random: use memmove instead of memcpy for remaining 32 bytes

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.

Reviewed-by: Dominik Brodowski <linux@dominikbrodowski.net>
Signed-off-by: Jason A. Donenfeld <Jason@zx2c4.com>

diff --git a/drivers/char/random.c b/drivers/char/random.c
index 6b01b2be9dd497cf3ea08eb16f324c41b427976c..3a293f919af97c5ed4e92b62c658727b5ee1e12b 100644 (file)
--- 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));
 }