Question

How to check if an integer contains an even or odd number of bits set in TypeScript

Answer 1

const num = 42; // 00101010
 
// Convert to binary string and count '1's
const bitCount: number = num.toString(2).split('').filter(ch => ch === '1').length;
const result: number = bitCount % 2;
 
console.log("0 = even number of bits set");
console.log("1 = odd number of bits set");
console.log("result: " + result);
 
  
     
/*
run:
      
"0 = even number of bits set" 
"1 = odd number of bits set" 
"result: 1" 
       
*/

 

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Answer 2

const num = 42; // 00101010
 
const result: number = num.toString(2)
                          .split('')
                          .reduce((parity: number, bit: any) => parity ^ bit, 0)
 
console.log("0 = even number of bits set");
console.log("1 = odd number of bits set");
console.log("result: " + result);
 
  
     
/*
run:
      
"0 = even number of bits set" 
"1 = odd number of bits set" 
"result: 1" 
       
*/

 

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