“Sort()” of confused

I was reviewing javaScript .sort() method, and I got confusion.

MDN Web Docs states (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort) following:

• If compareFunction(a, b) returns less than 0, leave a and b unchanged.
• If compareFunction(a, b) returns 0, leave a and b unchanged with respect to each other, but sorted with respect to all different elements. Note: the ECMAScript standard only started guaranteeing this behavior in 2019, thus, older browsers may not respect this.
• If compareFunction(a, b) returns greater than 0, sort b before a.

And above rule can be written as:

const array1 = [ 5, 3, 8 ];
// for Ascending order:
console.log( array1.sort((a, b) => a - b) );  // [3, 5, 8]
// for Descending order:
console.log( array1.sort((a, b) => b - a) );  // [8, 5, 3]

Perfect!!!

But when I threw in a console.log(), I found myself saying… huh?

In line 6, I added console.log() to see the values of “a” and “b”, and they weren’t what I expected.

This is what I expected: during first loop: “a”: 5(first element in array), “b”: 3.

But what I am seeing in console.log results are:

1st loop: “a”: 3, “b”: 5. a-b < 0, keep the order — Results [3, 5, 8]

2nd loop: “a”: 8, “b”: 3. a-b > 0, sort b before a — Results [3, 5, 8]

3rd loop: “a”: 8, “b”: 5. a-b > 0, sort b before a — Results [3, 5, 8]

If I sort [ 8, 5, 3 ], sort completes in two loop:

So, there must be a comparison algorithm that sort() method uses, and I was quite surprised how it was different than what I expected. So, threw in console.log() around. You might see something very interesting.