What is the difference between "def" and "val" to define a function

Question

What is the difference between:

def even: Int => Boolean = _ % 2 == 0

and

val even: Int => Boolean = _ % 2 == 0

Both can be called like even(10).

Answer 1

Method def even evaluates on call and creates new function every time (new instance of Function1).

def even: Int => Boolean = _ % 2 == 0
even eq even
//Boolean = false

val even: Int => Boolean = _ % 2 == 0
even eq even
//Boolean = true

With def you can get new function on every call:

val test: () => Int = {
  val r = util.Random.nextInt
  () => r
}

test()
// Int = -1049057402
test()
// Int = -1049057402 - same result

def test: () => Int = {
  val r = util.Random.nextInt
  () => r
}

test()
// Int = -240885810
test()
// Int = -1002157461 - new result

val evaluates when defined, def - when called:

scala> val even: Int => Boolean = ???
scala.NotImplementedError: an implementation is missing

scala> def even: Int => Boolean = ???
even: Int => Boolean

scala> even
scala.NotImplementedError: an implementation is missing

Note that there is a third option: lazy val.

It evaluates when called the first time:

scala> lazy val even: Int => Boolean = ???
even: Int => Boolean = <lazy>

scala> even
scala.NotImplementedError: an implementation is missing

But returns the same result (in this case same instance of FunctionN) every time:

lazy val even: Int => Boolean = _ % 2 == 0
even eq even
//Boolean = true

lazy val test: () => Int = {
  val r = util.Random.nextInt
  () => r
}

test()
// Int = -1068569869
test()
// Int = -1068569869 - same result

Performance

val evaluates when defined.

def evaluates on every call, so performance could be worse than val for multiple calls. You'll get the same performance with a single call. And with no calls you'll get no overhead from def, so you can define it even if you will not use it in some branches.

With a lazy val you'll get a lazy evaluation: you can define it even if you will not use it in some branches, and it evaluates once or never, but you'll get a little overhead from double check locking on every access to your lazy val.

As @SargeBorsch noted you could define method, and this is the fastest option:

def even(i: Int): Boolean = i % 2 == 0

But if you need a function (not method) for function composition or for higher order functions (like filter(even)) compiler will generate a function from your method every time you are using it as function, so performance could be slightly worse than with val.

With java 8+ lambda optimisations converting method to function is a cheap operation.