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@gmail.com] Sent: Monday, July 10, 2006 10:28 AM To: boost-users@lists.boost.org 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
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@physik.uni-erlangen.de mailto:Boris.Breidenbach@physik.uni-erlangen.de > 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. _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users