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x86: __memcpy_flushcache: fix wrong alignment if size > 2^32 

The first "if" condition in __memcpy_flushcache is supposed to align the
"dest" variable to 8 bytes and copy data up to this alignment.  However,
this condition may misbehave if "size" is greater than 4GiB.

The statement min_t(unsigned, size, ALIGN(dest, 8) - dest); casts both
arguments to unsigned int and selects the smaller one.  However, the
cast truncates high bits in "size" and it results in misbehavior.

For example:

	suppose that size == 0x100000001, dest == 0x200000002
	min_t(unsigned, size, ALIGN(dest, 8) - dest) == min_t(0x1, 0xe) == 0x1;
	...
	dest += 0x1;

so we copy just one byte "and" dest remains unaligned.

This patch fixes the bug by replacing unsigned with size_t.

Signed-off-by: Mikulas Patocka <mpatocka@redhat.com>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
This commit is contained in:
Mikulas Patocka 2022-04-19 09:56:23 -04:00 committed by Linus Torvalds
parent 559089e0a9
commit a6823e4e36
1 changed files with 1 additions and 1 deletions
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2
arch/x86/lib/usercopy_64.c
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@ -119,7 +119,7 @@ void __memcpy_flushcache(void *_dst, const void *_src, size_t size)
/* cache copy and flush to align dest */
if (!IS_ALIGNED(dest, 8)) {
unsigned len = min_t(unsigned, size, ALIGN(dest, 8) - dest);
size_t len = min_t(size_t, size, ALIGN(dest, 8) - dest);
memcpy((void *) dest, (void *) source, len);
clean_cache_range((void *) dest, len);