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JavaScript - Swap two bits at given position in an integer

JavaScript Bit Manipulation: Exercise-7 with Solution

Write a JavaScript program to swap two bits (from the right side, rightmost position is 0) in the binary representation of an integer at the given positions.

Test Data:
(245) -> 249

Explanation:
245 -> 11110101
Swap the 1st and 4th bits from the side of the said binary number.
11100111 -> 231
(137) -> 73
Explanation:
137 -> 10001001
Swap the 6th and 7th bits from the side of the said binary number.
01001001 -> 73

Sample Solution:

JavaScript Code:

const swap_bits = (n, pos1, pos2) => {
   if (typeof n!= "number") {
          return 'It must be number!'
      }
   if ((((n & (1 << pos1)) >> pos1) ^ ((n & (1 << pos2)) >> pos2)) == 1)
        {
            n ^= (1 << pos1);
            n ^= (1 << pos2);
        }
   return n;
}
console.log(swap_bits(245,1,4))
console.log(swap_bits(137,6,7))
console.log(swap_bits("16"))

Sample Output:

231
73
It must be number!

Flowchart:

Flowchart: JavaScript - Swap two bits at given position in an integer.

Live Demo:

See the Pen javascript-bit-manipulation-exercise-7 by w3resource (@w3resource) on CodePen.


* To run the code mouse over on Result panel and click on 'RERUN' button.*

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Previous: Check a number is a power of 4 or not.
Next: Binary logarithm using bitwise operators.

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JavaScript: Tips of the Day

Returns the sum of an array, after mapping each element to a value using the provided function

Example:

const tips_sumBy = (arr, fn) =>
  arr.map(typeof fn === 'function' ? fn : val => val[fn]).reduce((acc, val) => acc + val, 0);

console.log(tips_sumBy([{ n: 2 }, { n: 4 }, { n: 6 }, { n: 8 }], o => o.n));
console.log(tips_sumBy([{ n: 1 }, { n: 3 }, { n: 5 }, { n: 7 }], 'n'));

Output:

20
16