Wrote the same function in Ruby and JS, Ruby works but JS is undefined

Question

I wanted to try out recursion to calculate compounding interest rather than a loop.

In Ruby:

def compound balance, percentage
  balance + percentage / 100 * balance
end

def invest amount, years, rate
  return amount if years == 0
  invest compound(amount, rate), years - 1, rate
end

This works fine. $10,000 after 1 year at 5% is $10,500; after 10 years, $16,288.

Now the same logic in JavaScript (ES6).

function compound(balance, percentage) {
    return balance + percentage / 100 * balance;
}

function invest(amount, years, rate) {
    if (years === 0) {
        return amount;
    } else {
        invest(compound(amount, 5), years - 1, rate);
    }
}

This returns undefined, but I can't figure out why. It's calling invest the correct number of times, with the correct parameters, and the logic is the same. The compound function works, I tested that separately. So... what could be wrong?

T.J. Crowder · Accepted Answer · 2015-10-22 04:50:18Z

5

Ruby functions automatically return the last expression's value if the step-by-step code "falls off" the end of the function. JavaScript (and most other programming languages) don't do that, so you need to return the value in the else clause explicitly:

function invest(amount, years, rate) {
    if (years === 0) {
        return amount;
    } else {
       return invest(compound(amount, 5), years - 1, rate);
    }
}

Or using the conditional operator:

function invest(amount, years, rate) {
    return years === 0 ? amount : invest(compound(amount, 5), years - 1, rate);
}

edited Oct 22, 2015 at 4:50

T.J. Crowder

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answered Oct 22, 2015 at 4:42

daramao

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1 Comment

GreenTriangle Over a year ago

Ahhh, that solves it. Thank you for this. I did know JS lacked implicit returns, but I assumed the eventual return amount would return from the entire series of recursive calls. I guess it just returns to the stack frame that called it?

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