Changeset 784

Timestamp:

12/15/06 20:53:19 (2 years ago)

Author:

mfenniak

Message:

add documentation for encryption from PDF ref.

Files:

• pypdf/trunk/pyPdf/pdf.py (modified) (3 diffs)

pypdf/trunk/pyPdf/pdf.py

@@ -1079 +1079 @@
-        '\xa9\xfe\x64\x53\x69\x7a'
-
+# Implementation of algorithm 3.2 of the PDF standard security handler,
+# section 3.5.2 of the PDF 1.6 reference.
-def _alg32(password, rev, keylen, owner_entry, p_entry, id1_entry, metadata_encrypt=True):
+    # 1. Pad or truncate the password string to exactly 32 bytes.  If the
+    # password string is more than 32 bytes long, use only its first 32 bytes;
+    # if it is less than 32 bytes long, pad it by appending the required number
+    # of additional bytes from the beginning of the padding string
+    # (_encryption_padding).
+    password = (password + _encryption_padding)[:32]
+    # 2. Initialize the MD5 hash function and pass the result of step 1 as
+    # input to this function.
-    import md5, struct
-
-password = (password + _encryption_padding)[:32]
-m.update(password)
+password
+# 3. Pass the value of the encryption dictionary's /O entry to the MD5 hash
+# function.
-    m.update(owner_entry)
+    # 4. Treat the value of the /P entry as an unsigned 4-byte integer and pass
+    # these bytes to the MD5 hash function, low-order byte first.
-    p_entry = struct.pack('<i', p_entry)
-    m.update(p_entry)
+    # 5. Pass the first element of the file's file identifier array to the MD5
+    # hash function.
-    m.update(id1_entry)
+    # 6. (Revision 3 or greater) If document metadata is not being encrypted,
+    # pass 4 bytes with the value 0xFFFFFFFF to the MD5 hash function.
-    if rev >= 3 and not metadata_encrypt:
-        m.update("\xff\xff\xff\xff")
+    # 7. Finish the hash.
-    md5_hash = m.digest()
+    # 8. (Revision 3 or greater) Do the following 50 times: Take the output
+    # from the previous MD5 hash and pass the first n bytes of the output as
+    # input into a new MD5 hash, where n is the number of bytes of the
+    # encryption key as defined by the value of the encryption dictionary's
+    # /Length entry.
-    if rev >= 3:
-        for i in range(50):
-            md5_hash = md5.new(md5_hash[:keylen]).digest()
+    # 9. Set the encryption key to the first n bytes of the output from the
+    # final MD5 hash, where n is always 5 for revision 2 but, for revision 3 or
+    # greater, depends on the value of the encryption dictionary's /Length
+    # entry.
-    return md5_hash[:keylen]
-
+# Implementation of algorithm 3.3 of the PDF standard security handler,
+# section 3.5.2 of the PDF 1.6 reference.
-def _alg33(owner_pwd, user_pwd, rev, keylen):
+    # steps 1 - 4
-    key = _alg33_1(owner_pwd, rev, keylen)
+    # 5. Pad or truncate the user password string as described in step 1 of
+    # algorithm 3.2.
-    user_pwd = (user_pwd + _encryption_padding)[:32]
+    # 6. Encrypt the result of step 5, using an RC4 encryption function with
+    # the encryption key obtained in step 4.
-    val = utils.RC4_encrypt(key, user_pwd)
+    # 7. (Revision 3 or greater) Do the following 19 times: Take the output
+    # from the previous invocation of the RC4 function and pass it as input to
+    # a new invocation of the function; use an encryption key generated by
+    # taking each byte of the encryption key obtained in step 4 and performing
+    # an XOR operation between that byte and the single-byte value of the
+    # iteration counter (from 1 to 19).
-    if rev >= 3:
-        for i in range(1, 20):
@@ -1106 +1145 @@
-                new_key += chr(ord(key[l]) ^ i)
-            val = utils.RC4_encrypt(new_key, val)
+    # 8. Store the output from the final invocation of the RC4 as the value of
+    # the /O entry in the encryption dictionary.
-    return val
-
+# Steps 1-4 of algorithm 3.3
-def _alg33_1(password, rev, keylen):
+    # 1. Pad or truncate the owner password string as described in step 1 of
+    # algorithm 3.2.  If there is no owner password, use the user password
+    # instead.
+    password = (password + _encryption_padding)[:32]
+    # 2. Initialize the MD5 hash function and pass the result of step 1 as
+    # input to this function.
-    import md5
-
-password = (password + _encryption_padding)[:32]
-m.update(password)
+password
+# 3. (Revision 3 or greater) Do the following 50 times: Take the output
+# from the previous MD5 hash and pass it as input into a new MD5 hash.
-    md5_hash = m.digest()
-    if rev >= 3:
-        for i in range(50):
-            md5_hash = md5.new(md5_hash).digest()
+    # 4. Create an RC4 encryption key using the first n bytes of the output
+    # from the final MD5 hash, where n is always 5 for revision 2 but, for
+    # revision 3 or greater, depends on the value of the encryption
+    # dictionary's /Length entry.
-    key = md5_hash[:keylen]
-    return key
-
+# Implementation of algorithm 3.4 of the PDF standard security handler,
+# section 3.5.2 of the PDF 1.6 reference.
-def _alg34(password, owner_entry, p_entry, id1_entry):
+    # 1. Create an encryption key based on the user password string, as
+    # described in algorithm 3.2.
-    key = _alg32(password, 2, 5, owner_entry, p_entry, id1_entry)
+    # 2. Encrypt the 32-byte padding string shown in step 1 of algorithm 3.2,
+    # using an RC4 encryption function with the encryption key from the
+    # preceding step.
-    U = utils.RC4_encrypt(key, _encryption_padding)
+    # 3. Store the result of step 2 as the value of the /U entry in the
+    # encryption dictionary.
-    return U, key
-
+# Implementation of algorithm 3.4 of the PDF standard security handler,
+# section 3.5.2 of the PDF 1.6 reference.
-def _alg35(password, rev, keylen, owner_entry, p_entry, id1_entry, metadata_encrypt):
+    # 1. Create an encryption key based on the user password string, as
+    # described in Algorithm 3.2.
+    key = _alg32(password, rev, keylen, owner_entry, p_entry, id1_entry)
+    # 2. Initialize the MD5 hash function and pass the 32-byte padding string
+    # shown in step 1 of Algorithm 3.2 as input to this function. 
-    import md5
-    m = md5.new()
-    m.update(_encryption_padding)
+    # 3. Pass the first element of the file's file identifier array (the value
+    # of the ID entry in the document's trailer dictionary; see Table 3.13 on
+    # page 73) to the hash function and finish the hash.  (See implementation
+    # note 25 in Appendix H.) 
-    m.update(id1_entry)
-    md5_hash = m.digest()
-
+
+
-    val = utils.RC4_encrypt(key, md5_hash)
+    # 5. Do the following 19 times: Take the output from the previous
+    # invocation of the RC4 function and pass it as input to a new invocation
+    # of the function; use an encryption key generated by taking each byte of
+    # the original encryption key (obtained in step 2) and performing an XOR
+    # operation between that byte and the single-byte value of the iteration
+    # counter (from 1 to 19). 
-    for i in range(1, 20):
-        new_key = ''
@@ -1138 +1217 @@
-            new_key += chr(ord(key[l]) ^ i)
-        val = utils.RC4_encrypt(new_key, val)
+    # 6. Append 16 bytes of arbitrary padding to the output from the final
+    # invocation of the RC4 function and store the 32-byte result as the value
+    # of the U entry in the encryption dictionary. 
+    # (implementator note: I don't know what "arbitrary padding" is supposed to
+    # mean, so I have used null bytes.  This seems to match a few other
+    # people's implementations)
-    return val + ('\x00' * 16), key
-