Re: the question about raid0_make_request

2006-06-20 Thread liu yang
I understand what you mean now. Valid layerout is not always the 'right' layerout. Thanks for you help. Regards. YangLiu - To unsubscribe from this list: send the line "unsubscribe linux-raid" in the body of a message to [EMAIL PROTECTED] More majordomo info at http://vger.kernel.org/majordomo

Re: the question about raid0_make_request

2006-06-19 Thread liu yang
Thanks,Neil. I noticed that the whole codes of calculating the underlying device is below { sector_t x = (block - zone->zone_offset) >> chunksize_bits; sector_div(x, zone->nb_dev); chunk = x; BUG_ON(x != (sector_t)chunk);

the question about raid0_make_request

2006-06-18 Thread liu yang
When I read the code of raid0_make_request,I meet some questions. 1\ block = bio->bi_sector >> 1,it's the device offset in kilotytes. so why do we use block substract zone->zone_offset? The zone->zone_offset is the zone offset relative the mddev in sectors. 2\ the codes below: x = block >> chu

Re: to understand the logic of raid0_make_request

2006-06-16 Thread liu yang
Neil,thanks for you help.Your explanation is helpful really. I went through the code of raid0_make_request again this evening, and I still had some questions. 1\ block = bio->bi_sector >> 1,it's the device offset in kilotytes. so why do we use block substract zone->zone_offset? The zone->zone_off

Re: to understand the logic of raid0_make_request

2006-06-15 Thread liu yang
2006/6/13, Neil Brown <[EMAIL PROTECTED]>: On Tuesday June 13, [EMAIL PROTECTED] wrote: > hello,everyone. > I am studying the code of raid0.But I find that the logic of > raid0_make_request is a little difficult to understand. > Who can tell me what the function of raid0_make_request will do even

to understand the logic of raid0_make_request

2006-06-12 Thread liu yang
hello,everyone. I am studying the code of raid0.But I find that the logic of raid0_make_request is a little difficult to understand. Who can tell me what the function of raid0_make_request will do eventually? Regards. Thanks! YangL