Remarks#

As val are semantically static, they are initialized “in-place” wherever they appear in the code. This can produce surprising and undesirable behavior when used in abstract classes and traits.

For example, let’s say we would like to make a trait called PlusOne that defines an increment operation on a wrapped Int. Since Ints are immutable, the value plus one is known at initialization and will never be changed afterwards, so semantically it’s a val. However, defining it this way will produce an unexpected result.

trait PlusOne {
    val i:Int

    val incr = i + 1
}

class IntWrapper(val i: Int) extends PlusOne

No matter what value i you construct IntWrapper with, calling .incr on the returned object will always return 1. This is because the val incr is initialized in the trait, before the extending class, and at that time i only has the default value of 0. (In other conditions, it might be populated with Nil, null, or a similar default.)

The general rule, then, is to avoid using val on any value that depends on an abstract field. Instead, use lazy val, which does not evaluate until it is needed, or def, which evaluates every time it is called. Note however that if the lazy val is forced to evaluate by a val before initialization completes, the same error will occur.

A fiddle (written in Scala-Js, but the same behavior applies) can be found here.

Var, Val, and Def

var

A var is a reference variable, similar to variables in languages like Java. Different objects can be freely assigned to a var, so long as the given object has the same type that the var was declared with:

scala> var x = 1
x: Int = 1

scala> x = 2
x: Int = 2

scala> x = "foo bar"
<console>:12: error: type mismatch;
 found   : String("foo bar")
 required: Int
       x = "foo bar"
       ^

Note in the example above the type of the var was inferred by the compiler given the first value assignment.

val

A val is a constant reference. Thus, a new object cannot be assigned to a val that has already been assigned.

scala> val y = 1
y: Int = 1

scala> y = 2
<console>:12: error: reassignment to val
       y = 2
         ^

However, the object that a val points to is not constant. That object may be modified:

scala> val arr = new Array[Int](2)
arr: Array[Int] = Array(0, 0)

scala> arr(0) = 1

scala> arr
res1: Array[Int] = Array(1, 0)

def

A def defines a method. A method cannot be re-assigned to.

scala> def z = 1
z: Int

scala> z = 2
<console>:12: error: value z_= is not a member of object $iw
       z = 2
       ^

In the above examples, val y and def z return the same value. However, a def is evaluated when it is called, whereas a val or var is evaluated when it is assigned. This can result in differing behavior when the definition has side effects:

scala> val a = {println("Hi"); 1}
Hi
a: Int = 1

scala> def b = {println("Hi"); 1}
b: Int

scala> a + 1
res2: Int = 2

scala> b + 1
Hi
res3: Int = 2

Functions

Because functions are values, they can be assigned to val/var/defs. Everything else works in the same manner as above:

scala> val x = (x: Int) => s"value=$x"
x: Int => String = <function1>

scala> var y = (x: Int) => s"value=$x"
y: Int => String = <function1>

scala> def z = (x: Int) => s"value=$x"
z: Int => String

scala> x(1)
res0: String = value=1

scala> y(2)
res1: String = value=2

scala> z(3)
res2: String = value=3

Lazy val

lazy val is a language feature where the initialization of a val is delayed until it is accessed for the first time. After that point, it acts just like a regular val.

To use it add the lazy keyword before val. For example, using the REPL:

scala> lazy val foo = {
     |   println("Initializing")
     |   "my foo value"
     | }
foo: String = <lazy>

scala> val bar = {
     |   println("Initializing bar")
     |   "my bar value"
     | }
Initializing bar
bar: String = my bar value

scala> foo
Initializing
res3: String = my foo value

scala> bar
res4: String = my bar value

scala> foo
res5: String = my foo value

This example demonstrates the execution order. When the lazy val is declared, all that is saved to the foo value is a lazy function call that hasn’t been evaluated yet. When the regular val is set, we see the println call execute and the value is assigned to bar. When we evalute foo the first time we see println execute - but not when it’s evaluated the second time. Similarly, when bar is evaluated we don’t see println execute - only when it is declared.

When To Use ‘lazy’

1. Initialization is computationally expensive and usage of val is rare.

   lazy val tiresomeValue = {(1 to 1000000).filter(x => x % 113 == 0).sum}
   if (scala.util.Random.nextInt > 1000) {
     println(tiresomeValue)
   }

   tiresomeValue takes a long time to calculate, and it’s not always used. Making it a lazy val saves unnecessary computation.

2. Resolving cyclic dependencies

   Let’s look at an example with two objects that need to be declared at the same time during instantiation:

   object comicBook {
     def main(args:Array[String]): Unit = {
       gotham.hero.talk()
       gotham.villain.talk()
     }
   }
   
   class Superhero(val name: String) {
     lazy val toLockUp = gotham.villain
     def talk(): Unit = {
       println(s"I won't let you win ${toLockUp.name}!")
     }
   }
   
   class Supervillain(val name: String) {
     lazy val toKill = gotham.hero
     def talk(): Unit = {
       println(s"Let me loosen up Gotham a little bit ${toKill.name}!")
     }
   }
   
   object gotham {
     val hero: Superhero = new Superhero("Batman")
     val villain: Supervillain = new Supervillain("Joker")
   }

   Without the keyword lazy, the respective objects can not be members of an object. Execution of such a program would result in a java.lang.NullPointerException. By using lazy, the reference can be assigned before it is initialized, without fear of having an uninitialized value.

Overloading Def

You can overload a def if the signature is different:

def printValue(x: Int) {
  println(s"My value is an integer equal to $x")
}

def printValue(x: String) {
  println(s"My value is a string equal to '$x'")
}

printValue(1)  // prints "My value is an integer equal to 1"
printValue("1") // prints "My value is a string equal to '1'"

This functions the same whether inside classes, traits, objects or not.

Named Parameters

When invoking a def, parameters may be assigned explicitly by name. Doing so means they needn’t be correctly ordered. For example, define printUs() as:

// print out the three arguments in order.
def printUs(one: String, two: String, three: String) = 
   println(s"$one, $two, $three")

Now it can be called in these ways (amongst others):

printUs("one", "two", "three") 
printUs(one="one", two="two", three="three")
printUs("one", two="two", three="three")
printUs(three="three", one="one", two="two") 

This results in one, two, three being printed in all cases.

If not all arguments are named, the first arguments are matched by order. No positional (non-named) argument may follow a named one:

printUs("one", two="two", three="three") // prints 'one, two, three'
printUs(two="two", three="three", "one") // fails to compile: 'positional after named argument'