Javascript: Reduce function with ||

Question

Can someone explain the following code? inputWords is supposed to be an array containing various words and this function is supposed to return an array containing the number of times a word appears in inputWords.

ie. var inputWords = ['Apple', 'Banana', 'Apple', 'Durian', 'Durian', 'Durian']

console.log(countWords(inputWords))
// =>
// {
//   Apple: 2,
//   Banana: 1,
//   Durian: 3
// }

I understand what the Reduce function does, but what is resultObj[word] = ++resultObj[word] || 1; doing?

Thanks so much :)

function countWords(inputWords) {
  return inputWords.reduce(function(resultObj, word) {
    resultObj[word] = ++resultObj[word] || 1;
    return resultObj;
  }, {});
}

module.exports = countWords;

Answer 1

The code attempts to assign to a key (that may not yet exist) an incremented value (of a key that may not yet exist), of if that it that is falsey, it assigns a 1.

This is called short-circuit evaluation. Given a = b || c, if b is truthy, c never gets evaluated so a takes on the value of b. If b is falsey, c is evaluated and assigned to a instead. In your case, when the key doesn't exist, ++resultObj[word] is falsey.

In my humble opinion, I think that it would have been a clearer statement of the author's intention if they had instead done:

if (word in resultObj) {
    ++resultObj[word];
}
else {
    resultObj[word] = 1;
}

or even:

resultObj[word] = word in resultObj ? resultObj[word] + 1 : 1;

either of which would have saved you the bother of asking this question.

Answer 2

It's 'defaulting' a non-existent key in resultObj to 1.

Javascript's || will actually return the first (leftmost) truthy value out of a comparison, so for a nonexistant value added one by the preincrement operator such as ++resultObj['banana'] returns NaN, which is falsey, and the || operator will replace it with 1.

Edit: See http://nfriedly.com/techblog/2009/07/advanced-javascript-operators-and-truthy-falsy/

Answer 3

I would write:

resultObj[word] = (resultObj[word] || 0) + 1;