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| author | Rémi Verschelde <remi@verschelde.fr> | 2022-07-18 15:38:13 +0200 |
|---|---|---|
| committer | GitHub <noreply@github.com> | 2022-07-18 15:38:13 +0200 |
| commit | 5ca5381f2c8d910e32cac078529e6a35cc40c136 (patch) | |
| tree | b6729671dfe327ccc6b058e6f7e9a643e8298128 /thirdparty/mbedtls/library/rsa.c | |
| parent | 4e9d3130f509472cde6eeaacaa0062d841cbebef (diff) | |
| parent | 9403a68853784e542bbff51a84e6dc5c89241d2b (diff) | |
Merge pull request #63146 from Faless/mbedtls/2.18.1
Diffstat (limited to 'thirdparty/mbedtls/library/rsa.c')
| -rw-r--r-- | thirdparty/mbedtls/library/rsa.c | 4 |
1 files changed, 2 insertions, 2 deletions
diff --git a/thirdparty/mbedtls/library/rsa.c b/thirdparty/mbedtls/library/rsa.c index 8a5d40ff1e..d1f6ddb177 100644 --- a/thirdparty/mbedtls/library/rsa.c +++ b/thirdparty/mbedtls/library/rsa.c @@ -832,10 +832,10 @@ cleanup: * the more bits of the key can be recovered. See [3]. * * Collecting n collisions with m bit long blinding value requires 2^(m-m/n) - * observations on avarage. + * observations on average. * * For example with 28 byte blinding to achieve 2 collisions the adversary has - * to make 2^112 observations on avarage. + * to make 2^112 observations on average. * * (With the currently (as of 2017 April) known best algorithms breaking 2048 * bit RSA requires approximately as much time as trying out 2^112 random keys. |