Undefined returned from recursive function

Question

I have the following functions defined in JavaScript:

        function rfactorial(n, f) {
            if (n == 0) {
                return f;
            }
            rfactorial(n-1, f*n);
        }

        function dofactorial(){
            var n = prompt("Enter number");
            var f = rfactorial(parseInt(n), 1);
            alert("Factorial = " + f);

            if (confirm("Do another one?")) {
                dofactorial();
            }
        }

The problem is that in dofactorial(), f is undefined. This is strange since if I check the value of f in rfactorial() it is correctly calculated right before the return. What am I doing wrong?

Did you perhaps forget the return at the rfactorial(n-1, f*n); ? — Wolph
– Wolph, Commented Nov 16, 2013 at 12:48

mohamed-ibrahim · Accepted Answer · 2013-11-16 12:50:36Z

2

function rfactorial(n, f) {
        if (n == 0) {
            return f;
        }
       return rfactorial(n-1, f*n);
    }

    function dofactorial(){
        var n = prompt("Enter number");
        var f = rfactorial(parseInt(n), 1);
        alert("Factorial = " + f);

        if (confirm("Do another one?")) {
            dofactorial();
        }
    }

    dofactorial()

Note return rfactorial(n-1, f*n); You forget return

answered Nov 16, 2013 at 12:50

mohamed-ibrahim

11.2k4 gold badges42 silver badges52 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

linkyndy Over a year ago

Oh my, how couldn't have I notice that...Thank you!

Progo · Accepted Answer · 2013-11-16 12:53:30Z

0

Try modifying your code to this:

01| function rfactorial(n, f) {
02|    if (n == 0) {
03|       return f;
04|    }
05|    return new rfactorial(n-1, f*n);
06| }
07| function dofactorial(){
08|    var n = prompt("Enter number");
09|    var f = rfactorial(parseInt(n), 1);
10|   alert("Factorial = " + f);
11|    if (confirm("Do another one?")) {
12|    dofactorial();
13|   }
14| }

On line 3 you have 2 options return new rfactorial(n-1, f*n);, or return new rfactorial(n-1, f*n);. Both work.

answered Nov 16, 2013 at 12:53

Progo

3,4887 gold badges31 silver badges44 bronze badges

Comments

nrsharma · Accepted Answer · 2013-11-16 12:55:45Z

The point is you forgot to write return before calling the function recursively

To call a function recursively it's needed to assign it's value to some variable or some parameter (it's depends on the logic). In your case you are calling rfactorial() but when it's comes in a part of recursive call it's not returning value. So you have to put return over there.

function rfactorial(n, f) {
        if (n == 0) {
            return f;
        }
       return rfactorial(n-1, f*n);  // here you forgot to include return keyword
    }

    function dofactorial(){
        var n = prompt("Enter number");
        var f = rfactorial(parseInt(n), 1);
        alert("Factorial = " + f);

        if (confirm("Do another one?")) {
            dofactorial();
        }
    }

    dofactorial();

Here is working FIDDLE.

Collectives™ on Stack Overflow

Undefined returned from recursive function

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related