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Can anyone explain why the last result in these statements is not the bit-flipped version of arr?

(Debug) In[189]:= arr = {0, 0, 1, 0, 0, 0, 1, 0}

(Debug) Out[189]= {0, 0, 1, 0, 0, 0, 1, 0}

(Debug) In[190]:= FromDigits[%, 2]

(Debug) Out[190]= 34

(Debug) In[191]:= BitNot[%]

(Debug) Out[191]= -35

(Debug) In[192]:= IntegerDigits[%, 2, 8]

(Debug) Out[192]= {0, 0, 1, 0, 0, 0, 1, 1}
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10
  • 5
    $\begingroup$ "IntegerDigits[n] discards the sign of n." $\endgroup$
    – kglr
    Commented Mar 12, 2019 at 21:37
  • $\begingroup$ Is there a work around? $\endgroup$
    – bc888
    Commented Mar 12, 2019 at 21:43
  • $\begingroup$ not any I know of. $\endgroup$
    – kglr
    Commented Mar 12, 2019 at 21:44
  • 10
    $\begingroup$ Integers can have arbitrary length, so how many leading zeros should be flipped? The documentation clarifies: "Integers are assumed to be represented in two's complement form, with an unlimited number of digits, so that BitNot[n] is simply equivalent to -1-n." $\endgroup$
    – Greg Hurst
    Commented Mar 12, 2019 at 22:00
  • 2
    $\begingroup$ If you just want to flip "bits" in an array of 1/0 elements without the need to go between integer representation, just use BitXor[1,array]... $\endgroup$
    – ciao
    Commented Mar 14, 2019 at 6:34

4 Answers 4

Reset to default
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I don't think there is a built-in function to generate the two's complement representation. Easy to implement though.

twosComplement[x_, n_] := IntegerDigits[2^x - n, 2, n]
twosComplement[35, 8]
(* {1, 1, 0, 1, 1, 1, 0, 1} *)
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$\begingroup$ 
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]

{1, 1, 0, 1, 1, 1, 0, 1}

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Without using IntegerDigits[]:

With[{n = 34}, 
     {n, BitXor[BitShiftLeft[1, BitLength[n]] - 1, n]} // BaseForm[#, 2] &]
   {100010₂, 11101₂}

With[{n = 34, p = 8},
     {n, BitXor[BitShiftLeft[1, p] - 1, n]} // BaseForm[#, 2] &]
   {100010₂, 11011101₂}
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$\begingroup$ 
FlipBits[num_Integer, len_.] := 
 Module[{arr}, arr = IntegerDigits[num, 2, len];
  1 - arr]
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