scala/scala

A relaxed chat room about all things Scala. Beginner questions welcome. https://scala-lang.org/conduct/ applies

Peter Aaser

@PeterAaser

at least asm is pretty easy

Christopher Davenport

@ChristopherDavenport

Here is my cormorant csv parser if that helps at all.

Peter Aaser

@PeterAaser

might be, thanks

Maybe I should read the parser chapter in the red book to get a more fundamental approach

Christopher Davenport

@ChristopherDavenport

I love atto. It's a fun tool.

Peter Aaser

@PeterAaser

How can I cause a parser to fail based on predicate?

Thinking flatMap( x => if(pred(x)) err("reason") else ??? )

Christopher Davenport

@ChristopherDavenport

Failure is one of the options of a parser at any point.

Peter Aaser

@PeterAaser

yes, but I want to have a predicate failure too

I think ok is what I want

Abdhesh Kumar

@abdheshkumar

Anyone has an idea how to override common configuration in included configuration that referenced with environment variable? For example, I have common.conf that has all common configurations and local.conf that override common configurations as per local environment.

common.conf

myapp {
    server-address = ${SERVER_HOSTNAME}
    server-port = ${SERVER_PORT}
}
----------------------
local.conf

include "common"
# override default (Common) settings
myapp {
    server-address = "localhost"
    server-port = 9000
}

while loading local.conf. I am getting below error.

 Could not resolve substitution to a value: ${SERVER_HOSTNAME}

jonathan edwards

@_jedw_twitter

does anyone know what the difference between .contain() and .contains() is?

I believe they are a part of scalatest, but when i search for it I can’t find its definition

the context it’s being used is in this line: items => s"find $itemNum in the available list" >> items.map(_.itemNumber).should(contain(itemNum))

Abdhesh Kumar

@abdheshkumar

@_jedw_twitter contains must be inside scalatest Matchers

Vinayak Pathak

@vinayakpathak

I wasted a day struggling with this:

trait Model {
  type State
}

trait Algorithm {
  val model: Model
  type State = model.State

  def infer: State
}

val m: Model = new Model {
  type State = Int
}

val a = new Algorithm {
  val model = m

  def infer = 1
}

This throws an error:

ScalaFiddle.scala:21: error: type mismatch;
 found   : scala.this.Int(1)
 required: $anon.this.State
    (which expands to)  $anon.this.model.State
    def infer = 1
                ^

I stared at the longest time on the definition of val a because that's where the error was coming from, but eventually realized the cause was my definition of val m. If I remove the : Model and make it val m = new Model {...} it works.

Abdhesh Kumar

@abdheshkumar

@_jedw_twitter here you could find more about matchers http://www.scalatest.org/user_guide/using_matchers

Fabio Labella

@SystemFw

If I remove the : Model and make it val m = new Model {...} it works.

yes

when you do : Model you remove the refinement

which is inferred otherwise

it should bem: Model {type State = Int }

Vinayak Pathak

@vinayakpathak

yeah that's interesting

Fabio Labella

@SystemFw

you can look into the Aux pattern as well to see a shorthand

which is very commonly used in the shapeless world

Vinayak Pathak

@vinayakpathak

yeah I've seen the Aux pattern

Fabio Labella

@SystemFw

same thing, m: Model.Aux[Int]

I'd recommend using existentials only when you need them though

Vinayak Pathak

@vinayakpathak

if I had chosen to implement Model as trait Model[State] though, I would've been notified about the error at the line val m: Model = new Model[Int] {...} coz there would be a missing type parameter

Fabio Labella

@SystemFw

well : Model is a valid type there, so there's no error

the error is when you try to unify it with Int later

Vinayak Pathak

@vinayakpathak

ummm why is Model valid

Fabio Labella

@SystemFw

do you need State to be existential there, since you are already making the whole Model existential in Algorithm?

Vinayak Pathak

@vinayakpathak

i'm not sure what existential means

Fabio Labella

@SystemFw

ummm why is Model valid

Because scala allows you to have abstract (existential) types

well, replace that word with "abstract"

scala allows you to have abstract types

Vinayak Pathak

@vinayakpathak

oh i see ok

ok then i'm not sure when one would need a type to be existential as opposed to just a type parameter

Fabio Labella

@SystemFw

the two things, type parameters and abstract types (universals and existentials) are two different ways to enforce information boudaries

jonathan edwards

@_jedw_twitter

@abdheshkumar Thanks man, I guess i wasn’t reading enough of the page.

Fabio Labella

@SystemFw

well, one use case is to hide the specific type used there

for example, imagine List[Algorithm]

Vinayak Pathak

@vinayakpathak

i see

Fabio Labella

@SystemFw

assume that each algorithm has an input, output and internal state

and the in-out is the only thing that matters from the outside

if you had Algorithm[State] you woudn't be able to put them in a List with a proper type

Vinayak Pathak

@vinayakpathak

hmm i see

Fabio Labella

@SystemFw

but since in my hypothetical example the State is purely internal, you hide it by making it existential

in this case through an abstract type

does that make any sense?

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