scala - Does accessing type of a lazy val cause it to be evaluated? -

scala - Does accessing type of a lazy val cause it to be evaluated? -

as question title states, accessing type of lazy val fellow member result in evaluation of member? or utilize static type?

here illustration code in have implicit lazy val, , utilize type in method take implicit val type:

implicit lazy val nonspaces: array[(point, part)]

...

def randomnonspacecoordinate(implicit nonspaces: this.nonspaces.type): point = nonspaces(utils.random.randupto(nonspaces.length))._1

let's check out:

scala> object test { | lazy val test: string = {println("bang!"); "value"} | val p: this.test.type = null | def testdef(p: this.test.type) = println(p) | } defined module test scala> test.testdef(test.p) null scala> test.testdef(test.test) bang! value

so can see accessing type not require lazy val evaluated.

scala types lazy-evaluation