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authorYen Chi Hsuan <yan12125@gmail.com>2017-02-28 19:16:55 +0800
committerYen Chi Hsuan <yan12125@gmail.com>2017-02-28 22:10:31 +0800
commit0a5445ddbeb8d391bbac92e5fe9074c6aa2e1565 (patch)
tree61313a7ee99210ff44e2e9627a71ac91d7f2cf4c
parentf48409c7ac186fa38bbeb2df2b210e37a18eb04b (diff)
[utils] Add bytes_to_long() and long_to_bytes()
Used in daisuki.net (#4738) Both are adapted from public domain PyCrypto: https://github.com/dlitz/pycrypto/blob/master/lib/Crypto/Util/number.py
-rw-r--r--youtube_dl/utils.py51
1 files changed, 51 insertions, 0 deletions
diff --git a/youtube_dl/utils.py b/youtube_dl/utils.py
index 8bd075eaf..807183f4a 100644
--- a/youtube_dl/utils.py
+++ b/youtube_dl/utils.py
@@ -3319,6 +3319,57 @@ class PerRequestProxyHandler(compat_urllib_request.ProxyHandler):
self, req, proxy, type)
+# Both long_to_bytes and bytes_to_long are adapted from PyCrypto, which is
+# released into Public Domain
+# https://github.com/dlitz/pycrypto/blob/master/lib/Crypto/Util/number.py#L387
+
+def long_to_bytes(n, blocksize=0):
+ """long_to_bytes(n:long, blocksize:int) : string
+ Convert a long integer to a byte string.
+
+ If optional blocksize is given and greater than zero, pad the front of the
+ byte string with binary zeros so that the length is a multiple of
+ blocksize.
+ """
+ # after much testing, this algorithm was deemed to be the fastest
+ s = b''
+ n = int(n)
+ while n > 0:
+ s = compat_struct_pack('>I', n & 0xffffffff) + s
+ n = n >> 32
+ # strip off leading zeros
+ for i in range(len(s)):
+ if s[i] != b'\000'[0]:
+ break
+ else:
+ # only happens when n == 0
+ s = b'\000'
+ i = 0
+ s = s[i:]
+ # add back some pad bytes. this could be done more efficiently w.r.t. the
+ # de-padding being done above, but sigh...
+ if blocksize > 0 and len(s) % blocksize:
+ s = (blocksize - len(s) % blocksize) * b'\000' + s
+ return s
+
+
+def bytes_to_long(s):
+ """bytes_to_long(string) : long
+ Convert a byte string to a long integer.
+
+ This is (essentially) the inverse of long_to_bytes().
+ """
+ acc = 0
+ length = len(s)
+ if length % 4:
+ extra = (4 - length % 4)
+ s = b'\000' * extra + s
+ length = length + extra
+ for i in range(0, length, 4):
+ acc = (acc << 32) + compat_struct_unpack('>I', s[i:i + 4])[0]
+ return acc
+
+
def ohdave_rsa_encrypt(data, exponent, modulus):
'''
Implement OHDave's RSA algorithm. See http://www.ohdave.com/rsa/