Array<number>.sort() uses lexicographic comparator by default

TypeScript Version: 3.4.0-dev.201xxxxx

Search Terms:
typed arrays
default comparator
lexicographic comparator
numeric comparator

Code

let a: Array<number> = [12,4];
a.sort();
console.log(a);

Expected behavior:
[4, 12]
Actual behavior:
[12, 4]
Playground Link:
https://codepen.io/greddyatpt/pen/aepoaN?editors=1111

Related Issues:
#18286 is related but it recommends a breaking change. Instead, the default comparator should take the element type into consideration.

[MartinJohns]

This is unrelated to TypeScript, it's up to the runtime.

Also this is the correct behavior: MDN documentation of Array.prototype.sort()

The default sort order is built upon converting the elements into strings, then comparing their sequences of UTF-16 code units values.

Changing this would violate several of the TypeScript Design Goals:

• Goal 6.: Align with current and future ECMAScript proposals.
• Goal 7.: Preserve runtime behavior of all JavaScript code.
• Non-goal 5.: Add or rely on run-time type information in programs, or emit different code based on the results of the type system. Instead, encourage programming patterns that do not require run-time metadata.

[RyanCavanaugh]

TypeScript doesn't change the behavior of JavaScript code.

In that case, an emit-time warning would be very helpful.

[Rudxain]

What about enforcing something like require-array-sort-compare unconditionally? This would only require making the declaration signature stricter. Yes, this is a breaking change, but a compiler option for a single Array method seems like "nuking a mosquito".

Edit: I've noticed my request is a dupe of #18286