Re: [PATCH] btrfs: scrub: per-device bandwidth control
From: David Sterba <hidden>
Date: 2021-05-21 15:19:00
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On Thu, May 20, 2021 at 03:14:03PM +0200, Arnd Bergmann wrote:
On Thu, May 20, 2021 at 9:43 AM Geert Uytterhoeven [off-list ref] wrote:quoted
On Tue, 18 May 2021, David Sterba wrote:quoted
+ /* Start new epoch, set deadline */ + now = ktime_get(); + if (sctx->throttle_deadline == 0) { + sctx->throttle_deadline = ktime_add_ms(now, time_slice / div);ERROR: modpost: "__udivdi3" [fs/btrfs/btrfs.ko] undefined! div_u64(bwlimit, div)If 'time_slice' is in nanoseconds, the best interface to use is ktime_divns().
It's in miliseconds and the division above is int/int, the problematic one is below.
quoted
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+ sctx->throttle_sent = 0; + } + + /* Still in the time to send? */ + if (ktime_before(now, sctx->throttle_deadline)) { + /* If current bio is within the limit, send it */ + sctx->throttle_sent += sbio->bio->bi_iter.bi_size; + if (sctx->throttle_sent <= bwlimit / div) + return;Doesn't this also need to be changed?quoted
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+ /* We're over the limit, sleep until the rest of the slice */ + delta = ktime_ms_delta(sctx->throttle_deadline, now); + } else { + /* New request after deadline, start new epoch */ + delta = 0; + } + + if (delta) + schedule_timeout_interruptible(delta * HZ / 1000);ERROR: modpost: "__divdi3" [fs/btrfs/btrfs.ko] undefined! I'm a bit surprised gcc doesn't emit code for the division by the constant 1000, but emits a call to __divdi3(). So this has to become div_u64(), too.There is schedule_hrtimeout(), which takes a ktime_t directly but has slightly different behavior. There is also an msecs_to_jiffies helper that should produce a fast division.
I'll use msecs_to_jiffies, thanks. If 'hr' in schedule_hrtimeout stands for high resolution, it's not necessary here.