4 ms·

My favourite JS WATism: let a = NaN; let b = null; [a,b].indexOf(a); [a,b].indexOf(b); FYI Array.prototype.Includes was made to "fix" this pro

My favourite JS WATism: let a = NaN; let b = null; [a,b].indexOf(a); [a,b].indexOf(b); FYI Array.prototype.Includes was made to "fix" this problem.

leodriesch 5y ago
I suppose this is because `NaN === NaN` is false. Comparison via `Object.is(NaN, NaN)` works as expected though.
- xnorswap 5y ago
  Indeed, and not just NaN === NaN is false but even NaN == NaN is false too.
- wdroz 5y ago
  The opposite would lead to `(0 * Infinity) === ("foo" / 3)` to be true. I don't know which is better...
masklinn 5y ago
I don't think that beats ["1", "7", "11"].map(parseInt) returning [1, NaN, 3]
- deleted 5y ago
  [deleted]