Bits of .NET

⚡ sort() Isn’t Sorting Numbers the Way You Think

Calling .sort() on an array of numbers with no arguments doesn’t sort numerically — it converts every number to a string first and compares them alphabetically, and nothing throws an error to warn you it happened.

🐞 The Problem

const values = [10, 1, 2, 21, 100];
values.sort();
console.log(values);
// [ 1, 10, 100, 2, 21 ]  <- NOT numerically sorted at all
// "10" alphabetically comes before "2" because '1' < '2',
// exactly like a word starting with "1" beats one starting with "2".

🔍 Why This Happens

Array.prototype.sort() with no comparator converts every element
to a string and sorts by UTF-16 code unit order - i.e. alphabetical
order. It was designed this way because sort() has to work for
ANY array (strings, mixed types, objects), so it needs a universal
default - numeric order was never it.

✅ The Fix: Always Pass a Comparator for Numbers

• (a, b) => a – b sorts ascending; (a, b) => b – a sorts descending.
• There is no safe way to sort numbers correctly with the zero-argument form — always pass the comparator, every time, without exception.

🔢 Correct Version

const values = [10, 1, 2, 21, 100];
values.sort((a, b) => a - b);
console.log(values); // [ 1, 2, 10, 21, 100 ] - correct

⚠️ This Bug Hides Especially Well When

• All your test data happens to be single-digit or already-ordered numbers — the bug is invisible until a two-digit value sorts against a one-digit one.
• You’re sorting an array of objects by a numeric property using .sort((a, b) => a.value – b.value) is correct; forgetting the subtraction and comparing raw values isn’t.