BitNot does not flip bits in the way I expected












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}





binary







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$endgroup$








  • 4




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












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










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










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    3 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
    3 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}





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.







$endgroup$








  • 4




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












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










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










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    3 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
    3 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}





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.







$endgroup$




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




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




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share|improve this question




share|improve this question








edited 2 hours ago









m_goldberg

87.4k872198




87.4k872198






New contributor




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asked 3 hours ago









bc888bc888

363




363




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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.








  • 4




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












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










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










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    3 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
    3 hours ago














  • 4




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












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










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










  • $begingroup$
    BitNot should yield 221
    $endgroup$
    – bc888
    3 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
    3 hours ago








4




4




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






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














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




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












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




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












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




$begingroup$
BitNot should yield 221
$endgroup$
– bc888
3 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
3 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
3 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









$endgroup$





















    4












    $begingroup$

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



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







    share|improve this answer









    $endgroup$





















      1












      $begingroup$

      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









        $endgroup$













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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

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          active

          oldest

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          active

          oldest

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          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 3 hours ago









              Rohit NamjoshiRohit Namjoshi

              1,3061213




              1,3061213























                  4












                  $begingroup$

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



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







                  share|improve this answer









                  $endgroup$


















                    4












                    $begingroup$

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



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







                    share|improve this answer









                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      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 3 hours ago









                      Okkes DulgerciOkkes Dulgerci

                      5,3441918




                      5,3441918























                          1












                          $begingroup$

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





                          share|improve this answer








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                          $endgroup$


















                            1












                            $begingroup$

                            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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                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              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






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                              answered 3 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.
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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 2 hours ago









                                      J. M. is slightly pensive♦J. M. is slightly pensive

                                      97.8k10304464




                                      97.8k10304464






















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