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Filipe Manana authored
When inserting a new key, we release the write lock on the leaf's parent only after doing the binary search on the leaf. This is because if the key ends up at slot 0, we will have to update the key at slot 0 of the parent node. The same reasoning applies to any other upper level nodes when their slot is 0. We also need to keep the parent locked in case the leaf does not have enough free space to insert the new key/item, because in that case we will split the leaf and we will need to add a new key to the parent due to a new leaf resulting from the split operation. However if the leaf has enough space for the new key and the key does not end up at slot 0 of the leaf we could release our write lock on the parent before doing the binary search on the leaf to figure out the destination slot. That leads to reducing the amount of time other tasks are blocked waiting to lock the parent, therefore increasing parallelism when there are other tasks that are trying to access other leaves accessible through the same parent. This also applies to other upper nodes besides the immediate parent, when their slot is 0, since we keep locks on them until we figure out if the leaf slot is slot 0 or not. In fact, having the key ending at up slot 0 when is rare. Typically it only happens when the key is less than or equals to the smallest, the "left most", key of the entire btree, during a split attempt when we try to push to the right sibling leaf or when the caller just wants to update the item of an existing key. It's also very common that a leaf has enough space to insert a new key, since after a split we move about half of the keys from one into the new leaf. So unlock the parent, and any other upper level nodes, when during a key insertion we notice the key is greater then the first key in the leaf and the leaf has enough free space. After unlocking the upper level nodes, do the binary search using a low boundary of slot 1 and not slot 0, to figure out the slot where the key will be inserted (or where the key already is in case it exists and the caller wants to modify its item data). This extra comparison, with the first key, is cheap and the key is very likely already in a cache line because it immediately follows the header of the extent buffer and we have recently read the level field of the header (which in fact is the last field of the header). The following fs_mark test was run on a non-debug kernel (debian's default kernel config), with a 12 cores intel CPU, and using a NVMe device: $ cat run-fsmark.sh #!/bin/bash DEV=/dev/nvme0n1 MNT=/mnt/nvme0n1 MOUNT_OPTIONS="-o ssd" MKFS_OPTIONS="-O no-holes -R free-space-tree" FILES=100000 THREADS=$(nproc --all) FILE_SIZE=0 echo "performance" | \ tee /sys/devices/system/cpu/cpu*/cpufreq/scaling_governor mkfs.btrfs -f $MKFS_OPTIONS $DEV mount $MOUNT_OPTIONS $DEV $MNT OPTS="-S 0 -L 10 -n $FILES -s $FILE_SIZE -t $THREADS -k" for ((i = 1; i <= $THREADS; i++)); do OPTS="$OPTS -d $MNT/d$i" done fs_mark $OPTS umount $MNT Before this change: FSUse% Count Size Files/sec App Overhead 0 1200000 0 165273.6 5958381 0 2400000 0 190938.3 6284477 0 3600000 0 181429.1 6044059 0 4800000 0 173979.2 6223418 0 6000000 0 139288.0 6384560 0 7200000 0 163000.4 6520083 1 8400000 0 57799.2 5388544 1 9600000 0 66461.6 5552969 2 10800000 0 49593.5 5163675 2 12000000 0 57672.1 4889398 After this change: FSUse% Count Size Files/sec App Overhead 0 1200000 0 167987.3 (+1.6%) 6272730 0 2400000 0 198563.9 (+4.0%) 6048847 0 3600000 0 197436.6 (+8.8%) 6163637 0 4800000 0 202880.7 (+16.6%) 6371771 1 6000000 0 167275.9 (+20.1%) 6556733 1 7200000 0 204051.2 (+25.2%) 6817091 1 8400000 0 69622.8 (+20.5%) 5525675 1 9600000 0 69384.5 (+4.4%) 5700723 1 10800000 0 61454.1 (+23.9%) 5363754 3 12000000 0 61908.7 (+7.3%) 5370196 Reviewed-by: Josef Bacik <josef@toxicpanda.com> Signed-off-by: Filipe Manana <fdmanana@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
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