linux-kernel - Writing a driver for a legacy serial device

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Message-ID: <20070619190502.05d4e0c8@hyperion.delvare>
Date:	Tue, 19 Jun 2007 19:05:02 +0200
From:	Jean Delvare <khali@...ux-fr.org>
To:	LKML <linux-kernel@...r.kernel.org>
Cc:	Vojtech Pavlik <vojtech@....cz>
Subject: Writing a driver for a legacy serial device

Hi all,

I want to write a Linux kernel driver for a device which connects to
the legacy serial port. I started writing a driver, however I am
already stuck at the very beginning. The .connect function of my serial
driver is never called, and I just don't get why. I couldn't find any
documentation about writing such a legacy driver in Documentation nor
in LDD3. Is there anyone out there which could lend a helping hand?

I know that the device and my serial port both work. I can talk to the
device using minicom just fine. I have the following drivers loaded: 

$ lsmod | grep 8250
8250_pnp               11648  0
8250                   23464  1 8250_pnp
serial_core            19392  1 8250

Serial: 8250/16550 driver $Revision: 1.90 $ 2 ports, IRQ sharing disabled
00:08: ttyS0 at I/O 0x3f8 (irq = 4) is a 16550A

But I need to implement my driver in kernel space. My code looks like
this:

static struct serio_device_id taos_serio_ids[] = {
	{
		.type	= SERIO_RS232,
		.proto	= SERIO_ANY,
		.id	= SERIO_ANY,
		.extra	= SERIO_ANY,
	},
	{ 0 }
};
MODULE_DEVICE_TABLE(serio, taos_serio_ids);

static struct serio_driver taos_drv = {
	.driver		= {
		.name	= "taos-evm",
	},
	.description	= "TAOS evaluation module driver",
	.id_table	= taos_serio_ids,
	.connect	= taos_connect,
	.disconnect	= taos_disconnect,
	.interrupt	= taos_interrupt,
};

static int __init taos_init(void)
{
	return serio_register_driver(&taos_drv);
}

static void __exit taos_exit(void)
{
	serio_unregister_driver(&taos_drv);
}

The problem is that taos_connect is never called. I suppose that I need
different values for .type, .proto or .id, except that I just don't
know what to put there. I tried a few random values without success.
What's the trick?

Thanks,
-- 
Jean Delvare
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