/*
* linux/fs/ext4/indirect.c
*
* from
*
* linux/fs/ext4/inode.c
*
* Copyright (C) 1992, 1993, 1994, 1995
* Remy Card (card@masi.ibp.fr)
* Laboratoire MASI - Institut Blaise Pascal
* Universite Pierre et Marie Curie (Paris VI)
*
* from
*
* linux/fs/minix/inode.c
*
* Copyright (C) 1991, 1992 Linus Torvalds
*
* Goal-directed block allocation by Stephen Tweedie
* (sct@redhat.com), 1993, 1998
*/
#include <linux/aio.h>
#include "ext4_jbd2.h"
#include "truncate.h"
#include <trace/events/ext4.h>
typedef struct {
__le32 *p;
__le32 key;
struct buffer_head *bh;
} Indirect;
static inline void add_chain(Indirect *p, struct buffer_head *bh, __le32 *v)
{
p->key = *(p->p = v);
p->bh = bh;
}
/**
* ext4_block_to_path - parse the block number into array of offsets
* @inode: inode in question (we are only interested in its superblock)
* @i_block: block number to be parsed
* @offsets: array to store the offsets in
* @boundary: set this non-zero if the referred-to block is likely to be
* followed (on disk) by an indirect block.
*
* To store the locations of file's data ext4 uses a data structure common
* for UNIX filesystems - tree of pointers anchored in the inode, with
* data blocks at leaves and indirect blocks in intermediate nodes.
* This function translates the block number into path in that tree -
* return value is the path length and @offsets[n] is the offset of
* pointer to (n+1)th node in the nth one. If @block is out of range
* (negative or too large) warning is printed and zero returned.
*
* Note: function doesn't find node addresses, so no IO is needed. All
* we need to know is the capacity of indirect blocks (taken from the
* inode->i_sb).
*/
/*
* Portability note: the last comparison (check that we fit into triple
* indirect block) is spelled differently, because otherwise on an
* architecture with 32-bit longs and 8Kb pages we might get into trouble
* if our filesystem had 8Kb blocks. We might use long long, but that would
* kill us on x86. Oh, well, at least the sign propagation does not matter -
* i_block would have to be negative in the very beginning, so we would not
* get there at all.
*/
static int ext4_block_to_path(struct inode *inode,
ext4_lblk_t i_block,
ext4_lblk_t offsets[4], int *boundary)
{
int ptrs = EXT4_ADDR_PER_BLOCK(inode->i_sb);
int ptrs_bits = EXT4_ADDR_PER_BLOCK_BITS(inode->i_sb);
const long direct_blocks = EXT4_NDIR_BLOCKS,
indirect_blocks = ptrs,
double_blocks = (1 << (ptrs_bits * 2));
int n = 0;
int final = 0;
if (i_block < direct_blocks) {
offsets[n++] = i_block;
final 