lock bit and erase errors

Josh Boyer jdub at us.ibm.com
Mon Mar 14 14:08:12 EST 2005

Caveat:  I am using some pretty old code in this case, but the current
code is also questionable so bear with me.

I can't for the life of me figure out what the following code from
do_erase_oneblock cfi_cmdset_0001.c is doing:

	/* check for lock bit */
	if (status & CMD(0x3a)) {
		unsigned char chipstatus = status;
		if (status != CMD(status & 0xff)) {
			int i = cfi->interleave;
			for (i = 1; i<cfi->interleave; i++) {
				      chipstatus |= status >> (cfi->device_type * 8);
			printk(KERN_WARNING "Status is not identical for all chips: 0x%x. Merging to give 0x%02x\n", status, chipstatus);

My device is a 16-bit device, the interleave is 1, and I have a few
blocks with the lock bit on.  When I do a write to /dev/mtdblock/0/0, an
erase is attempted.  When I run an older version of this code, I get:

"Status not identical for all chips: 0xa200.  Merging to give 0x00"

What I don't understand is how any of the above code is expected to
work.  A 1 byte variable (chipstatus) is assigned a 4 byte value
(status), which gets truncated.  Then status is shifted by 16 bits (in
my case), which still means chipstatus is still 0.  When all is said and
done, I get the merging printk but none of the checks below this code
match.  That doesn't seem right...

I hacked the older code for my specific instance to do:

chipstatus = status >> 8;

and I get the proper return code for my situation (-EROFS).  I still
don't get an error in the application writing to /dev/mtdblock/0/0, but
at least the chip driver is actually returning an error.

Any ideas?


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