Hello... Unfortunately, /dev/lp1 is not equivalent to LPT1: under DOS... /dev/lp0 is the equivalent to LPT1: Why? Anyone here ever program in nearly any computer language/shell language in the past twenty years? Well, traditionally, arrays, and other type of sturctures (BASIC's DIM comes to mind) have always started with element 0 being the first real element number 1. And, many things and people have been developed since then which this type of system in mind... Thus, /dev/lp0 = LPT1: /dev/lp1 = LPT2: /dev/lp2 = LPT3: Understand? My explanation above is not meant to fully explain the reasoning behind this scheme, only to give a real world example, and show how in one possible way it might have been carried on down though tradition... Cheers! Brian - Moriarty ----- Original Message ----- From: Jyry Kuukkanen <jyryq@luukku.com> To: Simeó <simeo@nil.fut.es> Cc: <suse-linux-e@suse.com> Sent: Tuesday, July 27, 1999 5:44 AM Subject: [suse-linux-e] Vs: [suse-linux-e] I can't print with suse 6.1
on 27.7.1999 05:48 Simeó wrote:
Hi friends, I was able to print with my Hp500C since linux SuSE 5.3 to 6.0, but now, I can't do it with my new suse 6.1. Why? My Printer is in /dev/lp1 (lpt1 in win) and works fine in windows. I made /etc/printcap with yast. I believe that linux don't see my printer because I have this message in status "waiting for lp to become ready (offline)?". Thanks
PD: lpd is running, and I have the original kernel
/dev/lp1 is equal to lpt1 under WinDos, so should you be using /dev/lp0 instead?
-- -- * Jyry *
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