def _decrypt(self, password): encrypt = self.trailer['/Encrypt'].getObject() if encrypt['/Filter'] != '/Standard': raise NotImplementedError( "only Standard PDF encryption handler is available") if not (encrypt['/V'] in (1, 2)): raise NotImplementedError( "only algorithm code 1 and 2 are supported") user_password, key = self._authenticateUserPassword(password) if user_password: self._decryption_key = key return 1 else: rev = encrypt['/R'].getObject() if rev == 2: keylen = 5 else: keylen = encrypt['/Length'].getObject() // 8 key = _alg33_1(password, rev, keylen) real_O = encrypt["/O"].getObject() if rev == 2: userpass = utils.RC4_encrypt(key, real_O) else: val = real_O for i in range(19, -1, -1): new_key = b_('') for l in range(len(key)): new_key += b_(chr(utils.ord_(key[l]) ^ i)) val = utils.RC4_encrypt(new_key, val) userpass = val owner_password, key = self._authenticateUserPassword(userpass) if owner_password: self._decryption_key = key return 2 return 0
def decode_pdfdocencoding(byte_array): retval = u_('') for b in byte_array: c = _pdfDocEncoding[ord_(b)] if c == u_('\u0000'): raise UnicodeDecodeError("pdfdocencoding", utils.barray(b), -1, -1, "does not exist in translation table") retval += c return retval
def writeToStream(self, stream, encryption_key): # Try to write the string out as a PDFDocEncoding encoded string. It's # nicer to look at in the PDF file. Sadly, we take a performance hit # here for trying... try: bytearr = encode_pdfdocencoding(self) except UnicodeEncodeError: bytearr = codecs.BOM_UTF16_BE + self.encode("utf-16be") if encryption_key: bytearr = RC4_encrypt(encryption_key, bytearr) obj = ByteStringObject(bytearr) obj.writeToStream(stream, None) else: stream.write(b_("(")) for c in bytearr: if not chr_(c).isalnum() and c != b_(' '): stream.write(b_("\\%03o" % ord_(c))) else: stream.write(b_(chr_(c))) stream.write(b_(")"))
def decode(data, decodeParms): data = decompress(data) predictor = 1 if decodeParms: try: predictor = decodeParms.get("/Predictor", 1) except AttributeError: pass # usually an array with a null object was read # predictor 1 == no predictor if predictor != 1: columns = decodeParms["/Columns"] # PNG prediction: if predictor >= 10 and predictor <= 15: output = StringIO() # PNG prediction can vary from row to row rowlength = columns + 1 assert len(data) % rowlength == 0 prev_rowdata = (0,) * rowlength for row in range(len(data) // rowlength): rowdata = [ord_(x) for x in data[(row*rowlength):((row+1)*rowlength)]] filterByte = rowdata[0] if filterByte == 0: pass elif filterByte == 1: for i in range(2, rowlength): rowdata[i] = (rowdata[i] + rowdata[i-1]) % 256 elif filterByte == 2: for i in range(1, rowlength): rowdata[i] = (rowdata[i] + prev_rowdata[i]) % 256 else: # unsupported PNG filter raise PdfReadError("Unsupported PNG filter %r" % filterByte) prev_rowdata = rowdata output.write(''.join([chr(x) for x in rowdata[1:]])) data = output.getvalue() else: # unsupported predictor raise PdfReadError("Unsupported flatedecode predictor %r" % predictor) return data
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. m = md5() 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.original_bytes) md5_hash = m.digest() # 4. Encrypt the 16-byte result of the hash, using an RC4 encryption # function with the encryption key from step 1. 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 = b_('') for l in range(len(key)): new_key += b_(chr(utils.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 + (b_('\x00') * 16), key