the || and NaN operator. How do they really work?

Question

Let's say I have an array of words, and I would like to use an object to get the count of it. I tried this:

const results = {};

const words = ["word", "hello", "code", "five", "hello", "word", "new", "code"];

words.forEach( word => {
  results[word] = results[word] + 1 || 1;  
});

results in this case returns:

{ word: 2, hello: 2, code: 2, five: 1, new: 1 }

NaN === false is false and results[word] + 1 => NaN. I don't quite understand why results wouldn't be:

{ word: NaN, hello: NaN, code: NaN, five: NaN, new: NaN }

Someone care to explain? :)

Possible duplicate of [Is NaN falsy? Why NaN === false returns false](http://stackoverflow.com/questions/22600248/is-nan-falsy-why-nan-false-returns-false) — Raymond Chen, May 11 '17 at 02:32

score 3 · Answer 1 · answered May 11 '17 at 02:31

3

results[word] when results doesn't contain word is undefined. Though NaN === false is false, it's still a falsy value (!!NaN is false).

Therefore, results[word] + 1 || 1 = undefined + 1 || 1 = NaN || 1 = 1.

answered May 11 '17 at 02:31

Karin

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Sterling Archer · Answer 2 · 2017-05-11T02:35:06.243

-1

undefined+1 = NaN

results[word] = results[word] + 1 || 1;

You're initializing the value at one where the || (or) operation sets the default to 1 instead of 0. The way I solved this was wrapping the condition like so:

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

edited May 11 '17 at 02:35

answered May 11 '17 at 02:33

Sterling Archer

22,070
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81
118

the || and NaN operator. How do they really work?

2 Answers2