(j3.2006) question about deallocation

[Bill Long]

Robert Corbett wrote:
> Bill Long wrote:
>>
>> Robert Corbett wrote:
>>
>>> Consider the code fragment
>>>
>>>       MODULE M
>>>         . . .
>>>         TYPE BASE
>>>           INTEGER I
>>>         CONTAINS
>>>           FINAL SUBR1
>>>         END TYPE
>>>         TYPE, EXTENDS(BASE) :: EXTENDED
>>>         CONTAINS
>>>           FINAL SUBR2
>>>         END TYPE
>>>         . . .
>>>       END MODULE
>>>
>>>       PROGRAM MAIN
>>>         USE M
>>>         TYPE(EXTENDED), POINTER :: P
>>>         TYPE(BASE), POINTER :: Q
>>>         . . .
>>>         ALLOCATE(P)
>>>         Q => P%BASE
>>>         . . .
>>>         DEALLOCATE(Q)
>>>         . . .
>>>       END
>>>
>>> Is the DEALLOCATE statement standard conforming?  
>>
>>
>> The extended type does not add any components, so it looks OK to me.
>
> Your answer is consistent with what I have heard from members of the
> committee regarding similar cases, as I noted in my response to
> Reinhold Bader.
>
>>> If so,
>>> are both final subroutines invoked, or is only SUBR1
>>> invoked? 
>>
>>
>> The type of the object being deallocated is BASE, and a final routine 
>> is bound to the type, so I would say SUBR1 (only) is executed.
>
> Your answer here is what I think a literal reading of the standard
> would yield.  I fear that interpretation will prove troublesome for
> users.

I fear that final subroutines, in general, will prove troublesome for 
users.   In this case, Q is not polymorphic and has only one possible 
type. The type EXTENDED is nowhere visible to the finalization process 
for Q.  I don't see how SUBR2 would be (or should be) found in this case.

Cheers,
Bill



>
> Bob Corbett

-- 
Bill Long                                   longb at cray.com
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