comparing 2 strings alphabetically for sorting purposes

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You do say that the comparison is for sorting purposes. Then I suggest localeCompare instead:

"a".localeCompare("b");

It returns -1 since "a" < "b", 1 or 0 otherwise, like you need for Array.prototype.sort()

Keep in mind that sorting is locale dependent. E.g. in German, ä is a variant of a, so "ä".localeCompare("b", "de-DE") returns -1. In Swedish, ä is one of the last letters in the alphabet, so "ä".localeCompare("b", "se-SE") returns 1.

Without the second parameter to localeCompare, the browser’s locale is used. Which in my experience is never what I want, because then it’ll sort differently than the server, which has a fixed locale for all users.

Also, if what you are sorting contains numbers, you may want:

"a5b".localeCompare("a21b", undefined, { numeric: true })

This returns -1, recognizing that 5 as a number is less than 21. Without { numeric: true } it returns 1, since “2” sorts before “5”. In many real-world applications, users expect “a5b” to come before “a21b”.

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