How to manipulate bits in Javascript?
-2
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
asked Nov 10 at 13:35
Badis Merabet
2,2501630
add a comment |
-2
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
asked Nov 10 at 13:35
Badis Merabet
2,2501630
1
You can set a given bitito0withx & (~(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 one1too 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
add a comment |
-2
-2
-2
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
asked Nov 10 at 13:35
Badis Merabet
2,2501630
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
asked Nov 10 at 13:35
Badis Merabet
2,2501630
asked Nov 10 at 13:35
Badis Merabet
2,2501630
edited Nov 10 at 13:54
asked Nov 10 at 13:35
Badis Merabet
2,2501630
asked Nov 10 at 13:35
Badis Merabet
2,2501630
asked Nov 10 at 13:35
Badis Merabet
2,2501630
2,2501630
1
You can set a given bitito0withx & (~(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 one1too 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
add a comment |
1
You can set a given bitito0withx & (~(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 one1too 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
add a comment |
1 Answer
1
active
oldest
votes
4
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));answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
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
add a comment |
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1 Answer
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oldest
votes
4
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));answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
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
add a comment |
4
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));answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
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
add a comment |
4
4
4
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));answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
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));answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
edited Nov 10 at 13:49
answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
answered Nov 10 at 13:41
T.J. Crowder
676k12011961292
676k12011961292
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
add a comment |
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
add a comment |
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1
You can set a given bit
ito0withx & (~(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
1too 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