RE: dealloc (Correction)

Statement one deallocates the storage and set the pointer to null. 

I don't believe this is true.  It won't set the pointer to null.  In
order to do that, you need the (n) extender:

Dealloc(n) myField_P;


Kurt Anderson
Application Developer
Highsmith Inc

-----Original Message-----
From: rpg400-l-bounces@xxxxxxxxxxxx
[mailto:rpg400-l-bounces@xxxxxxxxxxxx] On Behalf Of Alan Campin
Sent: Wednesday, December 06, 2006 12:43 PM
To: rpg400-l@xxxxxxxxxxxx
Subject: RE:dealloc (Correction) 

<snip>
Assuming that myField_P is the based of myField, does these two
statements yield same result?
 
Statement 1 ==>   dealloc myField_P;
 
Statement 2 ==>   myField_P = *null; 

</snip>

No, statement 2 sets the value of the pointer to null but does not
deallocate the storage. It's possible that the OS would generate a
runtime error if you tried but I have never tried. You could prove by
going into a loop, allocating a block or storage and then moving in
null. At some point, the program is going to crash because it is out of
heap storage. 

Statement one deallocates the storage and set the pointer to null. 



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