Can you delete a pointer without newing it up?

[asimorio]

Hi all,
If I don't new up a pointer, can I delete it?
see code below:

void A::foo()
{
char* buffer;
delete buffer;
return;
}
regards,
Vynce

[Rolf Magnus]

asimorio wrote:

Hi all,
If I don't new up a pointer, can I delete it?
Well, you can, but the behavior will be undefined unless the pointer was a
null pointer. In that case, nothing is done.
see code below:

void A::foo()
{
char* buffer;
delete buffer;
return;
}

[Ian Collins]

asimorio wrote:

Hi all,
If I don't new up a pointer, can I delete it?
see code below:

void A::foo()
{
char* buffer;
delete buffer;

You can try, but your toilet might explode.

--
Ian Collins.

[Jens Müller]

asimorio wrote:

Hi all,
If I don't new up a pointer, can I delete it?
see code below:

void A::foo()
{
char* buffer;
delete buffer;
return;
}

No you can't, the behaviour of your

delete buffer;

is undefined.

What exactly is the purpose of that code?

[uraniumore235]

Now lets think about it. When you instantiate a pointer variable, you
obviously want to point it to some defined memory location. Since you
did not new-up or NULL-up a memory location, the pointer basically has
an undefined property. When you delete this pointer variable, which has
an undefined property, you are basically telling the compiler to return
the pointed to location ??? to the heap stack. Does this even make
sense? No!

asimorio wrote:

Hi all,
If I don't new up a pointer, can I delete it?
see code below:

void A::foo()
{
char* buffer;
delete buffer;
return;
}
regards,
Vynce

[asimorio]

Well, now i modify a bit my code:
Will it make sense?

void A::Foo(void* a)
{
char* buffer = (char*) a;
delete buffer;
return;
}
regards,
Vynce

[Rolf Magnus]

ur***********@h otmail.com wrote:

Now lets think about it. When you instantiate a pointer variable, you
obviously want to point it to some defined memory location. Since you
did not new-up or NULL-up a memory location, the pointer basically has
an undefined property. When you delete this pointer variable, which has
an undefined property, you are basically telling the compiler to return
the pointed to location ??? to the heap stack. Does this even make
sense? No!

That reminds me of the Chewbacca defense :-)

[Rolf Magnus]

asimorio wrote:

Well, now i modify a bit my code:
Will it make sense?

void A::Foo(void* a)
{
char* buffer = (char*) a;
delete buffer;
return;
}

Yes, as long as A::Foo is only called with a pointer that points to a single
char allocated with new.

[Ian Collins]

asimorio wrote:

Well, now i modify a bit my code:
Will it make sense?

void A::Foo(void* a)
{
char* buffer = (char*) a;
delete buffer;
return;
}

Even if your toilet doesn't explode, you program will surely crash. You
are telling delete to free memory that doesn't belong to the heap.

--
Ian Collins.

[Ian Collins]

Ian Collins wrote:

asimorio wrote:

Well, now i modify a bit my code:
Will it make sense?

void A::Foo(void* a)
{
char* buffer = (char*) a;
delete buffer;
return;
}

Even if your toilet doesn't explode, you program will surely crash. You
are telling delete to free memory that doesn't belong to the heap.

Oops, didn't see the pointer passed in as a parameter.

--
Ian Collins.