"You still perform double penetration, but you hide it from the user" - - Tsoding

Tsoding trolling us by asking us to implement the composition of monads 🤣

the bottom of the screen says "Porn Folder: 43.9 GiB"...that's not nearly enough, man.

@quaternaryyy

4 жыл бұрын

too smol PepeHands

@LucianoRobino

4 жыл бұрын

Well there are a lot of penetrations and double penetrations in the video

Tsoding is the king of penetration

@nilp0inter2

3 жыл бұрын

And double penetrantion as well

@EidosGaming

3 жыл бұрын

@@nilp0inter2 fmap . fmap so?

I love how you use the word "secret" to make this video more appealing to click on

@objectobject6172

3 жыл бұрын

Sure appealed to me

"Injecting" things is how we get endofunctors. Truly the mathematical way of doing compositions and metaprogramming.

@thinkbait8363

4 жыл бұрын

pun intended?

Rewatching your old videos, and this intro hit as hard as it did 4 years ago :D

Rust values can be mapped. It's pretty easy to implement a functor trait. Unwrap is actually a code smell

@weirdblackcat

4 жыл бұрын

You can't do it generically though because Rust doesn't support higher-kinded types, which means no fmap. There's a big GH issue to bring HKTs to Rust one day so the Rust community can start writing monad tutorials like "A monad is just a shell exchanging crabs bro what's the problem???"

@YuriAlbuquerque

3 жыл бұрын

it's impossible to implement a functor trait in Rust, as of now.

Ah hell yea so that's how you use it! How about triple without fmap? Maybe one on transformers / free monads or effects to help "compose" monads?

"In Rust you unwrap everything" Oh yeah? fmap (\x -> x + 1) (Just 5) is written Some(5).map(|x| x + 1); in Rust no unwrap required.

@jethrolarson

3 жыл бұрын

I dont think he was saying you can't but that devs often don't. I've seen this in other languages like java

@NikolajLepka

3 жыл бұрын

@Sydney Bean the question mark operator is just monadic bind in the context of optionals. It performs the exact same function

@NikolajLepka

3 жыл бұрын

@Sydney Bean Point is: fn test1() -> Option { let a = may_fail()?; let b = may_also_fail()?; Some(a + b) } is the same as fn test2() -> Option { may_fail().and_then(|a| may_also_fail().map(|b| a + b)) } play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=d5e6fa175cdba9bc18b108c98f828124

@chaoky

3 жыл бұрын

@@NikolajLepka how would this look like in Haskell?

@NikolajLepka

3 жыл бұрын

@@chaoky the link I posted in the comment has the Haskell versions written in comments

text version of would be called "apply". if you're talking about the specific operator I call it the spaceship operator

@variadicism1047

4 жыл бұрын

That might be confused with the spaceship operator in languages like Perl, .

@chromosundrift

2 жыл бұрын

@@variadicism1047 not likely!

Thanks for another great video, maybe a tut about haskell memoization for us mere mortals. This mortal is struggling.

nice. glad i subscribed half a year ago

Yay! Haskell it is!

I know you are partiality joking about Rust developers unwraping everything, but there do exist functor and monadic functions in rust for Option, Result, and other types and should be used along with the try operator when you need use wrapped values (much like do notation).

You can (and should) use map on Option in Rust

inb4 this gets demonetized for "double penetration"

why not newtype Compose , bljat'?!

That intro tho...

yay i wanna be on the list of people at the end

2:31 "b-but alexey w-we haven't even been on a date yet..."

I'm following allowing in VS Code, and I'm curious why the linter included with the haskell extension is recommending `newtype` instead of `data`. Do you have an opinion Tsoding?

@aaoth4689

2 жыл бұрын

in this case newtype is better because it essentially creates a wrapper for your types which exists only in design time and is stripped in runtime. so by using newtype you aren't creating new entities, just referencing the same types and implementing different interfaces for them

Another great tutorial. The only thing; my brain hurts.

What the text editor are you using? It looks so cool.

@Egzvorg

Жыл бұрын

That's emacs

amazing intro 😂

This is a very nice and elegant solution. However it's a little bit unsatisfactory that it only wraps two functors at a time... is there a way to declare a recursive datatype that fmaps over an arbitray length of composition? A little bit like the free monad

@JaycenGiga

4 жыл бұрын

You could nest the compose. Then the outer compose uses the fmap of the inner one in its fmap, so everything would work automatically.

@kriswright5112

4 жыл бұрын

((+1) ) (map Just [0..4]) gives [Just 1,Just 2,Just 3,Just 4,Just 5]

@SomeMrMindism

4 жыл бұрын

@@JaycenGiga Yes, but then to get the inner value you'd need to know how long the chain is.

@SomeMrMindism

4 жыл бұрын

@@kriswright5112 I'm sorry but I don't understand what you're trying to say. You've just switched the two functors

@kriswright5112

4 жыл бұрын

@@SomeMrMindism i don't know what I'm trying to say either. the only recursive datatype I know in Haskell is Tree. Maybe you could do something like the Functor Tree example from learnyouahaskell but instead of left and right branches, just do a right branch. don't know. I'm just learning Haskell.

That was great, thank you! Now, I thought `bind` was *exactly* a composition of monads? The comments here suggest otherwise.

2:26 succ

...Next train to "JSON Parser" station departs in 26 minutes 44 seconds from "Tsoding Channel" platform...

this guy is the best

If this channel needs one thing, it's regular scheduling

@user-qt4gz1dl6w

Жыл бұрын

See tsoding daily

@user-qt4gz1dl6w

Жыл бұрын

I know that you wrote it 3 years before

20:13 g that contains b, that just made so much things click!

"Rust developers must constantly unwrap everything." Uh... we have Option::map and Either::map_ok. Sure, they're not typeclasses, but they're still quite composable.

I thought monads do not compose in general.. edit ok just saw the description of the vid lol

it bothers me soo much that the type signature of is () :: Functor f => (a -> b) -> f a -> f b and not () :: Functor f => f a -> (a -> b) -> f b with the second signature we could've written beautiful code like foo :: Functor a -> Functor b foo x = x func1 func2 func 3 just like the monad >>= operator

@drdilyor

9 ай бұрын

you can use for exactly that. is defined that way to be similar to $. is also similar to &. is defined in Data.Functor, & is defined in Data.Function

Monads don't compose lol

damn - I really want to share this with my co-workers but it's clearly NSFW

fmap = fmap . fmap pure = pure . pure 😜

@Bratjuuc

2 жыл бұрын

liftA2 = liftA2 . liftA2

defining liftA2 instead of () for Compose is much easier and intuitive instance (Applicative f, Applicative g) => Applicative (Compose f g) where pure = Compose . pure . pure liftA2 f (Compose a) (Compose b) = Compose $ (liftA2 . liftA2) f a b