Why does [NaN].includes(NaN) return true in JavaScript?

Question

I'm familiar with NaN being "weird" in JavaScript, i.e., NaN === NaN always returns false, as described here. So one should not make === comparisons to check for NaN, but use isNaN(..) instead.

So I was surprised to discover that

> [NaN].includes(NaN)
true

This seems inconsistent. Why have this behavior?

How does it even work? Does the includes method specifically check isNaN?

Answer 1

According to MDN's document say that

Note: Technically speaking, includes() uses the sameValueZero algorithm to determine whether the given element is found.

const x = NaN, y = NaN;
console.log(x == y); // false                -> using ‘loose’ equality
console.log(x === y); // false               -> using ‘strict’ equality
console.log([x].indexOf(y)); // -1 (false)   -> using ‘strict’ equality
console.log(Object.is(x, y)); // true        -> using ‘Same-value’ equality
console.log([x].includes(y)); // true        -> using ‘Same-value-zero’ equality

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More detailed explanation:

1. Same-value-zero equality similar to same-value equality, but +0 and −0 are considered equal.
2. Same-value equality is provided by the Object.is() method: The only difference between Object.is() and === is in their treatment of signed zeroes and NaNs.

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Additional resources:

• Which equals operator (== vs ===) should be used in JavaScript comparisons?
• Array.prototype.includes vs. Array.prototype.indexOf
• Are +0 and -0 the same?