing object " << id
          << ", size = " << size << endl;
   }
   ~Object() {
     cout << "Destructing object " << id << endl;
     delete []data; 
   }
};
int main() {
   Object* a = new Object(40, 'a');
   delete a;
   void* b = new Object(40, 'b');
   delete b;
}

I just want to know that on deleting pointer b what will happen ????

This program is running not calling b destructor. Then what is deleted by "delete b " ???

Thanks

**Dino** · 07-18-2008

Pay attention to your warnings:

**Bargi** · 07-18-2008

Ya...... I know that............
But I want to know that what b is deleting ?????
Surely it is not freeing the m/m since destructor is not called .....

Thanks

**matsp** · 07-18-2008

Originally Posted by Bargi

Ya...... I know that............
But I want to know that what b is deleting ?????
Surely it is not freeing the m/m since destructor is not called .....

Thanks

It would free the memory itself.

Just like:

Code:

char *p = new char [20];
... 
delete p;

Since your void * doesn't have a destructor, there is no call to the destructor.

--
Mats

**anon** · 07-18-2008

If it's undefined as the warning says, how can you be sure that it indeed frees the memory?

**matsp** · 07-18-2008

Originally Posted by anon

If it's undefined as the warning says, how can you be sure that it indeed frees the memory?

Well, to be technical about it, I can't be SURE. But I do expect delete to actually free the memory related to the memory allocation.

However, there's very rarely a good need to use void pointers to point to objects, and not being able to run the destructor is of course a hugely bad thing.

--
Mats

**cpjust** · 07-18-2008

Originally Posted by Bargi

Hi,

I have a following program:

Code:

#include <iostream>
using namespace std;
class Object {
   void* data; // Some storage
   const int size;
   const char id;
public:
   Object(int sz, char c) : size(sz), id(c) {
     data = new char[size];
     cout << "Constructing object " << id
          << ", size = " << size << endl;
   }
   ~Object() {
     cout << "Destructing object " << id << endl;
     delete []data; 
   }
};
int main() {
   Object* a = new Object(40, 'a');
   delete a;
   void* b = new Object(40, 'b');
   delete b;
}

I just want to know that on deleting pointer b what will happen ????

This program is running not calling b destructor. Then what is deleted by "delete b " ???

Thanks

Is there any particular reason why you're using a void* instead of a char* ?

**Elysia** · 07-18-2008

Well, as we would have it, operator delete takes a void* (although I'm not sure if the standard says this).
Therefore, technically it would be able to free the memory of the object since it is allocated by operator new.
The only thing missing is the destructor.

**matsp** · 07-18-2008

Originally Posted by Elysia

Well, as we would have it, operator delete takes a void* (although I'm not sure if the standard says this).
Therefore, technically it would be able to free the memory of the object since it is allocated by operator new.
The only thing missing is the destructor.

I believe the standard says that operator ::delete() would take a void pointer. It is the keyword delete, not operator function delete that causes the call to the destructor - by this, I meant hat the compiler generates code that essentially does this:

Code:

T *p = new T;
... 
delete p;
// becomes:
p->~T();
::delete(p);

The blue part is not code that we see in the source, but what the compiler actually generates when it sees the blue delete p;

Of course, if T is not a class type, then T doesn't have a ~T(), so there's no code for the compiler to call to delete the object. I had the same question as cpjust on my mind when I started reading this thread, and it's still unanswered.

--
Mats

**Bargi** · 07-18-2008

I had the same question as cpjust on my mind when I started reading this thread, and it's still unanswered.

Well, reason for posting is that I want to know that whether the memory is freed or there is a m/m leak.

**Elysia** · 07-19-2008

The behavior is undefined.
Never call delete on a void pointer.
Instead, if you need to work with generic types, use templates.

**King Mir** · 07-19-2008

You can use ::delete(ptr) to free the memory pointed to by a void pointer. However, unless the value pointed to is a POD type, a primitive, or has been previously destroyed, there is likely to be a problem resulting from an object contained in your object using dynamic memory or other resource.

**CornedBee** · 07-19-2008

No, you can't. That's still undefined behaviour.

You can use ::operator delete(ptr) to just call the memory deallocation function.

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