0
 l = [10,11,12,13,14,15,16,17]
 bstr = b'fedora'
 to_xor = bstr[0]

How would I XOR the 8 bits of to_xor with the 8 values in the list in an efficient manner?

i.e. the binary value of to_xor,'f', is 102: 1100110.

I would like to XOR the first bit of 'f' with the LSB of 10, second bit with the LSB of 11, third bit with LSB of 12, and so on.

1 ^ LSB of 10
1 ^ LSB of 11
0 ^ LSB of 12

This post gives some tips on converting bytes to bits, but not on XOR-ing by individual bits.

Cheetaiean
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  • The LSBs of 10,11,12,13,14 etc are 0,1,0,1,0 and so on. Is that really what you meant? Of course if you XOR any value with zero, it will be unchanged – DarkKnight Mar 05 '22 at 19:43
  • Yeah that is the goal. If the particular bit in the letter 'f' is 0, then yes it is fine if the corresponding number in the array is unchanged. – Cheetaiean Mar 05 '22 at 19:57

1 Answers1

1

If you want to XOR every bit of bstr[0] with the LSB of the corresponding integer in l,

[((bstr[0] >> i) & 1) ^ (l[i] & 1) for i in range(8)]

(bstr[0] >> i) & 1) extracts the ith bit of bstr[0], (l[i] & 1) extracts the LSB of the integer l[i], and we know there are 8 bits in a byte, hence the range(8).

What you want seems to be the inverse of this (first integer in l XORs the MSB in bstr[0]), so if the solution above is inverted, try

_l = l[::-1]
XORs = [((bstr[0] >> i) & 1) ^ (_l[i] & 1) for i in range(8)][::-1]
kwsp
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  • Neat, but it appears this XOR's each letter in the bstr with its respective number in the list. However, I just want to XOR the bits of the first letter, 'f', or bstr[0] – Cheetaiean Mar 05 '22 at 19:27
  • @Cheetaiean I've edit my answer – kwsp Mar 05 '22 at 20:00
  • Thanks that is the rough idea. [((bstr[0] >> i) & 1) ^ (l[i] ) for i in range(8)]. However it appears to be in reversed order (it's XOR'ing bstr[0] from right to left instead of left to right, so processing 0110011 instead of 1100110 ). I can compensate for that though, just scan the list from right to left too. – Cheetaiean Mar 05 '22 at 20:08