FreeBSD: Fix integer conversion for vnlru_free{,_vfsops}() #13882
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When reviewing openzfs#13875, I noticed that our FreeBSD code has an issue where it converts from `int64_t` to `int` when calling `vnlru_free{,_vfsops}()`. The result is that if the int64_t is `1 << 36`, the int will be 0, since the low bits are 0. Even when some low bits are set, a value such as `((1 << 36) + 1)` would truncate to 1, which is wrong. There is protection against this on 32-bit platforms, but on 64-bit platforms, there is no check to protect us, so we add a check. Signed-off-by: Richard Yao <richard.yao@alumni.stonybrook.edu>
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When reviewing openzfs#13875, I noticed that our FreeBSD code has an issue where it converts from `int64_t` to `int` when calling `vnlru_free{,_vfsops}()`. The result is that if the int64_t is `1 << 36`, the int will be 0, since the low bits are 0. Even when some low bits are set, a value such as `((1 << 36) + 1)` would truncate to 1, which is wrong. There is protection against this on 32-bit platforms, but on 64-bit platforms, there is no check to protect us, so we add a check. Reviewed-by: Alexander Motin <mav@FreeBSD.org> Reviewed-by: Ryan Moeller <ryan@iXsystems.com> Signed-off-by: Richard Yao <richard.yao@alumni.stonybrook.edu> Closes openzfs#13882
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When reviewing openzfs#13875, I noticed that our FreeBSD code has an issue where it converts from `int64_t` to `int` when calling `vnlru_free{,_vfsops}()`. The result is that if the int64_t is `1 << 36`, the int will be 0, since the low bits are 0. Even when some low bits are set, a value such as `((1 << 36) + 1)` would truncate to 1, which is wrong. There is protection against this on 32-bit platforms, but on 64-bit platforms, there is no check to protect us, so we add a check. Reviewed-by: Alexander Motin <mav@FreeBSD.org> Reviewed-by: Ryan Moeller <ryan@iXsystems.com> Signed-off-by: Richard Yao <richard.yao@alumni.stonybrook.edu> Closes openzfs#13882
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When reviewing openzfs#13875, I noticed that our FreeBSD code has an issue where it converts from `int64_t` to `int` when calling `vnlru_free{,_vfsops}()`. The result is that if the int64_t is `1 << 36`, the int will be 0, since the low bits are 0. Even when some low bits are set, a value such as `((1 << 36) + 1)` would truncate to 1, which is wrong. There is protection against this on 32-bit platforms, but on 64-bit platforms, there is no check to protect us, so we add a check. Reviewed-by: Alexander Motin <mav@FreeBSD.org> Reviewed-by: Ryan Moeller <ryan@iXsystems.com> Signed-off-by: Richard Yao <richard.yao@alumni.stonybrook.edu> Closes openzfs#13882
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When reviewing openzfs#13875, I noticed that our FreeBSD code has an issue where it converts from `int64_t` to `int` when calling `vnlru_free{,_vfsops}()`. The result is that if the int64_t is `1 << 36`, the int will be 0, since the low bits are 0. Even when some low bits are set, a value such as `((1 << 36) + 1)` would truncate to 1, which is wrong. There is protection against this on 32-bit platforms, but on 64-bit platforms, there is no check to protect us, so we add a check. Reviewed-by: Alexander Motin <mav@FreeBSD.org> Reviewed-by: Ryan Moeller <ryan@iXsystems.com> Signed-off-by: Richard Yao <richard.yao@alumni.stonybrook.edu> Closes openzfs#13882
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Oct 1, 2022
When reviewing openzfs#13875, I noticed that our FreeBSD code has an issue where it converts from `int64_t` to `int` when calling `vnlru_free{,_vfsops}()`. The result is that if the int64_t is `1 << 36`, the int will be 0, since the low bits are 0. Even when some low bits are set, a value such as `((1 << 36) + 1)` would truncate to 1, which is wrong. There is protection against this on 32-bit platforms, but on 64-bit platforms, there is no check to protect us, so we add a check. Reviewed-by: Alexander Motin <mav@FreeBSD.org> Reviewed-by: Ryan Moeller <ryan@iXsystems.com> Signed-off-by: Richard Yao <richard.yao@alumni.stonybrook.edu> Closes openzfs#13882
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Motivation and Context
When reviewing #13875, I noticed that our FreeBSD code has an issue where it converts from
int64_ttointwhen callingvnlru_free{,_vfsops}(). The result is that if the int64_t is1 << 36, the int will be 0, since the low bits are 0. Even when some low bits are set, a value such as((1 << 36) + 1)would truncate to 1, which is wrong.Description
There is protection against this on 32-bit platforms, but on 64-bit platforms, there is no check to protect us, so we add a check.
How Has This Been Tested?
The buildbot is expected to test for us.
Types of changes
Checklist:
Signed-off-by.