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>
This commit is contained in:
Jason A. Donenfeld 2022-04-14 01:50:38 +02:00
parent b0c3e796f2
commit 35a33ff380

View File

@ -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));
}