CODEWITHSUNDEEP
javascriptrecursion
msmith1114
msmith1114

Reputation: 3229

Javascript recursive function not returning value?

Im solving a codewars problem and im pretty sure i've got it working:

function digital_root(n) {
    // ...
    n = n.toString();
    if (n.length === 1) {
        return parseInt(n);
    } else {
        let count = 0;
        for (let i = 0; i < n.length; i++) {
            //console.log(parseInt(n[i]))
            count += parseInt(n[i]);
        }
        //console.log(count);
        digital_root(count);
    }
}

console.log(digital_root(942));

Essentially it's supposed to find a "digital root":

A digital root is the recursive sum of all the digits in a number. Given n, take the sum of the digits of n. If that value has two digits, continue reducing in this way until a single-digit number is produced. This is only applicable to the natural numbers.

So im actually getting the correct answer at the end but for whatever reason on the if statement (which im watching the debugger run and it does enter that statement it will say the return value is the correct value.

But then it jumps out of the if statement and tries to return from the main digital_root function?

Why is this? shouldn't it break out of this when it hits the if statement? Im confused why it attempt to jump out of the if statement and then try to return nothing from digital_root so the return value ends up being undefined?

Upvotes: 3

Views: 4021

Answers (3)

Mulan
Mulan

Reputation: 135197

Code review

My remarks is code comments below

// javascript generally uses camelCase for function names
// so this should be digitalRoot, not digital_root
function digital_root(n) {
    // variable reassignment is generally frowned upon
    // it's somewhat silly to convert a number to a string if you're just going to parse it again
    n = n.toString();
    if (n.length === 1) {
        // you should always specify a radix when using parseInt
        return parseInt(n);
    } else {
        let count = 0;
        for (let i = 0; i < n.length; i++) {
            //console.log(parseInt(n[i]))
            count += parseInt(n[i]);
        }
        // why are you looping above but then using recursion here?
        // missing return keyword below
        digital_root(count);
    }
}

console.log(digital_root(942));

Simple recursive solution

With some of those things in mind, let's simplify our approach to digitalRoot...

const digitalRoot = n =>
  n < 10 ? n : digitalRoot(n % 10 + digitalRoot((n - n % 10) / 10))
  
console.log(digitalRoot(123))         //          => 6
console.log(digitalRoot(1234))        //       10 => 1
console.log(digitalRoot(12345))       //       15 => 6
console.log(digitalRoot(123456))      //       21 => 3
console.log(digitalRoot(99999999999)) // 99 => 18 => 9

Using reduce

A digital root is the recursive sum of all the digits in a number. Given n, take the sum of the digits of n. If that value has two digits, continue reducing in this way until a single-digit number is produced. This is only applicable to the natural numbers.

If you are meant to use an actual reducing function, I'll show you how to do that here. First, we'll make a toDigits function which takes an integer, and returns an Array of its digits. Then, we'll implement digitalRoot by reducing those those digits using an add reducer initialized with the empty sum, 0

// toDigits :: Int -> [Int]
const toDigits = n =>
  n === 0 ? [] :  [...toDigits((n - n % 10) / 10), n % 10]

// add :: (Number, Number) -> Number
const add = (x,y) => x + y

// digitalRoot :: Int -> Int
const digitalRoot = n =>
  n < 10 ? n : digitalRoot(toDigits(n).reduce(add, 0))
      
console.log(digitalRoot(123))         //          => 6
console.log(digitalRoot(1234))        //       10 => 1
console.log(digitalRoot(12345))       //       15 => 6
console.log(digitalRoot(123456))      //       21 => 3
console.log(digitalRoot(99999999999)) // 99 => 18 => 9

Upvotes: 2

ibrahim mahrir
ibrahim mahrir

Reputation: 31682

You're not returning anything inside else. It should be:

return digital_root(count);
^^^^^^^

Why?

digital_root is supposed to return something. If we call it with a one digit number, then the if section is executed, and since we return from that if, everything works fine. But if we provide a number composed of more than one digit then the else section get executed. Now, in the else section we calculate the digital_root of the count but we don't use that value (the value that should be returned). The line above could be split into two lines of code that makes it easy to understand:

var result = digital_root(count); // get the digital root of count (may or may not call digital_root while calculating it, it's not owr concern)
return result;                    // return the result of that so it can be used from the caller of digital_root

Upvotes: 10

subramanian
subramanian

Reputation: 1305

its a recursive function the code should be somewhat like this

function digital_root(n) {
    // ...
    n=n.toString();
    if(n.length === 1){
         return parseInt(n);
    }
    else
    {
    let count = 0;
    for(let i = 0; i<n.length;i++)
    {
    //console.log(parseInt(n[i]))
    count+=parseInt(n[i]);
    }
   //console.log(count);
   return digital_root(count);
   }
}

you should return the same function instead of just calling it to get the correct call stack

Upvotes: 0

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