From mboxrd@z Thu Jan 1 00:00:00 1970 From: Stas Sergeev Subject: Re: Can't access I/O ports with dosemu 1.0.2 Date: Sat, 28 Dec 2002 01:14:55 +0300 Sender: linux-msdos-owner@vger.kernel.org Message-ID: <3E0CD0DF.7060509@yahoo.com> Mime-Version: 1.0 Content-Transfer-Encoding: 7bit Return-path: List-Id: Content-Type: text/plain; charset="us-ascii"; format="flowed" To: linux-msdos@vger.kernel.org Hello. Brad Rodriguez wrote: > I see that COM1, COM2, and LPT1 are all listed in /proc/ioports, so > I'll have to learn how to disable the kernel support. Probably you can simply hack the dosemu's ports.c and replace a /proc/ioports there with /dev/null. I don't know for sure will such a cheat work or will it be safe, but at least you can try. As for disabling the in-kernel support, you have to reconfigure/recompile your kernel without a parport support, or, which is better, if the support is modular, just unload the relevant modules and that should do the trick. > But 0x320-0x327 (the > EPROM emulator) is not listed in /proc/ioports and I don't seem able to > access those ports either. Yes, I also think it is a permissions problem. >> Yes, LPT is a weak part of dosemu unless >> you do a simple printing. > Unfortunately, I'm not using it to print. And unfortunately dosemu doesn't have a proper LPT support. >> Why would the one require a direct >> access to the COM ports? The support >> for those is fairly complete under >> dosemu. > Perhaps I don't understand how these are supported under DOSEMU. The Yep. > programs I am using are not using BIOS calls to access the serial port. > As far as I know, they include their own serial I/O drivers that talk > directly > to the bare metal. (Come to think of it, they probably also speak to > the interrupt controller chip, and I don't know *how* I can allow that > without completely corrupting the system.) Please just stop making problems where they don't exist. The LPT is problematic, yes, but the PIC and Serials are fully supported (except that the speed of the emulated serial is not always sufficient).