Why is my recursive function call not returning a value?

Question

I have the following code, trying to return the range between two numbers, and the inner function is not returning a value. It simply returns [Function]. Can anyone tell me why?

var range = function(num1, num2) {
    var output = [];  
    return function range2 () {
        if (num2 - num1 === 2) {
            return [num1+1];
        }
        else if (output.length + 1 === num2 - num1) {
            return output;
        }
        else if (output.length + 1 !== num2 - num1) {
            output.push(num1 + 1);
        }
        return range2(num1 + 1, num2);
    };
};

//UPDATE: This is my code that ended up working, FWIW:

var range = function(num1, num2) {
    var list = [];
    var range2 = function(num1, num2) {
        list.push(num1 + 1);
        if (num2 - num1 === 2) {
            return list;
        }
        else {
            return range2(num1 + 1, num2);
        }
    };
    return range2(num1, num2);
};

return function ... returns the function, it doesn't call it. — Barmar
– Barmar, Commented Jan 27, 2016 at 21:45
you need to execute the function,, so have: return function range2(){}(); — Fallenreaper
– Fallenreaper, Commented Jan 27, 2016 at 21:45
You need parentheses around the definition to call it inline, @Fallenreaper. return (function range2(){...})();. — Mark Reed
– Mark Reed, Commented Jan 27, 2016 at 21:53

Fallenreaper · Accepted Answer · 2016-01-27 21:46:49Z

1

Your return will return the definition. If you are trying to return the execution of the function you will need to add a ()

See here:

var range = function(num1, num2) {
  var output = [];  
  return function range2 () {
    if (num2 - num1 === 2) {
        return [num1+1];
    }
    else if (output.length + 1 === num2 - num1) {
        return output;
    }
    else if (output.length + 1 !== num2 - num1) {
        output.push(num1 + 1);
    }
    return range2(num1 + 1, num2);
  }();  // <----  this will execute the defined function.
};

answered Jan 27, 2016 at 21:46

Fallenreaper

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3 Comments

user5369036 Over a year ago

This returns an array of the correct length, but each value === num1+1. So the arguments aren't getting updated somehow?

Barmar Over a year ago

@user5369036 range2() doesn't take any arguments. So when you call range2(num1 + 1, num2), those parameters have no effect.

user5369036 Over a year ago

@Barmar Ahh, gotcha.

Barmar · Accepted Answer · 2016-01-27 21:47:41Z

0

The range() function never calls range2(). It simply returns the function expression.

Also, range2() needs to take arguments.

var range = function(num1, num2) {
  var output = [];

  function range2(num1, num2) {
    if (num2 - num1 === 2) {
      return [num1 + 1];
    } else if (output.length + 1 === num2 - num1) {
      return output;
    } else if (output.length + 1 !== num2 - num1) {
      output.push(num1 + 1);
    }
    return range2(num1 + 1, num2);
  };
  return range2(num1, num2);
};

alert(range(1, 10));

answered Jan 27, 2016 at 21:47

Barmar

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4 Comments

user5369036 Over a year ago

So this returns an array of a single value, num2-1. Probably has to do with the closure?

Barmar Over a year ago

It only does that when num2 - num1 === 2. That's the first line of range2.

Barmar Over a year ago

range(1, 10) returns [2, 3, 4, 5]

Barmar Over a year ago

I don't really understand the logic of your algorithm, I ignored that. I just showed how to fix the syntax so that the recursive function is called.

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