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authorSteven Miao <realmz6@gmail.com>2011-06-06 00:09:14 -0400
committerArtem Bityutskiy <Artem.Bityutskiy@nokia.com>2011-06-06 15:36:25 +0300
commit69b83b9c043d3c1e5e05726f3ec7abff3f418d70 (patch)
tree0fa0a89b299b7027a237ac5320826dd78f9bff12 /lib/libmtd_legacy.c
parentb94485374c836450e66537494347386351ab887c (diff)
flash_{lock, unlock}: fix off-by-one error for "entire device" length
This basically reverts commit 43feb39f35a9ee0ed3 which changed the full length calculation to be one less than one sector. I don't understand the logic in the commit message where it states that the length should be one sector smaller as this results in misbehavior at runtime. For example, with a mtd device with total size 0x400000 and erase block size of 0x20000 (which gives us a total of 32 sectors), this new logic results in: mtdLockInfo.start = 0; mtdLockInfo.length = 0x3e0000; /* (32 - 1) * 0x20000 */ Calling MEMLOCK/MEMUNLOCK on the device with this range leaves the last sector unchanged which is certainly not what we want. So drop this -1 part of the calculation. To look at it another way, if we only attempt to lock one sector, this calculation would end up with the .length set to 0. Calling MEMLOCK with a length of 0 does not lock the sector as this simple code shows: int main(int argc, char *argv[]) { erase_info_t e0 = { 0, 0 }, e1 = { 0, 0x20000 }; int fd = open(argv[1], O_RDONLY); ioctl(fd, MEMUNLOCK, &e1); printf("%i\n", ioctl(fd, MEMISLOCKED, &e1)); ioctl(fd, MEMLOCK, &e0); printf("%i\n", ioctl(fd, MEMISLOCKED, &e1)); } MEMISLOCKED returns 0 both times. If we change the argument to MEMLOCK to e1, then MEMISLOCKED returns 1. Signed-off-by: Steven Miao <realmz6@gmail.com> Signed-off-by: Mike Frysinger <vapier@gentoo.org> Signed-off-by: Artem Bityutskiy <Artem.Bityutskiy@nokia.com>
Diffstat (limited to 'lib/libmtd_legacy.c')
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