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BitNot does not flip bits in the way I expected


Bitwise operators - Hamlet for MathematicaHow does Mathematica decide that Log[2,8] is integer?How is the mysterious Raw function used?Graph from binary matrix (not adjacency) respecting the original matrix positionsHow to replace the selected value in matrix with a specific digit?Performance-driven approach to using Big Data without killing the hard-driveUsing the binary search algorithm in a sorted grid













3












$begingroup$


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









share|improve this question









New contributor




bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 4




    $begingroup$
    "IntegerDigits[n] discards the sign of n."
    $endgroup$
    – kglr
    7 hours ago











  • $begingroup$
    Is there a work around?
    $endgroup$
    – bc888
    7 hours ago










  • $begingroup$
    not any I know of.
    $endgroup$
    – kglr
    7 hours ago










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    7 hours ago






  • 6




    $begingroup$
    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."
    $endgroup$
    – Chip Hurst
    7 hours ago















3












$begingroup$


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









share|improve this question









New contributor




bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 4




    $begingroup$
    "IntegerDigits[n] discards the sign of n."
    $endgroup$
    – kglr
    7 hours ago











  • $begingroup$
    Is there a work around?
    $endgroup$
    – bc888
    7 hours ago










  • $begingroup$
    not any I know of.
    $endgroup$
    – kglr
    7 hours ago










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    7 hours ago






  • 6




    $begingroup$
    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."
    $endgroup$
    – Chip Hurst
    7 hours ago













3












3








3


1



$begingroup$


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









share|improve this question









New contributor




bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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






binary






share|improve this question









New contributor




bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 6 hours ago









m_goldberg

87.4k872198




87.4k872198






New contributor




bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 7 hours ago









bc888bc888

363




363




New contributor




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New contributor





bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 4




    $begingroup$
    "IntegerDigits[n] discards the sign of n."
    $endgroup$
    – kglr
    7 hours ago











  • $begingroup$
    Is there a work around?
    $endgroup$
    – bc888
    7 hours ago










  • $begingroup$
    not any I know of.
    $endgroup$
    – kglr
    7 hours ago










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    7 hours ago






  • 6




    $begingroup$
    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."
    $endgroup$
    – Chip Hurst
    7 hours ago












  • 4




    $begingroup$
    "IntegerDigits[n] discards the sign of n."
    $endgroup$
    – kglr
    7 hours ago











  • $begingroup$
    Is there a work around?
    $endgroup$
    – bc888
    7 hours ago










  • $begingroup$
    not any I know of.
    $endgroup$
    – kglr
    7 hours ago










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    7 hours ago






  • 6




    $begingroup$
    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."
    $endgroup$
    – Chip Hurst
    7 hours ago







4




4




$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
7 hours ago





$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
7 hours ago













$begingroup$
Is there a work around?
$endgroup$
– bc888
7 hours ago




$begingroup$
Is there a work around?
$endgroup$
– bc888
7 hours ago












$begingroup$
not any I know of.
$endgroup$
– kglr
7 hours ago




$begingroup$
not any I know of.
$endgroup$
– kglr
7 hours ago












$begingroup$
BitNot should yield 221
$endgroup$
– bc888
7 hours ago




$begingroup$
BitNot should yield 221
$endgroup$
– bc888
7 hours ago




6




6




$begingroup$
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."
$endgroup$
– Chip Hurst
7 hours ago




$begingroup$
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."
$endgroup$
– Chip Hurst
7 hours ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

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 *)





share|improve this answer









$endgroup$




















    5












    $begingroup$

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



    1, 1, 0, 1, 1, 1, 0, 1







    share|improve this answer









    $endgroup$




















      2












      $begingroup$

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





      share|improve this answer








      New contributor




      bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




















        1












        $begingroup$

        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





        share|improve this answer









        $endgroup$












          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          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 *)





          share|improve this answer









          $endgroup$

















            2












            $begingroup$

            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 *)





            share|improve this answer









            $endgroup$















              2












              2








              2





              $begingroup$

              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 *)





              share|improve this answer









              $endgroup$



              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 *)






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 7 hours ago









              Rohit NamjoshiRohit Namjoshi

              1,3061213




              1,3061213





















                  5












                  $begingroup$

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



                  1, 1, 0, 1, 1, 1, 0, 1







                  share|improve this answer









                  $endgroup$

















                    5












                    $begingroup$

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



                    1, 1, 0, 1, 1, 1, 0, 1







                    share|improve this answer









                    $endgroup$















                      5












                      5








                      5





                      $begingroup$

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



                      1, 1, 0, 1, 1, 1, 0, 1







                      share|improve this answer









                      $endgroup$



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



                      1, 1, 0, 1, 1, 1, 0, 1








                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 7 hours ago









                      Okkes DulgerciOkkes Dulgerci

                      5,3541918




                      5,3541918





















                          2












                          $begingroup$

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





                          share|improve this answer








                          New contributor




                          bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$

















                            2












                            $begingroup$

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





                            share|improve this answer








                            New contributor




                            bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            $endgroup$















                              2












                              2








                              2





                              $begingroup$

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





                              share|improve this answer








                              New contributor




                              bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                              $endgroup$



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






                              share|improve this answer








                              New contributor




                              bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                              share|improve this answer



                              share|improve this answer






                              New contributor




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                              answered 7 hours ago









                              bc888bc888

                              363




                              363




                              New contributor




                              bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                              New contributor





                              bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                              bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.





















                                  1












                                  $begingroup$

                                  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





                                  share|improve this answer









                                  $endgroup$

















                                    1












                                    $begingroup$

                                    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





                                    share|improve this answer









                                    $endgroup$















                                      1












                                      1








                                      1





                                      $begingroup$

                                      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





                                      share|improve this answer









                                      $endgroup$



                                      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






                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 6 hours ago









                                      J. M. is slightly pensiveJ. M. is slightly pensive

                                      97.8k10304464




                                      97.8k10304464




















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