Date: Thu, 04 Dec 2008 15:08:14 -0800 From: Marcel Moolenaar <xcllnt@mac.com> To: John Baldwin <jhb@freebsd.org> Cc: FreeBSD Arch <arch@freebsd.org> Subject: Re: RFC: making gpart default Message-ID: <5783CEB0-6163-429E-8B28-2F9D6FBCF4A8@mac.com> In-Reply-To: <200812041313.34565.jhb@freebsd.org> References: <e7db6d980811291356w54256e6du82350baf3c57d591@mail.gmail.com> <e7db6d980812011605h18b40700v1043e376ef392365@mail.gmail.com> <20081203.193714.693830802.imp@bsdimp.com> <200812041313.34565.jhb@freebsd.org>
next in thread | previous in thread | raw e-mail | index | archive | help
On Dec 4, 2008, at 10:13 AM, John Baldwin wrote: > No, the way GPT works, you have a PMBR at sector 0, then immediately > following > that you have the Primary partition table in the next N sectors (the > first > sector in the table has a header that contains the size of the > table). Then > you have a backup Secondary partition table in the last N sectors of > the disk > as well. At least with the old gpt(8) tool you could actually tell > it how > big of a table to make when you created a GPT, and I imagine gpart > probably > can do the same. Yes. For schemes that support it, you can specify how many entries to allocate. The 34 corresponds to 128 entries for GPT (4 entries per sector)... -- Marcel Moolenaar xcllnt@mac.com
Want to link to this message? Use this URL: <https://mail-archive.FreeBSD.org/cgi/mid.cgi?5783CEB0-6163-429E-8B28-2F9D6FBCF4A8>