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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
$begingroup$
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
edited 1 hour ago
m_goldberg
87.4k872198
asked 2 hours ago
bc888bc888
213
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.
$endgroup$
|
show 4 more comments
$begingroup$
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
edited 1 hour ago
m_goldberg
87.4k872198
asked 2 hours ago
bc888bc888
213
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.
$endgroup$
3
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
2 hours ago
$begingroup$
Is there a work around?
$endgroup$
– bc888
2 hours ago
$begingroup$
not any I know of.
$endgroup$
– kglr
2 hours ago
$begingroup$
BitNot should yield 221
$endgroup$
– bc888
2 hours ago
5
$begingroup$
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 thatBitNot[n]is simply equivalent to-1-n."
$endgroup$
– Chip Hurst
2 hours ago
|
show 4 more comments
$begingroup$
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
edited 1 hour ago
m_goldberg
87.4k872198
asked 2 hours ago
bc888bc888
213
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
edited 1 hour ago
m_goldberg
87.4k872198
asked 2 hours ago
bc888bc888
213
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.
edited 1 hour ago
m_goldberg
87.4k872198
asked 2 hours ago
bc888bc888
213
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.
edited 1 hour ago
m_goldberg
87.4k872198
edited 1 hour ago
m_goldberg
87.4k872198
edited 1 hour ago
m_goldberg
87.4k872198
87.4k872198
asked 2 hours ago
bc888bc888
213
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.
asked 2 hours ago
bc888bc888
213
asked 2 hours ago
bc888bc888
213
213
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.
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.
3
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
2 hours ago
$begingroup$
Is there a work around?
$endgroup$
– bc888
2 hours ago
$begingroup$
not any I know of.
$endgroup$
– kglr
2 hours ago
$begingroup$
BitNot should yield 221
$endgroup$
– bc888
2 hours ago
5
$begingroup$
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 thatBitNot[n]is simply equivalent to-1-n."
$endgroup$
– Chip Hurst
2 hours ago
|
show 4 more comments
3
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
2 hours ago
$begingroup$
Is there a work around?
$endgroup$
– bc888
2 hours ago
$begingroup$
not any I know of.
$endgroup$
– kglr
2 hours ago
$begingroup$
BitNot should yield 221
$endgroup$
– bc888
2 hours ago
5
$begingroup$
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 thatBitNot[n]is simply equivalent to-1-n."
$endgroup$
– Chip Hurst
2 hours ago
3
3
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
2 hours ago
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
2 hours ago
$begingroup$
Is there a work around?
$endgroup$
– bc888
2 hours ago
$begingroup$
Is there a work around?
$endgroup$
– bc888
2 hours ago
$begingroup$
not any I know of.
$endgroup$
– kglr
2 hours ago
$begingroup$
not any I know of.
$endgroup$
– kglr
2 hours ago
$begingroup$
BitNot should yield 221
$endgroup$
– bc888
2 hours ago
$begingroup$
BitNot should yield 221
$endgroup$
– bc888
2 hours ago
5
5
$begingroup$
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
2 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
2 hours ago
|
show 4 more comments
4 Answers
4
active
oldest
votes
$begingroup$
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} *)
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
$endgroup$
add a comment |
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
$endgroup$
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered 2 hours ago
bc888bc888
213
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.
$endgroup$
add a comment |
$begingroup$
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}
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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} *)
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
$endgroup$
add a comment |
$begingroup$
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} *)
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
$endgroup$
add a comment |
$begingroup$
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} *)
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
$endgroup$
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} *)
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
answered 2 hours ago
Rohit NamjoshiRohit Namjoshi
1,2861213
1,2861213
add a comment |
add a comment |
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
$endgroup$
add a comment |
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
$endgroup$
add a comment |
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
$endgroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
answered 1 hour ago
Okkes DulgerciOkkes Dulgerci
5,3341918
5,3341918
add a comment |
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered 2 hours ago
bc888bc888
213
New contributor
bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered 2 hours ago
bc888bc888
213
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.
$endgroup$
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered 2 hours ago
bc888bc888
213
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.
$endgroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered 2 hours ago
bc888bc888
213
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.
answered 2 hours ago
bc888bc888
213
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.
answered 2 hours ago
bc888bc888
213
answered 2 hours ago
bc888bc888
213
213
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.
Check out our Code of Conduct.
bc888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
$begingroup$
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}
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
$endgroup$
add a comment |
$begingroup$
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}
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
$endgroup$
add a comment |
$begingroup$
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}
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
$endgroup$
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}
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
answered 1 hour ago
J. M. is slightly pensive♦J. M. is slightly pensive
97.8k10304464
97.8k10304464
add a comment |
add a comment |
bc888 is a new contributor. Be nice, and check out our Code of Conduct.
bc888 is a new contributor. Be nice, and check out our Code of Conduct.
bc888 is a new contributor. Be nice, and check out our Code of Conduct.
bc888 is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
2 hours ago
$begingroup$
Is there a work around?
$endgroup$
– bc888
2 hours ago
$begingroup$
not any I know of.
$endgroup$
– kglr
2 hours ago
$begingroup$
BitNot should yield 221
$endgroup$
– bc888
2 hours ago
5
$begingroup$
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
2 hours ago