BitNot does not flip bits in the way I expected
$begingroup$
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
New contributor
$endgroup$
|
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
New contributor
$endgroup$
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 thatBitNot[n]
is simply equivalent to-1-n
."
$endgroup$
– Chip Hurst
3 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
New contributor
$endgroup$
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
New contributor
New contributor
edited 2 hours ago
m_goldberg
87.4k872198
87.4k872198
New contributor
asked 3 hours ago
bc888bc888
363
363
New contributor
New contributor
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 thatBitNot[n]
is simply equivalent to-1-n
."
$endgroup$
– Chip Hurst
3 hours ago
|
show 4 more comments
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 thatBitNot[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
|
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} *)
$endgroup$
add a comment |
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
$endgroup$
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
New contributor
$endgroup$
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}
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
bc888 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193136%2fbitnot-does-not-flip-bits-in-the-way-i-expected%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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} *)
$endgroup$
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} *)
$endgroup$
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} *)
$endgroup$
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 3 hours ago
Rohit NamjoshiRohit Namjoshi
1,3061213
1,3061213
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}
$endgroup$
add a comment |
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
$endgroup$
add a comment |
$begingroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
$endgroup$
twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n]
twosComplement[35, 8]
{1, 1, 0, 1, 1, 1, 0, 1}
answered 3 hours ago
Okkes DulgerciOkkes Dulgerci
5,3441918
5,3441918
add a comment |
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
New contributor
$endgroup$
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
New contributor
$endgroup$
add a comment |
$begingroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
New contributor
$endgroup$
FlipBits[num_Integer, len_.] :=
Module[{arr}, arr = IntegerDigits[num, 2, len];
1 - arr]
New contributor
New contributor
answered 3 hours ago
bc888bc888
363
363
New contributor
New contributor
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}
$endgroup$
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}
$endgroup$
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}
$endgroup$
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 2 hours 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.
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193136%2fbitnot-does-not-flip-bits-in-the-way-i-expected%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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