[PATCH] random: use memmove instead of memcpy for remaining 32 bytes

From: Jason A. Donenfeld
Date: Wed Apr 13 2022 - 19:57:11 EST


In order to immediately overwrite the old key on the stack, before
servicing a userspace request for bytes, we use the remaining 32 bytes
of block 0 as the key. This means moving indices 8,9,a,b,c,d,e,f ->
4,5,6,7,8,9,a,b. Since 4 < 8, for the kernel implementations of
memcpy(), this doesn't actually appear to be a problem in practice. But
relying on that characteristic seems a bit brittle. So let's change that
to a proper memmove(), which is the by-the-books way of handling
overlapping memory copies.

Cc: Dominik Brodowski <linux@xxxxxxxxxxxxxxxxxxxx>
Signed-off-by: Jason A. Donenfeld <Jason@xxxxxxxxx>
---
drivers/char/random.c | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/drivers/char/random.c b/drivers/char/random.c
index 6b01b2be9dd4..3a293f919af9 100644
--- a/drivers/char/random.c
+++ b/drivers/char/random.c
@@ -333,7 +333,7 @@ static void crng_fast_key_erasure(u8 key[CHACHA_KEY_SIZE],
chacha20_block(chacha_state, first_block);

memcpy(key, first_block, CHACHA_KEY_SIZE);
- memcpy(random_data, first_block + CHACHA_KEY_SIZE, random_data_len);
+ memmove(random_data, first_block + CHACHA_KEY_SIZE, random_data_len);
memzero_explicit(first_block, sizeof(first_block));
}

--
2.35.1