ext4: fix stripe-unaligned allocations

When a filesystem is created using:

	mkfs.ext4 -b 4096 -E stride=512 <dev>

and we try to allocate 64MB extent, we will end up directly in
ext4_mb_complex_scan_group(). This is because the request is detected
as power-of-two allocation (so we start in ext4_mb_regular_allocator()
with ac_criteria == 0) however the check before
ext4_mb_simple_scan_group() refuses the direct buddy scan because the
allocation request is too large. Since cr == 0, the check whether we
should use ext4_mb_scan_aligned() fails as well and we fall back to
ext4_mb_complex_scan_group().

Fix the problem by checking for upper limit on power-of-two requests
directly when detecting them.

Reported-by: Ross Zwisler <ross.zwisler@linux.intel.com>
Signed-off-by: Jan Kara <jack@suse.cz>
Signed-off-by: Theodore Ts'o <tytso@mit.edu>

fs/ext4/mballoc.c

@@ -2146,8 +2146,10 @@ ext4_mb_regular_allocator(struct ext4_allocation_context *ac)
 	 * We search using buddy data only if the order of the request
 	 * is greater than equal to the sbi_s_mb_order2_reqs
 	 * You can tune it via /sys/fs/ext4/<partition>/mb_order2_req
+	 * We also support searching for power-of-two requests only for
+	 * requests upto maximum buddy size we have constructed.
 	 */
-	if (i >= sbi->s_mb_order2_reqs) {
+	if (i >= sbi->s_mb_order2_reqs && i <= sb->s_blocksize_bits + 2) {
 		/*
 		 * This should tell if fe_len is exactly power of 2
 		 */
@@ -2217,7 +2219,7 @@ repeat:
 			}
 
 			ac->ac_groups_scanned++;
-			if (cr == 0 && ac->ac_2order < sb->s_blocksize_bits+2)
+			if (cr == 0)
 				ext4_mb_simple_scan_group(ac, &e4b);
 			else if (cr == 1 && sbi->s_stripe &&
 					!(ac->ac_g_ex.fe_len % sbi->s_stripe))