C++: destructor and delete

**Andrew Ward** · Jul 19 '05, 04:56 PM

Re: destructor and delete

Where are these statements from?
You have to differentiate between objects created dynamically on the heap
with the 'new' operator, and objects created on the stack in a scope.
Using the delete operator instructs the program to call the destructor of
the class and then free the memory for the object. Objects that simply go
out of scope implicitely have the same process performed on them.

"Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message
news:3f532593_1 @news.iprimus.c om.au...[color=blue]
> Hi,
>
> Statement 1: "A dynamically created local object will call it's destructor
> method when it goes out of scope when a procedure returms"
>
> Agree.
>
>
> Statement 2: "A dynamically created object will call it's destructor when[/color]
it[color=blue]
> is made a target of a delete".
>
> Does not make sense to me. Would not this result in an endless loop? (ie.
> delete calls destructor and within destructor it calls delete, which in[/color]
turn[color=blue]
> calls the destructor again and so on....)
>
> Any comments appreciated.
>
> asasas
>
>
>[/color]

**John Harrison** · Jul 19 '05, 04:56 PM

Re: destructor and delete

"Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message
news:3f532593_1 @news.iprimus.c om.au...[color=blue]
> Hi,
>
> Statement 1: "A dynamically created local object will call it's destructor
> method when it goes out of scope when a procedure returms"
>
> Agree.
>
>
> Statement 2: "A dynamically created object will call it's destructor when[/color]
it[color=blue]
> is made a target of a delete".
>
> Does not make sense to me. Would not this result in an endless loop? (ie.
> delete calls destructor and within destructor it calls delete, which in[/color]
turn[color=blue]
> calls the destructor again and so on....)
>
> Any comments appreciated.
>
> asasas
>[/color]

Destructor does not call delete, so there is no loop.

john

**Josephine Schafer** · Jul 19 '05, 04:56 PM

Re: destructor and delete

"Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message
news:3f532593_1 @news.iprimus.c om.au...[color=blue]
> Hi,
>
> Statement 1: "A dynamically created local object will call it's destructor
> method when it goes out of scope when a procedure returms"
>
> Agree.[/color]

Consider this code -

#include <iostream>
class A{
public:
A()
{
}
~A()
{
std::cout << "Inside D'tor";
}
};

void f()
{
A a;
A *ptrA = new A;
}

int main ()
{
f ();
}

If you mean that when f() returns two destructors will be called then it's
wrong.
Destructor would be called only for the object created on the heap.
In fact this is a memory leak.

[color=blue]
> Statement 2: "A dynamically created object will call it's destructor when[/color]
it[color=blue]
> is made a target of a delete".
>[/color]

That's true.

[color=blue]
> Does not make sense to me. Would not this result in an endless loop? (ie.
> delete calls destructor and within destructor it calls delete, which in[/color]
turn[color=blue]
> calls the destructor again and so on....)
>[/color]
Destructor does not call delete unless in your destructor you write -
delete this;

HTH,
J.Schafer

**Josephine Schafer** · Jul 19 '05, 04:56 PM

Re: destructor and delete

[color=blue]
> Destructor would be called only for the object created on the heap.[/color]

I meant stack ofcourse ;-).

**Newsnet Customer** · Jul 19 '05, 04:56 PM

Re: destructor and delete

> "Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message[color=blue]
> news:3f532593_1 @news.iprimus.c om.au...[color=green]
> > Hi,
> >
> > Statement 1: "A dynamically created local object will call it's[/color][/color]
destructor[color=blue][color=green]
> > method when it goes out of scope when a procedure returms"
> >
> > Agree.
> >
> >
> > Statement 2: "A dynamically created object will call it's destructor[/color][/color]
when[color=blue]
> it[color=green]
> > is made a target of a delete".
> >
> > Does not make sense to me. Would not this result in an endless loop?[/color][/color]
(ie.[color=blue][color=green]
> > delete calls destructor and within destructor it calls delete, which in[/color]
> turn[color=green]
> > calls the destructor again and so on....)
> >
> > Any comments appreciated.
> >
> > asasas
> >[/color]
>
> Destructor does not call delete, so there is no loop.
>
> john[/color]

Generally speaking, you have a delete in a destructor method. That said, a
delete calls the destructor and the endless cycles continues. I'm sure there
is something im missing.

sasas

**Newsnet Customer** · Jul 19 '05, 04:56 PM

Re: destructor and delete

[color=blue]
> Where are these statements from?
> You have to differentiate between objects created dynamically on the heap
> with the 'new' operator, and objects created on the stack in a scope.
> Using the delete operator instructs the program to call the destructor of
> the class and then free the memory for the object.[/color]

and whats USUALLY in the destructor method? a delete statement also,and
hence the loop.

sasasasas

**Peter van Merkerk** · Jul 19 '05, 04:56 PM

Re: destructor and delete

> > Destructor does not call delete, so there is no loop.[color=blue][color=green]
> >[/color]
> Generally speaking, you have a delete in a destructor method. That[/color]
said, a[color=blue]
> delete calls the destructor and the endless cycles continues. I'm sure[/color]
there[color=blue]
> is something im missing.[/color]

A destructor never calls the delete operator on itself (this), hence no
endless loop. In a destructor the delete operator (if any) is only used
for other objects owned by that class instance, not for itself.

--
Peter van Merkerk
peter.van.merke rk(at)dse.nl

**Scott Condit** · Jul 19 '05, 04:56 PM

Re: destructor and delete

Newsnet Customer wrote:[color=blue][color=green]
>>Where are these statements from?
>>You have to differentiate between objects created dynamically on the heap
>>with the 'new' operator, and objects created on the stack in a scope.
>>Using the delete operator instructs the program to call the destructor of
>>the class and then free the memory for the object.[/color]
>
>
> and whats USUALLY in the destructor method? a delete statement also,and
> hence the loop.[/color]

What loop?

delete statements can often be found in destructors, but they
do not target the object whose destructor it is.

S

**Newsnet Customer** · Jul 19 '05, 04:56 PM

Re: destructor and delete

[color=blue][color=green]
> > Destructor would be called only for the object created on the heap.[/color]
>
> I meant stack ofcourse ;-).[/color]

incorrect. as indicated in statement 1. an object's (dynamically created
....meaning HEAP) destructor is called when the object goes out of scope.

**Josephine Schafer** · Jul 19 '05, 04:56 PM

Re: destructor and delete

"Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message
news:3f534b75_1 @news.iprimus.c om.au...[color=blue]
>
>
>[color=green][color=darkred]
> > > Destructor would be called only for the object created on the heap.[/color]
> >
> > I meant stack ofcourse ;-).[/color]
>
> incorrect. as indicated in statement 1. an object's (dynamically created
> ...meaning HEAP) destructor is called when the object goes out of scope.[/color]

Why don't you run the code I posted and see for yourself if the destructor
gets called.
--
JS

**Karl Heinz Buchegger** · Jul 19 '05, 04:56 PM

Re: C++: destructor and delete

Newsnet Customer wrote:[color=blue]
>
> Hi,
>
> Statement 1: "A dynamically created local object will call it's destructor
> method when it goes out of scope when a procedure returms"
>[/color]

A dynamically created local object ...

int main()
{
std::string *p;

p = new std::string;

.... calls it's destructor method when it goes out of scope

well. p goes out of scope. But that does not mean, that the
dynamically created object, where p points to is destroyed.
[color=blue]
> Agree.
>
> Statement 2: "A dynamically created object will call it's destructor when it
> is made a target of a delete".[/color]

p = new std::string;

...

delete p;

the string which is pointed to by p is destroyed.
[color=blue]
>
> Does not make sense to me. Would not this result in an endless loop? (ie.
> delete calls destructor and within destructor it calls delete, which in turn
> calls the destructor again and so on....)[/color]

Why should this happen?

--
Karl Heinz Buchegger
kbuchegg@gascad .at

**John Harrison** · Jul 19 '05, 04:56 PM

Re: destructor and delete

"Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message
news:3f533d83_1 @news.iprimus.c om.au...[color=blue][color=green]
> > "Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message
> > news:3f532593_1 @news.iprimus.c om.au...[color=darkred]
> > > Hi,
> > >
> > > Statement 1: "A dynamically created local object will call it's[/color][/color]
> destructor[color=green][color=darkred]
> > > method when it goes out of scope when a procedure returms"
> > >
> > > Agree.
> > >
> > >
> > > Statement 2: "A dynamically created object will call it's destructor[/color][/color]
> when[color=green]
> > it[color=darkred]
> > > is made a target of a delete".
> > >
> > > Does not make sense to me. Would not this result in an endless loop?[/color][/color]
> (ie.[color=green][color=darkred]
> > > delete calls destructor and within destructor it calls delete, which[/color][/color][/color]
in[color=blue][color=green]
> > turn[color=darkred]
> > > calls the destructor again and so on....)
> > >
> > > Any comments appreciated.
> > >
> > > asasas
> > >[/color]
> >
> > Destructor does not call delete, so there is no loop.
> >
> > john[/color]
>
> Generally speaking, you have a delete in a destructor method. That said, a
> delete calls the destructor and the endless cycles continues. I'm sure[/color]
there[color=blue]
> is something im missing.
>
>[/color]

Sometimes you delete a subobject in the destructor. You don't delete the
original object.

class X
{
~X()
{
delete ptr; // delete Y object, calls Y destructor
}
Y* ptr;
};

X* p = new X;
delete p; // deletes an X object, calls X destructor

The only way there would be a loop was if you did 'delete this' in the
destructor.

john

**John Harrison** · Jul 19 '05, 04:56 PM

Re: destructor and delete

"Newsnet Customer" <nistygravy@ipr imus.com.au> wrote in message
news:3f534b75_1 @news.iprimus.c om.au...[color=blue]
>
>
>[color=green][color=darkred]
> > > Destructor would be called only for the object created on the heap.[/color]
> >
> > I meant stack ofcourse ;-).[/color]
>
> incorrect. as indicated in statement 1. an object's (dynamically created
> ...meaning HEAP) destructor is called when the object goes out of scope.
>[/color]

No it is not.

{
X* ptr = new X;
}

No destructors are called in the above code, because POINTERS do not have
destructors. ptr is a POINTER to X, its is not an X.

{
X x;
}

X destructor is called (assuming it has one).

You're very cocky considering, by your own admission, you don't get it.

john

**Karl Heinz Buchegger** · Jul 19 '05, 04:56 PM

Re: destructor and delete

Newsnet Customer wrote:[color=blue]
>[color=green][color=darkred]
> > > Destructor would be called only for the object created on the heap.[/color]
> >
> > I meant stack ofcourse ;-).[/color]
>
> incorrect. as indicated in statement 1. an object's (dynamically created
> ...meaning HEAP) destructor is called when the object goes out of scope.[/color]

The point is: a dynamically created object(that means on the heap) *never*
goes out of scope. The object is destroyed when you explicitely say so
with the help of delete.

{
std::string* p;
p = new std::string;
}

At this point, p goes out of scope and hence is destroyed.
But the dynamically allocated string object lives on, even
if there is no means to access it any longer.

--
Karl Heinz Buchegger
kbuchegg@gascad .at

C++: destructor and delete

C++: destructor and delete

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