BitNot does not flip bits in the way I expected
7
$begingroup$
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 Mar 12 at 22:45
m_goldberg
87.7k872198
asked Mar 12 at 21:30
bc888bc888
764
$endgroup$
|
show 5 more comments
7
$begingroup$
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 Mar 12 at 22:45
m_goldberg
87.7k872198
asked Mar 12 at 21:30
bc888bc888
764
$endgroup$
5
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
Mar 12 at 21:37
$begingroup$
Is there a work around?
$endgroup$
– bc888
Mar 12 at 21:43
$begingroup$
not any I know of.
$endgroup$
– kglr
Mar 12 at 21:44
9
$begingroup$
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
Mar 12 at 22:00
1
$begingroup$
If you just want to flip "bits" in an array of 1/0 elements without the need to go between integer representation, just useBitXor[1,array]...
$endgroup$
– ciao
Mar 14 at 6:34
|
show 5 more comments
7
7
7
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
edited Mar 12 at 22:45
m_goldberg
87.7k872198
asked Mar 12 at 21:30
bc888bc888
764
$endgroup$
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 Mar 12 at 22:45
m_goldberg
87.7k872198
asked Mar 12 at 21:30
bc888bc888
764
edited Mar 12 at 22:45
m_goldberg
87.7k872198
asked Mar 12 at 21:30
bc888bc888
764
edited Mar 12 at 22:45
m_goldberg
87.7k872198
edited Mar 12 at 22:45
m_goldberg
87.7k872198
edited Mar 12 at 22:45
m_goldberg
87.7k872198
87.7k872198
asked Mar 12 at 21:30
bc888bc888
764
asked Mar 12 at 21:30
bc888bc888
764
asked Mar 12 at 21:30
bc888bc888
764
764
5
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
Mar 12 at 21:37
$begingroup$
Is there a work around?
$endgroup$
– bc888
Mar 12 at 21:43
$begingroup$
not any I know of.
$endgroup$
– kglr
Mar 12 at 21:44
9
$begingroup$
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
Mar 12 at 22:00
1
$begingroup$
If you just want to flip "bits" in an array of 1/0 elements without the need to go between integer representation, just useBitXor[1,array]...
$endgroup$
– ciao
Mar 14 at 6:34
|
show 5 more comments
5
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
Mar 12 at 21:37
$begingroup$
Is there a work around?
$endgroup$
– bc888
Mar 12 at 21:43
$begingroup$
not any I know of.
$endgroup$
– kglr
Mar 12 at 21:44
9
$begingroup$
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
Mar 12 at 22:00
1
$begingroup$
If you just want to flip "bits" in an array of 1/0 elements without the need to go between integer representation, just useBitXor[1,array]...
$endgroup$
– ciao
Mar 14 at 6:34
5
5
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
Mar 12 at 21:37
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
Mar 12 at 21:37
$begingroup$
Is there a work around?
$endgroup$
– bc888
Mar 12 at 21:43
$begingroup$
Is there a work around?
$endgroup$
– bc888
Mar 12 at 21:43
$begingroup$
not any I know of.
$endgroup$
– kglr
Mar 12 at 21:44
$begingroup$
not any I know of.
$endgroup$
– kglr
Mar 12 at 21:44
9
9
$begingroup$
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
Mar 12 at 22:00
$begingroup$
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
Mar 12 at 22:00
1
1
$begingroup$
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
Mar 14 at 6:34
$begingroup$
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
Mar 14 at 6:34
|
show 5 more comments
4 Answers
4
active
oldest
votes
9
$begingroup$
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 Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
$endgroup$
add a comment |
10
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
$endgroup$
add a comment |
5
$begingroup$
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 Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
$endgroup$
add a comment |
4
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered Mar 12 at 21:55
bc888bc888
764
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
$begingroup$
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 Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
$endgroup$
add a comment |
9
$begingroup$
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 Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
$endgroup$
add a comment |
9
9
9
$begingroup$
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 Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
$endgroup$
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 Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
answered Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
answered Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
answered Mar 12 at 22:03
Rohit NamjoshiRohit Namjoshi
1,4701213
1,4701213
add a comment |
add a comment |
10
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
$endgroup$
add a comment |
10
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
$endgroup$
add a comment |
10
10
10
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
$endgroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
answered Mar 12 at 22:14
Okkes DulgerciOkkes Dulgerci
5,4141919
5,4141919
add a comment |
add a comment |
5
$begingroup$
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 Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
$endgroup$
add a comment |
5
$begingroup$
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 Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
$endgroup$
add a comment |
5
5
5
$begingroup$
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 Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
$endgroup$
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 Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
answered Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
answered Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
answered Mar 12 at 22:33
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10306465
98.1k10306465
add a comment |
add a comment |
4
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered Mar 12 at 21:55
bc888bc888
764
$endgroup$
add a comment |
4
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered Mar 12 at 21:55
bc888bc888
764
$endgroup$
add a comment |
4
4
4
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered Mar 12 at 21:55
bc888bc888
764
$endgroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
answered Mar 12 at 21:55
bc888bc888
764
answered Mar 12 at 21:55
bc888bc888
764
answered Mar 12 at 21:55
bc888bc888
764
answered Mar 12 at 21:55
bc888bc888
764
764
add a comment |
add a comment |
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5
$begingroup$
"IntegerDigits[n] discards the sign of n."
$endgroup$
– kglr
Mar 12 at 21:37
$begingroup$
Is there a work around?
$endgroup$
– bc888
Mar 12 at 21:43
$begingroup$
not any I know of.
$endgroup$
– kglr
Mar 12 at 21:44
9
$begingroup$
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
Mar 12 at 22:00
1
$begingroup$
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
Mar 14 at 6:34