Variable definition in Scala function literal

Question

I'm wondering the result of the piece of code

object localTest {
  def hello = {
    var t = 3
    () => {
      t = t + 3
      println(t)
    }
  }
}
object mainObj {
  def main(args: Array[String]): Unit = {
    val test = localTest.hello
    while (true) {
      Thread.sleep(1000)
      test()
    }
  }
}

Why is the variable t in hello function is only assigned once，and the result will be 6, 9, 12....

I guess this may be related to closure property，but why var t = 3 is only executed once?

Answer 1

This is not usual functional Scala code, where immutables and vals are prefered over mutability. The style reminds me of Javascript, where things like this are seen very often. Yes, you are correct, this is related to the closure:

• The hello method defines a scope. In this scope there are two things existing: the t variable and the lambda (the function literal () => {...} )
• The lambda is returned as a return value from the hello method, assigned into the test variable and repeatedly executed by the while loop
• The lambda is changing the t variable, which is captured in it.

The variable exists in the scope of hello , but as it is captured by the lambda, is it is the same variable used again and again. It is not the hello scope which gets executed from the while loop, rather the lambda body. The hello is executed only once to create the t variable and the lambda.

Expanding the hello definition might help you understand this easier:

val test = {
  var t = 3
  () => {
    t = t + 3
    println(t)
  }
}
while (true) {
  Thread.sleep(1000)
  test()
}

This could be transformed to following code with the same functionality, only the t scope would be expanded so that even the code other than the lambda can see it:

var t = 3
val test = () => {
    t = t + 3
    println(t)
  }
while (true) {
  Thread.sleep(1000)
  test()
}