How to manipulate bits in Javascript?









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Let's say I have a variable X = 4



How to create a number with a binary representation 1111 (ones with length X) using bitwise operators ( & | ~ << ^ ) and take a position and toggle the bit at that position to zero (0).



Example:



X = initial(4) // X should be : 1111
Y = solution(X, 2) // Y should be 1101
Z = solution(Y, 3) // Z should be 1001





javascript bit-manipulation







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  • 1




    You can set a given bit i to 0 with x & (~(1<<i)).
    – Willem Van Onsem
    Nov 10 at 13:37










  • ok, how to go from number 4 to 1111 ?
    – Badis Merabet
    Nov 10 at 13:38






  • 2




    @BadisMerabet: (1<<(n+1))-1.
    – Willem Van Onsem
    Nov 10 at 13:38






  • 1




    @WillemVanOnsem That's one 1 too much: console.log(((1<<(4+1))-1).toString(2)) // "11111"
    – Andreas
    Nov 10 at 13:41






  • 1




    @Andreas: argh, yes, (1<<4)-1.
    – Willem Van Onsem
    Nov 10 at 13:44














up vote
-2
down vote

favorite
1












Let's say I have a variable X = 4



How to create a number with a binary representation 1111 (ones with length X) using bitwise operators ( & | ~ << ^ ) and take a position and toggle the bit at that position to zero (0).



Example:



X = initial(4) // X should be : 1111
Y = solution(X, 2) // Y should be 1101
Z = solution(Y, 3) // Z should be 1001





javascript bit-manipulation







share|improve this question



















  • 1




    You can set a given bit i to 0 with x & (~(1<<i)).
    – Willem Van Onsem
    Nov 10 at 13:37










  • ok, how to go from number 4 to 1111 ?
    – Badis Merabet
    Nov 10 at 13:38






  • 2




    @BadisMerabet: (1<<(n+1))-1.
    – Willem Van Onsem
    Nov 10 at 13:38






  • 1




    @WillemVanOnsem That's one 1 too much: console.log(((1<<(4+1))-1).toString(2)) // "11111"
    – Andreas
    Nov 10 at 13:41






  • 1




    @Andreas: argh, yes, (1<<4)-1.
    – Willem Van Onsem
    Nov 10 at 13:44












up vote
-2
down vote

favorite
1









up vote
-2
down vote

favorite
1






1





Let's say I have a variable X = 4



How to create a number with a binary representation 1111 (ones with length X) using bitwise operators ( & | ~ << ^ ) and take a position and toggle the bit at that position to zero (0).



Example:



X = initial(4) // X should be : 1111
Y = solution(X, 2) // Y should be 1101
Z = solution(Y, 3) // Z should be 1001





javascript bit-manipulation







share|improve this question















Let's say I have a variable X = 4



How to create a number with a binary representation 1111 (ones with length X) using bitwise operators ( & | ~ << ^ ) and take a position and toggle the bit at that position to zero (0).



Example:



X = initial(4) // X should be : 1111
Y = solution(X, 2) // Y should be 1101
Z = solution(Y, 3) // Z should be 1001





javascript bit-manipulation




javascript bit-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 10 at 13:54

























asked Nov 10 at 13:35









Badis Merabet

1,9931529




1,9931529







  • 1




    You can set a given bit i to 0 with x & (~(1<<i)).
    – Willem Van Onsem
    Nov 10 at 13:37










  • ok, how to go from number 4 to 1111 ?
    – Badis Merabet
    Nov 10 at 13:38






  • 2




    @BadisMerabet: (1<<(n+1))-1.
    – Willem Van Onsem
    Nov 10 at 13:38






  • 1




    @WillemVanOnsem That's one 1 too much: console.log(((1<<(4+1))-1).toString(2)) // "11111"
    – Andreas
    Nov 10 at 13:41






  • 1




    @Andreas: argh, yes, (1<<4)-1.
    – Willem Van Onsem
    Nov 10 at 13:44












  • 1




    You can set a given bit i to 0 with x & (~(1<<i)).
    – Willem Van Onsem
    Nov 10 at 13:37










  • ok, how to go from number 4 to 1111 ?
    – Badis Merabet
    Nov 10 at 13:38






  • 2




    @BadisMerabet: (1<<(n+1))-1.
    – Willem Van Onsem
    Nov 10 at 13:38






  • 1




    @WillemVanOnsem That's one 1 too much: console.log(((1<<(4+1))-1).toString(2)) // "11111"
    – Andreas
    Nov 10 at 13:41






  • 1




    @Andreas: argh, yes, (1<<4)-1.
    – Willem Van Onsem
    Nov 10 at 13:44







1




1




You can set a given bit i to 0 with x & (~(1<<i)).
– Willem Van Onsem
Nov 10 at 13:37




You can set a given bit i to 0 with x & (~(1<<i)).
– Willem Van Onsem
Nov 10 at 13:37












ok, how to go from number 4 to 1111 ?
– Badis Merabet
Nov 10 at 13:38




ok, how to go from number 4 to 1111 ?
– Badis Merabet
Nov 10 at 13:38




2




2




@BadisMerabet: (1<<(n+1))-1.
– Willem Van Onsem
Nov 10 at 13:38




@BadisMerabet: (1<<(n+1))-1.
– Willem Van Onsem
Nov 10 at 13:38




1




1




@WillemVanOnsem That's one 1 too much: console.log(((1<<(4+1))-1).toString(2)) // "11111"
– Andreas
Nov 10 at 13:41




@WillemVanOnsem That's one 1 too much: console.log(((1<<(4+1))-1).toString(2)) // "11111"
– Andreas
Nov 10 at 13:41




1




1




@Andreas: argh, yes, (1<<4)-1.
– Willem Van Onsem
Nov 10 at 13:44




@Andreas: argh, yes, (1<<4)-1.
– Willem Van Onsem
Nov 10 at 13:44












1 Answer
1






active

oldest

votes

















up vote
4
down vote













Yes, you'd use Math.pow (or on modern browsers, the exponentiation operator, **) and the bitwise operators to do that.






function initial(digits) 
 return Math.pow(2, digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





Or — doh! — comments on the question point out that you can create the initial number without Math.pow or exponentiation:






function initial(digits) 
 return (1 << digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));








share|improve this answer






















  • Yes, I like the second solution. more down to earth.
    – Badis Merabet
    Nov 10 at 13:49






  • 1




    @BadisMerabet - Yeah, I wasn't thinking "bits" enough, but fortunately Willem Van Onsem was.
    – T.J. Crowder
    Nov 10 at 13:51










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1 Answer
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up vote
4
down vote













Yes, you'd use Math.pow (or on modern browsers, the exponentiation operator, **) and the bitwise operators to do that.






function initial(digits) 
 return Math.pow(2, digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





Or — doh! — comments on the question point out that you can create the initial number without Math.pow or exponentiation:






function initial(digits) 
 return (1 << digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));








share|improve this answer






















  • Yes, I like the second solution. more down to earth.
    – Badis Merabet
    Nov 10 at 13:49






  • 1




    @BadisMerabet - Yeah, I wasn't thinking "bits" enough, but fortunately Willem Van Onsem was.
    – T.J. Crowder
    Nov 10 at 13:51














up vote
4
down vote













Yes, you'd use Math.pow (or on modern browsers, the exponentiation operator, **) and the bitwise operators to do that.






function initial(digits) 
 return Math.pow(2, digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





Or — doh! — comments on the question point out that you can create the initial number without Math.pow or exponentiation:






function initial(digits) 
 return (1 << digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));








share|improve this answer






















  • Yes, I like the second solution. more down to earth.
    – Badis Merabet
    Nov 10 at 13:49






  • 1




    @BadisMerabet - Yeah, I wasn't thinking "bits" enough, but fortunately Willem Van Onsem was.
    – T.J. Crowder
    Nov 10 at 13:51












up vote
4
down vote










up vote
4
down vote









Yes, you'd use Math.pow (or on modern browsers, the exponentiation operator, **) and the bitwise operators to do that.






function initial(digits) 
 return Math.pow(2, digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





Or — doh! — comments on the question point out that you can create the initial number without Math.pow or exponentiation:






function initial(digits) 
 return (1 << digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));








share|improve this answer














Yes, you'd use Math.pow (or on modern browsers, the exponentiation operator, **) and the bitwise operators to do that.






function initial(digits) 
 return Math.pow(2, digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





Or — doh! — comments on the question point out that you can create the initial number without Math.pow or exponentiation:






function initial(digits) 
 return (1 << digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));








function initial(digits) 
 return Math.pow(2, digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





function initial(digits) 
 return Math.pow(2, digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





function initial(digits) 
 return (1 << digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));





function initial(digits) 
 return (1 << digits) - 1;

function solution(value, bit) 
 return value & ~(1 << (bit - 1)); // () around `bit - 1` aren't necessary,
 // but I find it clearer

var X = initial(4); // X should be : 1111
console.log(X.toString(2));
var Y = solution(X, 2); // Y should be 1101
console.log(Y.toString(2));
var Z = solution(Y, 3); // Z should be 1001
console.log(Z.toString(2));






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 10 at 13:49

























answered Nov 10 at 13:41









T.J. Crowder

667k11611771273




667k11611771273











  • Yes, I like the second solution. more down to earth.
    – Badis Merabet
    Nov 10 at 13:49






  • 1




    @BadisMerabet - Yeah, I wasn't thinking "bits" enough, but fortunately Willem Van Onsem was.
    – T.J. Crowder
    Nov 10 at 13:51
















  • Yes, I like the second solution. more down to earth.
    – Badis Merabet
    Nov 10 at 13:49






  • 1




    @BadisMerabet - Yeah, I wasn't thinking "bits" enough, but fortunately Willem Van Onsem was.
    – T.J. Crowder
    Nov 10 at 13:51















Yes, I like the second solution. more down to earth.
– Badis Merabet
Nov 10 at 13:49




Yes, I like the second solution. more down to earth.
– Badis Merabet
Nov 10 at 13:49




1




1




@BadisMerabet - Yeah, I wasn't thinking "bits" enough, but fortunately Willem Van Onsem was.
– T.J. Crowder
Nov 10 at 13:51




@BadisMerabet - Yeah, I wasn't thinking "bits" enough, but fortunately Willem Van Onsem was.
– T.J. Crowder
Nov 10 at 13:51

















 

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