On 03.11.2016 23:03, Eric Biggers wrote:
> With the new (in 4.9) option to use a virtually-mapped stack
> (CONFIG_VMAP_STACK), stack buffers cannot be used as input/output for
> the scatterlist crypto API because they may not be directly mappable to
> struct page. get_crypt_info() was using a stack buffer to hold the
See 1/2. :-)
> output from the encryption operation used to derive the per-file key.
> Fix it by using a heap buffer.
>
> This bug could most easily be observed in a CONFIG_DEBUG_SG kernel
> because this allowed the BUG in sg_set_buf() to be triggered.
>
> Signed-off-by: Eric Biggers <ebiggers@xxxxxxxxxx>
> ---
> fs/crypto/keyinfo.c | 16 +++++++++++++---
> 1 file changed, 13 insertions(+), 3 deletions(-)
>
> diff --git a/fs/crypto/keyinfo.c b/fs/crypto/keyinfo.c
> index 82f0285..67fb6d8 100644
> --- a/fs/crypto/keyinfo.c
> +++ b/fs/crypto/keyinfo.c
> @@ -185,7 +185,7 @@ int get_crypt_info(struct inode *inode)
> struct crypto_skcipher *ctfm;
> const char *cipher_str;
> int keysize;
> - u8 raw_key[FS_MAX_KEY_SIZE];
> + u8 *raw_key = NULL;
> int res;
>
> res = fscrypt_initialize();
> @@ -238,6 +238,15 @@ int get_crypt_info(struct inode *inode)
> if (res)
> goto out;
>
> + /*
> + * This cannot be a stack buffer because it is passed to the scatterlist
> + * crypto API as part of key derivation.
> + */
> + res = -ENOMEM;
> + raw_key = kmalloc(FS_MAX_KEY_SIZE, GFP_NOFS);
> + if (!raw_key)
> + goto out;
> +
> if (fscrypt_dummy_context_enabled(inode)) {
> memset(raw_key, 0x42, FS_AES_256_XTS_KEY_SIZE);
> goto got_key;
> @@ -276,7 +285,8 @@ int get_crypt_info(struct inode *inode)
> if (res)
> goto out;
>
> - memzero_explicit(raw_key, sizeof(raw_key));
> + kzfree(raw_key);
> + raw_key = NULL;
> if (cmpxchg(&inode->i_crypt_info, NULL, crypt_info) != NULL) {
> put_crypt_info(crypt_info);
> goto retry;
> @@ -287,7 +297,7 @@ int get_crypt_info(struct inode *inode)
> if (res == -ENOKEY)
> res = 0;
> put_crypt_info(crypt_info);
> - memzero_explicit(raw_key, sizeof(raw_key));
> + kzfree(raw_key);
> return res;
> }
Reviewed-by: Richard Weinberger <richard@xxxxxx>
Thanks,
//richard
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