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""" Python implementation of ASCII85/ASCIIHex decoder (Adobe version). This code is in the public domain. """ import re import struct # ascii85decode(data) def ascii85decode(data): """ In ASCII85 encoding, every four bytes are encoded with five ASCII letters, using 85 different types of characters (as 256**4 < 85**5). When the length of the original bytes is not a multiple of 4, a special rule is used for round up. The Adobe's ASCII85 implementation is slightly different from its original in handling the last characters. """ n = b = 0 out = b'' for i in iter(data): c = bytes((i,)) if b'!' <= c and c <= b'u': n += 1 b = b*85+(ord(c)-33) if n == 5: out += struct.pack('>L', b) n = b = 0 elif c == b'z': assert n == 0, str(n) out += b'\0\0\0\0' elif c == b'~': if n: for _ in range(5-n): b = b*85+84 out += struct.pack('>L', b)[:n-1] break return out # asciihexdecode(data) hex_re = re.compile(br'([a-f\d]{2})', re.IGNORECASE) trail_re = re.compile(br'^(?:[a-f\d]{2}|\s)*([a-f\d])[\s>]*$', re.IGNORECASE) def asciihexdecode(data): """ ASCIIHexDecode filter: PDFReference v1.4 section 3.3.1 For each pair of ASCII hexadecimal digits (0-9 and A-F or a-f), the ASCIIHexDecode filter produces one byte of binary data. All white-space characters are ignored. A right angle bracket character (>) indicates EOD. Any other characters will cause an error. If the filter encounters the EOD marker after reading an odd number of hexadecimal digits, it will behave as if a 0 followed the last digit. """ def decode(x): i = int(x, 16) return bytes((i,)) out = b'' for x in hex_re.findall(data): out += decode(x) m = trail_re.search(data) if m: out += decode(m.group(1)+b'0') return out