fs: jffs2: Move SYS_JFFS2_SORT_FRAGMENTS to Kconfig
Move the symbol SYS_JFFS2_SORT_FRAGMENTS to Kconfig and use the only remaining part of doc/README.JFFS2 that is still relevant and useful to the help for this option. Signed-off-by: Tom Rini <trini@konsulko.com> Reviewed-by: Simon Glass <sjg@chromium.org>
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JFFS2 options and usage.
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-----------------------
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JFFS2 in U-Boot is a read only implementation of the file system in
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Linux with the same name. To use JFFS2 define CONFIG_CMD_JFFS2.
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The module adds three new commands.
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fsload - load binary file from a file system image
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fsinfo - print information about file systems
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ls - list files in a directory
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chpart - change active partition
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If you do now need the commands, you can enable the filesystem separately
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with CONFIG_FS_JFFS2 and call the jffs2 functions yourself.
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If you boot from a partition which is mounted writable, and you
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update your boot environment by replacing single files on that
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partition, you should also define CONFIG_SYS_JFFS2_SORT_FRAGMENTS. Scanning
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the JFFS2 filesystem takes *much* longer with this feature, though.
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Sorting is done while inserting into the fragment list, which is
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more or less a bubble sort. That algorithm is known to be O(n^2),
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thus you should really consider if you can avoid it!
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---
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TODO.
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Remove the assumption that JFFS can dereference a pointer
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into the disk. The current code do not work with memory holes
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or hardware with a sliding window (PCMCIA).
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@ -17,3 +17,15 @@ config JFFS2_NAND
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depends on FS_JFFS2
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help
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Enable support for NAND flash as the backing store for JFFS2.
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config SYS_JFFS2_SORT_FRAGMENTS
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bool "Enable JFFS2 sorting of filesystem fragments (SLOW!)"
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depends on FS_JFFS2
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help
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If you boot from a partition which is mounted writable, and you
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update your boot environment by replacing single files on that
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partition, you should also define CONFIG_SYS_JFFS2_SORT_FRAGMENTS.
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Scanning the JFFS2 filesystem takes *much* longer with this feature,
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though. Sorting is done while inserting into the fragment list,
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which is more or less a bubble sort. That algorithm is known to be
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O(n^2), thus you should really consider if you can avoid it!
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