I tried to recursion those arrays to find odd/even numbers then push them to newArr but the result, not an array, that result is the string with numbers the result after found the odd/even numbers.

this is the code i wrote,

function odd(nums){
  var result = [];
  if(nums.length === 0) {
    return result;
  } else  if (nums[0] % 2 === 0){
    result.push(nums[0])
    // return odd(nums.slice(1))
  };
  return result + odd(nums.slice(1));
};

var arr = [1,8,3,4,4,5,9,13,13,9,10];
var print = odd(arr);
console.log(print)

I tried to recursion those arrays to find odd/even numbers then push them to newArr but the result, not an array, that result is the string with numbers the result after found the odd/even numbers.

this is the code i wrote,

function odd(nums){
  var result = [];
  if(nums.length === 0) {
    return result;
  } else  if (nums[0] % 2 === 0){
    result.push(nums[0])
    // return odd(nums.slice(1))
  };
  return result + odd(nums.slice(1));
};

var arr = [1,8,3,4,4,5,9,13,13,9,10];
var print = odd(arr);
console.log(print)

can anyone help me to explain why that result in a string, not an array I wanted

• javascript
• arrays
• loops
• if-statement
• recursion

5 Answers 5

You need to concat arrays. + does only work for strings or for numbers.

BTW, after block statements { ... }, you do not need a semicolon.

function odd(nums){
    var result = [];
    if (nums.length === 0) {
        return result;
    } 
    if (nums[0] % 2 === 0) {
        result.push(nums[0]);
    }
    return result.concat(odd(nums.slice(1)));
}

var arr = [1, 8, 3, 4, 4, 5, 9, 13, 13, 9, 10];
var print = odd(arr);
console.log(print);

Recursion is probably a bad idea in JS, you should use it unless there is no better way to solve a problem. It will be slower and possibly result in overflow, if too many calls of recursive function is made.

To get the result you needed, it is better to use array filter method. It is clearer and faster.

var arr = [1,8,3,4,4,5,9,13,13,9,10];
var odd = arr.filter(n => !(n % 2))
console.log(odd)

As for the original question, I'd do it this way:

function odd(nums, result = []){
  if (!nums.length) return result; // no need for else, this way clearer
  if (nums[0] % 2 === 0) result.push(nums[0])
  return odd(nums.slice(1), result)
};

var arr = [1,8,3,4,4,5,9,13,13,9,10];
console.log(odd(arr))

!!! Caution, next snippet WILL result in StackOverflow !!!

You may decrease array length to 10000 to check that it will work just fine.

function odd(nums, result = []){
  if (!nums.length) return result; // no need for else, this way clearer
  if (nums[0] % 2 === 0) result.push(nums[0])
  return odd(nums.slice(1), result)
};

const arr = Array.from({length: 100000}, (_,i) => i)
const result = odd(arr)
console.log(`Array length: ${arr.length}`)
console.log(`Result length: ${result.length}`)
console.log(result.slice(0,5))

You could use concat method.

function odd(nums){
  var result = [];
  if(nums.length === 0) {
    return result;
  } else  if (nums[0] % 2 === 0){
    result.push(nums[0])
    // return odd(nums.slice(1))
  }
  return result.concat(odd(nums.slice(1)));
}

var arr = [1,8,3,4,4,5,9,13,13,9,10];
var print = odd(arr);
console.log(print)

Another approach would be to pass the array containing the result as parameter, since it will be a reference, you don't have to return anything.

function odd(nums, result)
{
    if(nums.length != 0)
    {
        if (nums[0] % 2 === 0)
        {
            result.push(nums[0]);
        }
        odd(nums.slice(1), result);
    }
}

var arr = [1,8,3,4,4,5,9,13,13,9,10];
var result = [];
odd(arr, result);
console.log(result)

Instead of the concat you need in place of the string concatenation + you use now, better use filter

var odd = [], 
   even = [1,8,3,4,4,5,9,13,13,9,10]
  .filter(function(x) { if (x % 2 === 0) return true; else odd.push(x) } );
console.log(odd,even)