ctree.c (2d65685a4a6ff6ef6bfe42717222db92c04224c2) ctree.c (ce6ef5abe68251cb60c4585df2cc73e218ec0dba)
1// SPDX-License-Identifier: GPL-2.0
2/*
3 * Copyright (C) 2007,2008 Oracle. All rights reserved.
4 */
5
6#include <linux/sched.h>
7#include <linux/slab.h>
8#include <linux/rbtree.h>

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1496{
1497 if (blocknr < other && other - (blocknr + blocksize) < 32768)
1498 return 1;
1499 if (blocknr > other && blocknr - (other + blocksize) < 32768)
1500 return 1;
1501 return 0;
1502}
1503
1// SPDX-License-Identifier: GPL-2.0
2/*
3 * Copyright (C) 2007,2008 Oracle. All rights reserved.
4 */
5
6#include <linux/sched.h>
7#include <linux/slab.h>
8#include <linux/rbtree.h>

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1496{
1497 if (blocknr < other && other - (blocknr + blocksize) < 32768)
1498 return 1;
1499 if (blocknr > other && blocknr - (other + blocksize) < 32768)
1500 return 1;
1501 return 0;
1502}
1503
1504#ifdef __LITTLE_ENDIAN
1505
1504/*
1506/*
1507 * Compare two keys, on little-endian the disk order is same as CPU order and
1508 * we can avoid the conversion.
1509 */
1510static int comp_keys(const struct btrfs_disk_key *disk_key,
1511 const struct btrfs_key *k2)
1512{
1513 const struct btrfs_key *k1 = (const struct btrfs_key *)disk_key;
1514
1515 return btrfs_comp_cpu_keys(k1, k2);
1516}
1517
1518#else
1519
1520/*
1505 * compare two keys in a memcmp fashion
1506 */
1507static int comp_keys(const struct btrfs_disk_key *disk,
1508 const struct btrfs_key *k2)
1509{
1510 struct btrfs_key k1;
1511
1512 btrfs_disk_key_to_cpu(&k1, disk);
1513
1514 return btrfs_comp_cpu_keys(&k1, k2);
1515}
1521 * compare two keys in a memcmp fashion
1522 */
1523static int comp_keys(const struct btrfs_disk_key *disk,
1524 const struct btrfs_key *k2)
1525{
1526 struct btrfs_key k1;
1527
1528 btrfs_disk_key_to_cpu(&k1, disk);
1529
1530 return btrfs_comp_cpu_keys(&k1, k2);
1531}
1532#endif
1516
1517/*
1518 * same as comp_keys only with two btrfs_key's
1519 */
1520int __pure btrfs_comp_cpu_keys(const struct btrfs_key *k1, const struct btrfs_key *k2)
1521{
1522 if (k1->objectid > k2->objectid)
1523 return 1;

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1534/*
1535 * same as comp_keys only with two btrfs_key's
1536 */
1537int __pure btrfs_comp_cpu_keys(const struct btrfs_key *k1, const struct btrfs_key *k2)
1538{
1539 if (k1->objectid > k2->objectid)
1540 return 1;

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