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From: Andrew Holden (aholden_at_[hidden])
Date: 2006-07-10 14:42:56
You would manually call destructors when you do your own memory
management (like std::vector). In this case, you would also need to
manually call constructors too. Here is a modified version of your test
program that doesn't double-destruct the object.
class A
{
public:
A()
{
counter = 0;
}
~A()
{
counter++;
std::cout << "~A(): " << counter << std::endl;
}
int counter;
};
int main(int argc, char* argv[])
{
A *a;
//Allocate memory using C functions, which do not call
constructors
a = (A*) malloc (sizeof (A));
cout << a << ' ' << a->counter << std::endl;
//Manually call the constructor. The (a) tells the compiler which
pointer we are initializing
new (a) A ();
cout << a << ' ' << a->counter << std::endl;
//Manually call the destructor
a->~A();
cout << a << ' ' << a->counter << std::endl;
//Free memory using C functions, which do not call destructors
free (a);
return 0;
}
_____
From: bringiton bringiton [mailto:kneeride_at_[hidden]]
Sent: Monday, July 10, 2006 10:28 AM
To: boost-users_at_[hidden]
Subject: Re: [Boost-users] std::vector< boost::shared_ptr<int>
>::pop_back()
holy! you learn something every day. you can call the destructor of an
object (before destruction)
this is a trap for me because i dont clean up my variables in
destructors.
class A {
public:
A() {
counter = 0;
}
~A() {
counter++;
std::cout << "~A(): " << counter << std::endl;
}
int counter;
};
int main(int argc, char* argv[])
{
A a;
a.~A();
return 0;
}
NOTE: destructor gets called twice.
(so it's probably a good idea to NULL pointers in destructors)
On 7/11/06, bringiton bringiton <kneeride_at_[hidden]> wrote:
>>just look at the include files:
i looked at the header, but got a little confused.
i don't see how you can call a destruct of a a memory space that still
exists. ie a C version of a vector.
shared_ptr<int> list[1024];
int n = 0;
// add an item
shared_ptr<int> newItem(new int(1));
list[n++] = newItem;
// remove the tail item
n--;
// destructor never called
is it possible to call a destructor of an object that exists?
what then happens then the object goes out of scope? the destructor will
be called twice.
(or maybe i am completely missing the point)
BTW: not sure if above code compiles (did it in my head)
On 7/10/06, Boris Breidenbach < Boris.Breidenbach_at_[hidden]
<mailto:Boris.Breidenbach_at_[hidden]> > wrote:
On Mon, Jul 10, 2006 at 03:58:19PM +1000, bringiton bringiton wrote:
> This question is based on curiosity. How does std::vector::pop_back()
call
> the destructor of the item getting removed?
>
> i understood std::vector to be a contigious array of memory, therefore
an
> item's memory does not go out of scope when being popped. ie the item
goes
> out of scope when the entire array goes out of scope.
just look at the include files:
bits/stl_vector.h says:
void
pop_back()
{
--this->_M_impl._M_finish;
std::_Destroy(this->_M_impl._M_finish);
}
so the destructor is called in std::_Destroy. Which can be found in
bits/stl_construct.h:
/**
* @if maint
* Destroy the object pointed to by a pointer type.
* @endif
*/
template<typename _Tp>
inline void
_Destroy(_Tp* __pointer)
{ __pointer->~_Tp(); }
It just calls the destructor.
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