On Wed, Jan 27, 2010 at 02:06:22PM +0100, Ing. Jozef Goril wrote: > 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?
This code came from Linux; I'd try asking on linux-...@lists.infradead.org. -Scott _______________________________________________ U-Boot mailing list U-Boot@lists.denx.de http://lists.denx.de/mailman/listinfo/u-boot
