From: | "Dave Page" <dpage(at)pgadmin(dot)org> |
---|---|
To: | "Brendan Jurd" <direvus(at)gmail(dot)com> |
Cc: | "Heikki Linnakangas" <heikki(at)enterprisedb(dot)com>, PostgreSQL-development <pgsql-hackers(at)postgresql(dot)org> |
Subject: | Re: Free Space Map data structure |
Date: | 2008-04-08 08:59:08 |
Message-ID: | 937d27e10804080159i10e9bacckeeb7404ec5374835@mail.gmail.com |
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Lists: | pgsql-hackers |
On Tue, Apr 8, 2008 at 9:54 AM, Brendan Jurd <direvus(at)gmail(dot)com> wrote:
> If I understand your design correctly, this claim isn't true. If the
> topmost node reports 9 bytes free, could you not have seven pages each
> with 1 byte free, and an eighth page with 2 bytes free?
>
> 9
> 4 5
> 2 2 2 3
> 1 1 1 1 1 1 1 2
>
> So you'd actually end up walking two levels down the left hand side of
> the tree, discovering not enough space, and then walking two levels
> down the right hand side to again discover not enough space.
As I read it, each node takes the value of the largest child, not the
sum of the children.
--
Dave Page
EnterpriseDB UK: http://www.enterprisedb.com
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