Sure, you'll get a destructor called for temporary auto_ptr abjects because you passed the wp parameter by value, so a copy is made.
But this doesn't have anything to do with calling a destructor for a NULL pointer.
I will look my test code one more tomorrow.
there was such code (as I remember)
std::auto_ptr<Widget> pw1( new Widget );
std::auto_ptr<Widget> pw2;
std::auto_ptr<Widget> func( std::auto_ptr<Widget> wp );
and in the debug mode for the statement below I saw that a destructor of
Widget is called two times
pw2 = func( pw1 );
for (as I thought) temporary objects.
I will check the code anew.
Vladimir Grigoriev
"David Webber" <dave@musical-dot-demon-dot-co.uk> wrote in message
news:eKZ43$sOKHA.4964@TK2MSFTNGP06.phx.gbl...
"Vladimir Grigoriev" <vlad.moscow@mail.ru> wrote in message
news:Ox4ttysOKHA.1280@TK2MSFTNGP04.phx.gbl...
Should be destructor called always irrespective of that a pointer is
equal to 0?
No - how can it be? A NULL pointer means that nothing is being pointed
to, so there is nothing to destroy.
What does the C++ standard say about this?
Example
T *p = 0;
delete p; // will be a destructor of T object called here?
This is safe - it does nothing. That is so you can do
delete p;
when p may be pointing to a valid object or it may be NULL.
You can of course (if you wish) do
if( p ) delete p;
but the effect is the same.
It seems that operation 'delete p' when p == 0 is not safe.
It may *seem* so, but in fact it is explicitly defined to be safe, I think
in the standard, but certainly in the most recent versions of VC++.
Dave
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David Webber
Author of 'Mozart the Music Processor'
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