btrfs: don't add both copies of DUP to reada extent tree

Normally when there are 2 copies of a block, we add both to the
reada extent tree and prefetch only the one that is easier to reach.
This way we can better utilize multiple devices.
In case of DUP this makes no sense as both copies reside on the
same device.

Signed-off-by: Arne Jansen <sensille@gmx.net>

diff --git a/fs/btrfs/reada.c b/fs/btrfs/reada.c
index 8dec650099c8a3d924e778c4b6aa3e89b2c945fb..ac5d010858848d007e380d529476ad9eb4f6fb31 100644 (file)
--- a/fs/btrfs/reada.c
+++ b/fs/btrfs/reada.c
@@ -326,6 +326,7 @@ static struct reada_extent *reada_find_extent(struct btrfs_root *root,
        struct btrfs_mapping_tree *map_tree = &fs_info->mapping_tree;
        struct btrfs_bio *bbio = NULL;
        struct btrfs_device *dev;
+       struct btrfs_device *prev_dev;
        u32 blocksize;
        u64 length;
        int nzones = 0;
@@ -405,8 +406,20 @@ static struct reada_extent *reada_find_extent(struct btrfs_root *root,
                spin_unlock(&fs_info->reada_lock);
                goto error;
        }
+       prev_dev = NULL;
        for (i = 0; i < nzones; ++i) {
                dev = bbio->stripes[i].dev;
+               if (dev == prev_dev) {
+                       /*
+                        * in case of DUP, just add the first zone. As both
+                        * are on the same device, there's nothing to gain
+                        * from adding both.
+                        * Also, it wouldn't work, as the tree is per device
+                        * and adding would fail with EEXIST
+                        */
+                       continue;
+               }
+               prev_dev = dev;
                ret = radix_tree_insert(&dev->reada_extents, index, re);
                if (ret) {
                        while (--i >= 0) {