Hi,
On Tue, 11 Jan 2000, Richard Levitte - VMS Whacker wrote:
> From: Geoff Thorpe <[EMAIL PROTECTED]>
>
> geoff> plenty of *_copy() functions to deal with that ('twould seem that a copy
> geoff> ups the reference count on the new structure by definition so copy
> geoff> functions needn't have any form of switch between "r" and/or "i").
>
> I'm not sure if that parenthesis is telling the current state of
> affairs or how it should be. If it's the latter, the new copy of any
> structure should get a reference count of 1 and nothing else. The
> reason should be obvious.
The "by definition" bit was supposed to indicate what you have described.
Namely, you get a copy of the original structure and a reference to that
structure (ref-count goes to 1 making it "alive"). A ref-count of zero is
not supposed to occur except during the last _free() call as a condition
for deallocating the structure. So when making the copy you are returning
a reference (not just a pointer) to that new structure, hence my note
that there's not much point to discuss "r" and "i" versions of a _copy()
function - a copy operation is creating a new reference, it's just
adding the reference in the copy rather than the original. Perhaps think
of it this way - any call to a _copy() function will require a
corresponding _free() (whether or not other code grabs further references
on that new copy).
Have I further muddled things up? :-)
Regards,
Geoff
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Geoff Thorpe Email: [EMAIL PROTECTED]
Cryptographic Software Engineer, C2Net Europe http://www.int.c2.net
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