I'm trying to extend Array.prototype to include a square function. I have this:

Array.prototype.square = function(){
  return this.forEach(function(el){
    return (el * el)
  });
}

When I call this function on an array, say arr = [2, 2, 2] it returns undefined. If I add a console.log in there I can see that the callback function for the forEach function executes properly -- it logs 4 three times. Why is this function returning undefined instead of a new array of [4, 4, 4]?

I'm trying to extend Array.prototype to include a square function. I have this:

Array.prototype.square = function(){
  return this.forEach(function(el){
    return (el * el)
  });
}

When I call this function on an array, say arr = [2, 2, 2] it returns undefined. If I add a console.log in there I can see that the callback function for the forEach function executes properly -- it logs 4 three times. Why is this function returning undefined instead of a new array of [4, 4, 4]?

• javascript
• arrays
• prototype

Share Improve this question Follow asked Feb 12, 2014 at 16:20 Chris CloutenChris Clouten 1,08533 gold badges1212 silver badges2424 bronze badges 2

• 1 The .forEach() function does not return a value. – Pointy Commented Feb 12, 2014 at 16:22
• NB: if available, use Object.defineProperty(Array.prototype, 'square', { value: function() { ... } }) to prevent your function being an enumerable property of every array instance. – Alnitak Commented Feb 12, 2014 at 16:25

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 7

The forEach method does not return a value. You need to use map:

Array.prototype.square = function(){
  return this.map(function(el){
    return (el * el)
  });
}

console.log([2, 2, 2].square()); // [4, 4, 4]

As p.s.w.g. said, .map is the appropriate function, but in a ment you asked about using forEach. To get this to work, you'd have to create a temporary array:

Array.prototype.square = function(){
  var tmp = [];

  this.forEach(function(el){
    tmp.push(el * el)
  });

  return tmp;
}

console.log([2, 2, 2].square()); // [4, 4, 4]

.map() is better, though.