BitNot does not flip bits in the way I expected

Question

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}

I removed the bugs tag, since that requires verification by other users. — Chris K, Mar 12 '19 at 21:48
Welcome to Mathematica.SE, bc888! I suggest the following:

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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." — Greg Hurst, Mar 12 '19 at 22:00
Possible duplicate of Bitwise operators - Hamlet for Mathematica — Daniel Lichtblau, Mar 12 '19 at 23:33
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]... — ciao, Mar 14 '19 at 06:34

score 11 · Answer 1 · answered Mar 12 '19 at 22:14

11

twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]

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

answered Mar 12 '19 at 22:14

OkkesDulgerci

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score 10 · Accepted Answer · answered Mar 12 '19 at 22:03

10

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} *)

answered Mar 12 '19 at 22:03

Rohit Namjoshi

10,212
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score 7 · Answer 3 · answered Mar 12 '19 at 22:33

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}

score 4 · Answer 4 · answered Mar 12 '19 at 21:55

4

FlipBits[num_Integer, len_.] := 
 Module[{arr}, arr = IntegerDigits[num, 2, len];
  1 - arr]

answered Mar 12 '19 at 21:55

bc888

133
4

BitNot does not flip bits in the way I expected

4 Answers4