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64 lines
2.1 KiB
64 lines
2.1 KiB
#!/usr/bin/env python3
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""" Python implementation of ASCII85/ASCIIHex decoder (Adobe version).
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This code is in the public domain.
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"""
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import re
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import struct
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def ascii85decode(data):
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"""
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In ASCII85 encoding, every four bytes are encoded with five ASCII
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letters, using 85 different types of characters (as 256**4 < 85**5).
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When the length of the original bytes is not a multiple of 4, a special
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rule is used for round up.
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The Adobe's ASCII85 implementation is slightly different from
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its original in handling the last characters.
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The sample string is taken from:
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http://en.wikipedia.org/w/index.php?title=Ascii85
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"""
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if isinstance(data, str):
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data = data.encode('ascii')
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n = b = 0
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out = bytearray()
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for c in data:
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if ord('!') <= c and c <= ord('u'):
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n += 1
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b = b*85+(c-33)
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if n == 5:
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out += struct.pack(b'>L',b)
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n = b = 0
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elif c == ord('z'):
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assert n == 0
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out += b'\0\0\0\0'
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elif c == ord('~'):
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if n:
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for _ in range(5-n):
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b = b*85+84
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out += struct.pack(b'>L',b)[:n-1]
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break
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return bytes(out)
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hex_re = re.compile(r'([a-f\d]{2})', re.IGNORECASE)
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trail_re = re.compile(r'^(?:[a-f\d]{2}|\s)*([a-f\d])[\s>]*$', re.IGNORECASE)
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def asciihexdecode(data):
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"""
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ASCIIHexDecode filter: PDFReference v1.4 section 3.3.1
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For each pair of ASCII hexadecimal digits (0-9 and A-F or a-f), the
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ASCIIHexDecode filter produces one byte of binary data. All white-space
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characters are ignored. A right angle bracket character (>) indicates
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EOD. Any other characters will cause an error. If the filter encounters
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the EOD marker after reading an odd number of hexadecimal digits, it
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will behave as if a 0 followed the last digit.
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"""
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decode = (lambda hx: chr(int(hx, 16)))
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out = list(map(decode, hex_re.findall(data)))
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m = trail_re.search(data)
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if m:
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out.append(decode("%c0" % m.group(1)))
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return ''.join(out)
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