In JavaScript, you can sort an array using the `sort()`

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

as stated on MDN.

This means if you want to sort an array of *numbers* in ascending order, you *can't* just use the default sort order. It will give you an unexpected result:

`const numbers = [80, 1000, 9];`

numbers.sort(); // [1000, 80, 9]

Instead, you need to supply a callback function. If `a`

is less than `b`

, the function should return a negative number. If `a`

is greater than `b`

, the function should return a positive number. If `a`

is equal to `b`

, the function should return zero:

`numbers.sort(function (a, b) {`

// If a is less than b

if (a < b) {

return -1;

}

// If a is greater than b

if (a > b) {

return 1;

}

// If a is equal to b

return 0;

});

// Result: [9, 80, 1000]

You can shorten this by simply subtracting `b`

from `a`

:

`// Traditional function expression`

numbers.sort(function (a, b) {

return a - b;

});

// Or...

// Arrow function expression

numbers.sort((a, b) => a - b);

// Either way, the result is still [9, 80, 1000]

This makes sense when you think about it:

- If
`a === 1`

and`b === 2`

, then`(a - b) === (1 - 2) === -1`

. - If
`a === 2`

and`b === 1`

, then`(a - b) === (2 - 1) === 1`

. - If
`a === 1`

and`b === 1`

, then`(a - b) === (1 - 1) === 0`

.

It's the same as writing the function in the long form.

**If you have any questions, please email me at kieranbarker@outlook.com!**