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bcache: avoid oversize memory allocation by small stripe_size
Arraies bcache->stripe_sectors_dirty and bcache->full_dirty_stripes are used for dirty data writeback, their sizes are decided by backing device capacity and stripe size. Larger backing device capacity or smaller stripe size make these two arraies occupies more dynamic memory space. Currently bcache->stripe_size is directly inherited from queue->limits.io_opt of underlying storage device. For normal hard drives, its limits.io_opt is 0, and bcache sets the corresponding stripe_size to 1TB (1<<31 sectors), it works fine 10+ years. But for devices do declare value for queue->limits.io_opt, small stripe_size (comparing to 1TB) becomes an issue for oversize memory allocations of bcache->stripe_sectors_dirty and bcache->full_dirty_stripes, while the capacity of hard drives gets much larger in recent decade. For example a raid5 array assembled by three 20TB hardrives, the raid device capacity is 40TB with typical 512KB limits.io_opt. After the math calculation in bcache code, these two arraies will occupy 400MB dynamic memory. Even worse Andrea Tomassetti reports that a 4KB limits.io_opt is declared on a new 2TB hard drive, then these two arraies request 2GB and 512MB dynamic memory from kzalloc(). The result is that bcache device always fails to initialize on his system. To avoid the oversize memory allocation, bcache->stripe_size should not directly inherited by queue->limits.io_opt from the underlying device. This patch defines BCH_MIN_STRIPE_SZ (4MB) as minimal bcache stripe size and set bcache device's stripe size against the declared limits.io_opt value from the underlying storage device, - If the declared limits.io_opt > BCH_MIN_STRIPE_SZ, bcache device will set its stripe size directly by this limits.io_opt value. - If the declared limits.io_opt < BCH_MIN_STRIPE_SZ, bcache device will set its stripe size by a value multiplying limits.io_opt and euqal or large than BCH_MIN_STRIPE_SZ. Then the minimal stripe size of a bcache device will always be >= 4MB. For a 40TB raid5 device with 512KB limits.io_opt, memory occupied by bcache->stripe_sectors_dirty and bcache->full_dirty_stripes will be 50MB in total. For a 2TB hard drive with 4KB limits.io_opt, memory occupied by these two arraies will be 2.5MB in total. Such mount of memory allocated for bcache->stripe_sectors_dirty and bcache->full_dirty_stripes is reasonable for most of storage devices. Reported-by: Andrea Tomassetti <andrea.tomassetti-opensource@devo.com> Signed-off-by: Coly Li <colyli@suse.de> Reviewed-by: Eric Wheeler <bcache@lists.ewheeler.net> Link: https://lore.kernel.org/r/20231120052503.6122-2-colyli@suse.de Signed-off-by: Jens Axboe <axboe@kernel.dk>
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@ -265,6 +265,7 @@ struct bcache_device {
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#define BCACHE_DEV_WB_RUNNING 3
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#define BCACHE_DEV_WB_RUNNING 3
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#define BCACHE_DEV_RATE_DW_RUNNING 4
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#define BCACHE_DEV_RATE_DW_RUNNING 4
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int nr_stripes;
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int nr_stripes;
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#define BCH_MIN_STRIPE_SZ ((4 << 20) >> SECTOR_SHIFT)
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unsigned int stripe_size;
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unsigned int stripe_size;
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atomic_t *stripe_sectors_dirty;
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atomic_t *stripe_sectors_dirty;
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unsigned long *full_dirty_stripes;
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unsigned long *full_dirty_stripes;
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@ -905,6 +905,8 @@ static int bcache_device_init(struct bcache_device *d, unsigned int block_size,
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if (!d->stripe_size)
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if (!d->stripe_size)
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d->stripe_size = 1 << 31;
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d->stripe_size = 1 << 31;
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else if (d->stripe_size < BCH_MIN_STRIPE_SZ)
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d->stripe_size = roundup(BCH_MIN_STRIPE_SZ, d->stripe_size);
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n = DIV_ROUND_UP_ULL(sectors, d->stripe_size);
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n = DIV_ROUND_UP_ULL(sectors, d->stripe_size);
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if (!n || n > max_stripes) {
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if (!n || n > max_stripes) {
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