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

[ Upstream commit a6823e4 ]

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>
Signed-off-by: Sasha Levin <sashal@kernel.org>

arch/x86/lib/usercopy_64.c

@@ -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);