[U-Boot] bad block table stored in nand flash

[Ing. Jozef Goril <jgoril@local.elnec.sk>]

...sorry for confusing...

certainly, if reserved_block_code == 01b, then
rcode = 0x01;
msk[2] = ~rcode = ~0x01 = FE;

and I'm speaking about right side in parenthesis:
right side before negation: [FE,F8,E0,80]
right side after negation: [01,07,1F,7F]

But this doesn't change the fact that some blocks are irregularly set to invalid.

Thanks and best regards
Jozef


-------- Original Message --------
Subject: [U-Boot] bad block table stored in nand flash
From: Ing. Jozef Goril <jgoril@local.elnec.sk>
To: u-boot at lists.denx.de
Date: 27. 1. 2010 14:06

> Hi.
> 
> I'm new in U-Boot and need some help to understand how does BBT in nand flash work.
> I don't understand, how reserved blocks (those used for BBT storage) are marked 
> in BBT in flash. The file in concern is drivers/mtd/nand_bbt.c, of actual 
> release version (2009.11.1).
> 
> At file beginning, there is some note:
> 
> * The table uses 2 bits per block
> * 11b:	block is good
> * 00b:	block is factory marked bad
> * 01b, 10b:	block is marked bad due to wear
> *
> * The memory bad block table uses the following scheme:
> * 00b:		block is good
> * 01b:		block is marked bad due to wear
> * 10b:		block is reserved (to protect the bbt area)
> * 11b:		block is factory marked bad
> 
> Later, in function write_bbt, there is a code that converts RAM-based BBT to 
> flash-based one at lines 720-730.
> For line
> buf[offs + (i >> sft)] &= ~(msk[dat & 0x03] << sftcnt);
> 
> I cannot understand the case, if (dat&0x03) == 10b (reserved block).
> In that case, msk[2] value should be used.
> The value of msk[2] is set few lines above (line 649): msk[2] = ~rcode;
> The value of rcode is set at time of declaration:
> rcode = td->reserved_block_code;
> 
> Now, in case of reserved_block_code == 01b:
> rcode = 0x02;
> msk[2] = ~rcode = ~0x02 = FD;
> 
> Regarding to line
> buf[offs + (i >> sft)] &= ~(msk[dat & 0x03] << sftcnt);
> 
> it should be shifted left by sftcnt bits (sftcnt can be [0,2,4,6]).
> I.e. that the value in the parenthesis on the left side can be of
> [FD,F4,D0,40]. After negation: [02,0B,2F,BF].
> These are values, that original byte in buffer can be ANDed with. Since there 
> are zeros on higher bits position (over mask 11b), this ANDing will destroy the 
> block status information of some blocks using the same byte. 00b in flash means 
> invalid block...
> 
> Am I missing something important or is there a bug?
> 
> I understand the reserved_block_code can be:
> - zero : reserved blocks are not marked in flash-based BBT, the mechanism 
> rewrites rcode from 0x00 to 0xFF, later msk[2] = 0x00 and all works fine.
> - non zero : reserved blocks are marked in flash-based BBT. But the only 
> avoilable value is 01b (00b stands for invalid block, 11b for good block and 10b 
> will come from RAM-based 01b due to a procedure mentioned above).
> 
> But I cannot find how does reserved_block_code of 01b work...
> 
> Can you, please, make me clear?
> 
> Thanks a lot.
> Jozef Goril
> 
> 
> 
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