What Can Scala Puzzlers Teach Us?Daniel C. Sobral

Disclaimer #1: "Puzzlers"

● Java Puzzlers (the book/presentations):○ Weird behavior in Java that is difficult to predict or

explain● This presentation:

○ Weird behavior in Scala that confounds beginners■ (and sometimes experienced programmers)

○ FAQs, not Fun

I can't speak for any of the people who contributed to create and turn Scala into what it is. Furthermore, if you learn anything from this presentation, learn that a perfect understanding of Scala is an unlikely achievement...

Disclaimer #2: don't trust me.

Symbols, operators and punctuation

● What does it mean?● Where can I find information about it?● WTF?

The symbol puzzler

Source: Stack Overflow

The Origin of the Symbols

● Punctuation elements of the language○ @ # ( ) [ ] { } , . ; : ` ' "

● Keywords, reserved symbols and XML○ <- => <: >: <% // /* */ _* <? <!

● Normal definitions from libraries○ <:< =:= ☆ ★ η :/

● Underscore○ All of the above!

Symbols may be part of the language or come from external libraries; it can be difficult to tell where they begin and end; they are difficult to search for on the Internet, non-mnemonic and unpronounceable.

Then why?

def f(xs: Array[Int], a: Int, b: Int) =

for (i <- xs.indices)

xs(i) = a * i + b

So that libraries can extend the language seamlessly

def f(xs: M[T], a: T, b: T) =

for (i <- xs.indices)

xs(i) = a * i + bFor M and T library-types that support these operations

Which still doesn't really justify unicode...

Advice regarding Symbols

● Learn the language:○ What symbols are "built-in"○ The syntax rules

● If exploring code using unknown libraries,○ Use an IDE to explore it

harsh, indeed...

implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {

new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {

def id[A] = ☆(_ η)

def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g

}

}

Let's try to figure this one out

implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {

new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {

def id[A] = ☆(_ η)

def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g

}

}

Split Scala Syntax from Identifiers

implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {

new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {

def id[A] = ☆(_ η)

def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g

}

}

Identify what we are defining

implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {

new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {

def id[A] = ☆(_ η)

def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g

}

}

Where we are using our definitions

implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {

new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {

def id[A] = ☆(_ η)

def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g

}

}

What's left are outside identifiers

The Underscore

● What does it mean?● Why does it work here but not here?● WTF?● It's everywhere! It's the borg operator!

One underscore puzzler

Source: Stack overflow

The meanings of underscore

Source: myself, by way of Stack Overflow

The underscore is Scala's wildcard; it means "something" or "anything" and even "nothing". However, it's semantics can be widely different, even in similar contexts.

Why?

● ???● Picking a different symbol for each case

would stretch ASCII● Picking keywords is not in the spirit of Scala● They are somewhat intuitive

○ except when they bite you

Advice regarding underscore

● Intuition does go a long way● But, again, learn the language● And, in particular, know the difference

between these:○ f _○ f(_, y)○ f(x + _, y)

Eta, partial function application and placeholders

● f _○ an eta expansion: turn method f into an anonymous function.

● f(_, y)○ a partial function application: creates an anonymous function from

method f, whose parameters are the ones wholly replaced by underscores.

● f(x + _, y)○ a placeholder: creates an anonymous function from the expression "x

+ _", whose parameters are the underscores (and that function gets passed to f).

_ always picks the tightest non-degenerate scope it can

Seth Tisue

Erasure

● Where are my types?● How do I get them back?● WTF?

Some things that don't work

trait X {

def overload(f: Int => Int)

def overload(f: String => String)

}

Some things that don't work

def f[T](list: List[T]) = list match {

case _: List[Int] => "Ints"

case _: List[String] => "Strings"

case _ => "???"

}

Some things that don't work

class C[T] {

def f(obj: Any) = obj match {

case _: T => "Ours"

case _ => "Not Ours"

}

}

Why?

● JVM does not support "type parameters"● If Scala added metadata to "eliminate"

erasure, the interaction gap between it and Java would be much greater○ Though things like implicits and default parameters

already do it, to some extent○ At least, implicits and defaults are easy to avoid

Some things that don't work

trait X {

def overload(f: Function1[Int, Int])

def overload(f: Function1[String, String])

}

Erasure at work

trait X {

def overload(f: Function1)

def overload(f: Function1)

}

Erasure at work

def f(list: List) = list match {

case _: List => "Ints"

case _: List => "Strings"

case _ => "???"

}

Erasure at work

class C {

def f(obj: Any) = obj match {

case _: Any => "Ours"

case _ => "Not Ours"

}

}

A partial solution

class C[T : ClassTag] {

def f(obj: Any) = obj match {

case _: T => "Ours"

case _ => "Not Ours"

}

} though it will still ignore T's type parameters

Initialization Ordertrait A {

val a: Int

val b = a * 2

}

class B extends A {

val a = 5

println(b)

}

new B

All val’s are initialized in the order they are found.

Initialization Ordertrait A { val foo: Int val bar = 10 println("In A: foo: " + foo + ", bar: " + bar)}

class B extends A { val foo: Int = 25 println("In B: foo: " + foo + ", bar: " + bar)}

class C extends B { override val bar = 99 println("In C: foo: " + foo + ", bar: " + bar)}

new C

Source: Paul Phillips scala-faq by way of scala puzzlers

Initialization OrderIn A: foo: 0, bar: 0

In B: foo: 25, bar: 0

In C: foo: 25, bar: 99

Initialization Ordertrait A { val foo: Int val bar = 10 println("In A: foo: " + foo + ", bar: " + bar)}

class B extends A { val foo: Int = 25 println("In B: foo: " + foo + ", bar: " + bar)}

class C extends B { override val bar = 99 println("In C: foo: " + foo + ", bar: " + bar)}

new C

A val is initialized only once.

Initialization Ordertrait A { val x = 5 }

trait B {

val x: Int

println(x * 2)

}

trait C extends A with B

trait D extends B

class E extends D with C { println(x * 2) }

class F extends C with D { println(x * 2) }

new E

new F

Initialization Order0

10

10

10

Initialization Ordertrait A { val x = 5 }

trait B {

val x: Int

println(x * 2)

}

trait C extends A with B

trait D extends B

class E extends B with D with A with B with C { println(x * 2) }

class F extends A with B with C with B with D { println(x * 2) }

new E

new F

Trait linearization...

Advice regarding initialization order

● Learn the _________

Other things to look out for

● Pattern matching on val's and for's● All sorts of things implicit● Syntactic sugars, auto-tupling and the like● The return keyword

So, what did I learn?

It seems to me that there's a wide range of Scala features that depend on the user having deep knowledge of the language when something goes wrong. "What's going on?" seems to be much more common than "How do I do this?"

As for real Puzzlers...● Site:

○ http://scalapuzzlers.com/● Book:

○ http://www.artima.com/shop/scala_puzzlers