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<!-- name="Rogier Wolff" -->
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<!-- subject="Re: Two SCSI bus question..." -->
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<title>sane-devel: Re: Two SCSI bus question...</title>
<h1>Re: Two SCSI bus question...</h1>
<b>Rogier Wolff</b> (<a href="mailto:R.E.Wolff@BitWizard.nl"><i>R.E.Wolff@BitWizard.nl</i></a>)<br>
<i>Thu, 11 Dec 1997 00:55:08 +0100 (MET)</i>
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Matto Marjanovic wrote:<br>
<i>&gt; </i><br>
<i>&gt; </i><br>
<i>&gt; &gt; &gt; All detected devices are assigned a raw-device in the order in which they </i><br>
<i>&gt; &gt; &gt; are detected. Scanning the busses goes dev-id first, then bus-id.</i><br>
<i>&gt; &gt; &gt; </i><br>
<i>&gt; &gt; &gt; Thus you first get all devices on the first bus, then all on the second,</i><br>
<i>&gt; &gt; &gt; each in order of the SCSI IDs.</i><br>
<i>&gt; &gt; &gt; </i><br>
<i>&gt; &gt; &gt; Thus your scanner wanders whenever a device that is below it in that ordering</i><br>
<i>&gt; &gt; &gt; scheme is there or not.</i><br>
<i>&gt; &gt; &gt; </i><br>
<i>&gt; </i><br>
<i>&gt; Alright, out of curiousity then:</i><br>
<i>&gt; </i><br>
<i>&gt; Is there any particular reason why SCSI devices are assigned this way, rather</i><br>
<i>&gt; than having the X in /dev/sgX be a direct function of device id and bus id?</i><br>
<i>&gt; (e.g. /dev/sgA4 = device with ID=4 on the first bus)</i><br>
<i>&gt; </i><br>
<i>&gt; The way it is now, suppose one of 5 scsi disks blows out when you power up,</i><br>
<i>&gt; then the rest all get their device assignments shifted and your server is</i><br>
<i>&gt; *totally* screwed...</i><br>
<p>
There are just 8 bits to a minor. You need about 4 bits to code for<br>
the partition number, so you're left with just 4 bits to cram bus id<br>
and device ID into. With wide scsi, you're left with 0 bits for the<br>
bus ID. <br>
<p>
Roger.<br>
<p>
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