youtube-dl

commit 0a5445ddbeb8d391bbac92e5fe9074c6aa2e1565
parent f48409c7ac186fa38bbeb2df2b210e37a18eb04b
Author: Yen Chi Hsuan <yan12125@gmail.com>
Date:   Tue, 28 Feb 2017 19:16:55 +0800

[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

Diffstat:
M
youtube_dl/utils.py
 | 
51
+++++++++++++++++++++++++++++++++++++++++++++++++++

1 file changed, 51 insertions(+), 0 deletions(-)
diff --git 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/