btrfs: fix wrong calculation of stripe size
There are two tiny problem: - One is When we check the chunk size is greater than the max chunk size or not, we should take mirrors into account, but the original code didn't. - The other is btrfs shouldn't use the size of the residual free space as the length of of a dup chunk when doing chunk allocation. It is because the device space that a dup chunk needs is twice as large as the chunk size, if we use the size of the residual free space as the length of a dup chunk, we can not get enough free space. Fix it. Signed-off-by: Miao Xie <miaox@cn.fujitsu.com> Reviewed-by: Josef Bacik <josef@redhat.com> Signed-off-by: Chris Mason <chris.mason@oracle.com>
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@ -2177,6 +2177,7 @@ static int __btrfs_alloc_chunk(struct btrfs_trans_handle *trans,
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int num_stripes = 1;
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int min_stripes = 1;
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int sub_stripes = 0;
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int ncopies = 1;
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int looped = 0;
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int ret;
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int index;
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@ -2197,12 +2198,14 @@ static int __btrfs_alloc_chunk(struct btrfs_trans_handle *trans,
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if (type & (BTRFS_BLOCK_GROUP_DUP)) {
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num_stripes = 2;
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min_stripes = 2;
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ncopies = 2;
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}
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if (type & (BTRFS_BLOCK_GROUP_RAID1)) {
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if (fs_devices->rw_devices < 2)
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return -ENOSPC;
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num_stripes = 2;
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min_stripes = 2;
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ncopies = 2;
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}
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if (type & (BTRFS_BLOCK_GROUP_RAID10)) {
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num_stripes = fs_devices->rw_devices;
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@ -2210,6 +2213,7 @@ static int __btrfs_alloc_chunk(struct btrfs_trans_handle *trans,
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return -ENOSPC;
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num_stripes &= ~(u32)1;
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sub_stripes = 2;
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ncopies = 2;
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min_stripes = 4;
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}
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@ -2239,8 +2243,8 @@ again:
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map->num_stripes = num_stripes;
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}
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if (calc_size * num_stripes > max_chunk_size) {
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calc_size = max_chunk_size;
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if (calc_size * num_stripes > max_chunk_size * ncopies) {
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calc_size = max_chunk_size * ncopies;
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do_div(calc_size, num_stripes);
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do_div(calc_size, stripe_len);
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calc_size *= stripe_len;
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@ -2321,6 +2325,8 @@ again:
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if (!looped && max_avail > 0) {
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looped = 1;
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calc_size = max_avail;
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if (type & BTRFS_BLOCK_GROUP_DUP)
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do_div(calc_size, 2);
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goto again;
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}
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kfree(map);
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