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how to get sum of odd, even using reduce method, i have done as show in below code but returning undefined , @js-beginner

//code below

nums= [1,2,3,4,5,6,7,8,9]

function getOddEvenSum(numbers){     
    let{even,odd} = numbers.reduce((acc, cuu) => cuu%2 === 0?acc.even + cuu:acc.odd+cuu,{even:0, odd:0})

    return {even, odd}
  }

console.log(getOddEvenSum(nums)

//output i am getting below

{even:undefined, odd:undefined}

how to get sum of odd, even using reduce method, i have done as show in below code but returning undefined , @js-beginner

//code below

nums= [1,2,3,4,5,6,7,8,9]

function getOddEvenSum(numbers){     
    let{even,odd} = numbers.reduce((acc, cuu) => cuu%2 === 0?acc.even + cuu:acc.odd+cuu,{even:0, odd:0})

    return {even, odd}
  }

console.log(getOddEvenSum(nums)

//output i am getting below

{even:undefined, odd:undefined}

• javascript
• arrays
• reduce

Share Improve this question Follow edited Aug 27, 2021 at 11:39 sunilsalat asked Aug 27, 2021 at 11:26 sunilsalatsunilsalat 13611 silver badge1111 bronze badges 3

• You can use forEach and add the variables to respective conditions instead. – Dhana D. Commented Aug 27, 2021 at 11:31
• 1 You don't actually assign incremented value to acc properties, try to change acc.even + cuu:acc.odd+cuu for acc.even += cuu:acc.odd+=cuu and don't forget to return acc itself: numbers.reduce((acc, cuu) => (cuu%2 === 0?acc.even += cuu:acc.odd+=cuu, acc),{even:0, odd:0}) – Yevhen Horbunkov Commented Aug 27, 2021 at 11:33
• Furthermore, there's no point in destructuring and bining back the same object, you may simply do const getOddEvenSum = numbers => numbers.reduce(.. – Yevhen Horbunkov Commented Aug 27, 2021 at 11:38

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4 Answers 4

Sorted by: Reset to default 4

The value that you return from your reduce callback will be the value of acc upon the next invocation/iteration of your array of numbers. Currently, your acc starts off as an object, but as you're only returning a number from your first iteration, all subsequent iterations will use a number as acc, which don't have .even or .odd properties. You could instead return a new object with updated even/odd properties so that acc remains as an object through all iterations:

const nums = [1,2,3,4,5,6,7,8,9];

function getOddEven(numbers){     
  return numbers.reduce((acc, cuu) => cuu % 2 === 0
    ? {odd: acc.odd, even: acc.even + cuu}
    : {even: acc.even, odd: acc.odd+cuu},
  {even:0, odd:0});
}

console.log(getOddEven(nums));

You can use Array.prototype.reduce like this:

const nums = [1, 2, 3, 4, 5, 6, 7, 8, 9];

const [odds, evens] = nums.reduce(
  ([odds, evens], cur) =>
    cur % 2 === 0 ? [odds, evens + cur] : [odds + cur, evens],
  [0, 0]
);

console.log(odds);
console.log(evens);

This is not how the syntax of reduce works. One possible implementation:

function getOddEven(nums) {
  return nums.reduce(
    ({odd, even}, num) => num % 2 === 0 ?
      {odd, even: even + num} :
      {odd: odd + num, even},
    {odd: 0, even: 0},
  );
}

I would argue that this is not very clear. Since performance is probably not critical, a clearer alternative would be:

function getOddEven(nums) {
  return {
    odd: nums.filter(num => num % 2 == 1).reduce((acc, num) => acc + num),
    even: nums.filter(num => num % 2 == 0).reduce((acc, num) => acc + num),
  };
}

based on your code, you need to return acc

let nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]

    function getOddEven(numbers) {
      let {
        even,
        odd
      } = numbers.reduce((acc, cuu) => {
      
        if(cuu%2 == 0) 
        {
        acc.even += cuu    
        }
        else{
        acc.odd += cuu   
        }
        
        return acc
      
      }, 
      {
        even: 0,
        odd: 0
      })
    
      return {
        even,
        odd
      }
    }
    
    console.log(getOddEven(nums))

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