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Make one more llXorBase64[StringsCorrect] case computable.

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When the input string is at most 1 byte (2 base64 characters), the result is computable because the first byte is always zero.
This commit is contained in:
Sei Lisa 2017-01-19 02:58:42 +01:00
parent b8985adca9
commit 7c2b76949e
1 changed files with 14 additions and 8 deletions
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22
lslopt/lslbasefuncs.py
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@ -1800,11 +1800,14 @@ def llXorBase64(s, xor):
if L2 == 0:
# The input xor string starts with zero or one valid Base64 characters.
# This produces garbage bytes (the first byte is zero though).
# We don't produce a result in this case.
raise ELSLCantCompute
if L1 > 2:
# We don't produce a result in this case.
raise ELSLCantCompute
L2 = 2
xor = u'AA'
s = b64decode(s + u'='*(-L1 & 3))
xor = b64decode(xor + u'='*(-L2 & 3))
s = b64decode(s + u'=' * (-L1 & 3))
xor = b64decode(xor + u'=' * (-L2 & 3))
L2 = len(xor)
i = 0
@ -1887,11 +1890,14 @@ def llXorBase64StringsCorrect(s, xor):
if L2 == 0:
# The input xor string starts with zero or one valid Base64 characters.
# This produces garbage bytes (the first byte is zero though).
# We don't produce a result in this case.
raise ELSLCantCompute
if L1 > 2:
# We don't produce a result in this case.
raise ELSLCantCompute
L2 = 2
xor = u'AA'
s = b64decode(s + u'='*(-L1 & 3))
xor = b64decode(xor + u'='*(-L2 & 3)) + b'\x00'
s = b64decode(s + u'=' * (-L1 & 3))
xor = b64decode(xor + u'=' * (-L2 & 3)) + b'\x00'
i = 0
ret = b''