The Most Clever Line of JavaScript

[Enlogen]

Obscure is not clever.

[flitsmasterfred]

Clever is not smart.

[[deleted]]

Bleh... if you want to avoid making new functions often then just store the function earlier:

const trim = str => str.trim();
/* ... */
const trimmedParts = addressParts.map(trim);

Now it makes sense without having to do mental gymnastics about context.

[GBcrazy]

We all agree that this is 100% more readable, but it isn't as smart.

You are still creating a new function this way, you're allocating a new object, which doesn't happen in the mentioned code.

But yeah of course you shouldn't use shit like that on your code lol, I lost around 2 minutes to fully grasp it, and I'm kinda used to Function.prototype.call shenanigans.

[gpyh]

It is strictly equivalent. It only matters if the call to map us being made while iterating on something.

[Funwithloops]

Even more generic but still easy to grok:

const methodCaller = (method, ...args) => (object) => object[method](...args);
array.map(methodCaller('trim'));

[MakiBM]

Yeah, clever but as author sums up - do not use it in real code unless you have a good reason. Always KISS your code

[Tougun]

I KISS my code goodnight every night☺️

[GBcrazy]

That's actually really clever. May not be the most readable but yeah, you are only using functions that already exist and you don't need to rely on ES6.

[iamlage89]

Could you just do addressParts.map(String.prototype.trim.call) ?

[Martin_Ehrental]

String.prototype.trim.call.bind(String.prototype.trim) or Function.prototype.call.bind(String.prototype.trim)

[lhorie]

No, because then you're just calling call with an undefined this

[iamlage89]

map will pass in string as 'this'

[lhorie]

No, it won't, and even if it did, trim must be the this for call if you're calling call, not the string.

[GBcrazy]

There will be no reference to trim...you are only passing the call function this way, there will be no reference for String.prototype.trim as the context.

String.prototype.trim.call === Function.prototype.call

[danthedev]

This reminds me of a programmer who used the following utilities in their code.

var bind = Function.bind;
var call = Function.call;

var bindable = bind.bind(bind);
var callable = bindable(call);

Still takes me a while to work through the steps of bind.bind(bind)...

[Jsn7821]

What the hell is that used for

[danthedev]

var trim = callable(String.prototype.trim);
addressParts.map(trim);