wbsd: IRQ handler type mismatch

[Robert Hancock <hancockr@shaw.ca>]

I'm seeing this on bootup on my laptop with recent kernels (currently 
2.6.20-rc6-mm3):

Floppy drive(s): fd0 is 1.44M
IRQ handler type mismatch for IRQ 6
current handler: wbsd
 [<c0451fe7>] setup_irq+0x194/0x1ac
 [<e0ad7189>] floppy_hardint+0x0/0xc0 [floppy]
 [<c045207b>] request_irq+0x7c/0x98
 [<e0a5c67c>] init_module+0x546/0xe15 [floppy]
 [<c05ffe85>] _spin_unlock_irq+0x5/0x7
 [<c0442410>] __link_module+0x0/0x10
 [<c0442410>] __link_module+0x0/0x10
 [<c0443d6c>] sys_init_module+0x1781/0x18c8
 [<c04dbf10>] blk_init_queue+0x0/0x8
 [<c04734f6>] vfs_read+0xa6/0x152
 [<c0404e26>] sysenter_past_esp+0x5f/0x85
 [<c0600000>] error_code+0x68/0x84
 =======================
floppy0: Unable to grab IRQ6 for the floppy driver

It looks like wbsd is being loaded first, which being a nice PnP device 
sees that IRQ 6 is available and grabs it. Then the we try to load the 
floppy module (which is hard-coded to use IRQ 6) which fails because 
wbsd already owns IRQ6 (which is fine, there's no floppy controller 
anyway). The mismatch is occurring because wbsd passes IRQF_SHARED 
whereas floppy doesn't. I assume one of the two is likely wrong. Should 
wbsd really be passing IRQF_SHARED here? This is an LPC device which 
uses ISA-style edge triggered interrupts, is it safe to try and share an 
interrupt with this chip anyway?

--
Robert Hancock      Saskatoon, SK, Canada
To email, remove "nospam" from hancockr@nospamshaw.ca
Home Page: http://www.roberthancock.com/

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