From dae8ce0d85478e16f2368b2341632775313d41ed Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Sybren=20A=2E=20St=C3=BCvel?= <sybren@stuvel.eu>
Date: Sun, 15 Nov 2020 15:18:38 +0100
Subject: [PATCH] Fix #165: CVE-2020-25658 - Bleichenbacher-style timing oracle

Use as many constant-time comparisons as practical in the
`rsa.pkcs1.decrypt` function.

`cleartext.index(b'\x00', 2)` will still be non-constant-time. The
alternative would be to iterate over all the data byte by byte in
Python, which is several orders of magnitude slower. Given that a
perfect constant-time implementation is very hard or even impossible to
do in Python [1], I chose the more performant option here.

[1]: https://securitypitfalls.wordpress.com/2018/08/03/constant-time-compare-in-python/
---
 CHANGELOG.md |  5 +++++
 rsa/pkcs1.py | 12 ++++++++----
 2 files changed, 13 insertions(+), 4 deletions(-)

diff --git a/CHANGELOG.md b/CHANGELOG.md
index bc41f70..1838377 100644
--- a/CHANGELOG.md
+++ b/CHANGELOG.md
@@ -1,5 +1,10 @@
 # Python-RSA changelog
 
+## Version 4.7 - in development
+
+- Fix #165: CVE-2020-25658 - Bleichenbacher-style timing oracle in PKCS#1 v1.5
+  decryption code
+
 
 ## Version 4.4 & 4.6 - released 2020-06-12
 
diff --git a/rsa/pkcs1.py b/rsa/pkcs1.py
index 57b0276..19e24c7 100644
--- a/rsa/pkcs1.py
+++ b/rsa/pkcs1.py
@@ -30,6 +30,7 @@
 import os
 import sys
 import typing
+from hmac import compare_digest
 
 from . import common, transform, core, key
 
@@ -251,17 +252,20 @@ def decrypt(crypto: bytes, priv_key: key.PrivateKey) -> bytes:
     # Detect leading zeroes in the crypto. These are not reflected in the
     # encrypted value (as leading zeroes do not influence the value of an
     # integer). This fixes CVE-2020-13757.
-    if len(crypto) > blocksize:
-        raise DecryptionError('Decryption failed')
+    crypto_len_bad = len(crypto) > blocksize
 
     # If we can't find the cleartext marker, decryption failed.
-    if cleartext[0:2] != b'\x00\x02':
-        raise DecryptionError('Decryption failed')
+    cleartext_marker_bad = not compare_digest(cleartext[:2], b'\x00\x02')
 
     # Find the 00 separator between the padding and the message
     try:
         sep_idx = cleartext.index(b'\x00', 2)
     except ValueError:
+        sep_idx = -1
+    sep_idx_bad = sep_idx < 0
+
+    anything_bad = crypto_len_bad | cleartext_marker_bad | sep_idx_bad
+    if anything_bad:
         raise DecryptionError('Decryption failed')
 
     return cleartext[sep_idx + 1:]