CODEWITHSUNDEEP
scala
Aleksey N Yakushev
Aleksey N Yakushev

Reputation: 465

CallByName with 2 functions

I have the next code, where I study call by name. 

def callByName(v1: => Int) ={
  println(v1)
  println(v1)
}

val fn1 : (Int => Int) = {println("calling fn1 "); _ + 100}

callByName(fn1(1))

println("=================")

val fn2 = (x: Int) => {println("calling fn2 "); x + 100 }

callByName(fn2(1))

I expect the same output, but it is:

calling fn1
fn1: Int => Int = <function>89c5c0

101
101

=================

fn2: Int => Int = <function>d15e781

calling fn2
101
calling fn2
101

I know that the second case is a "function literal (also known as an anonymous function)".

Could you tell me please where I can read about these 2 cases or explain? Thanks.

And the third case:

val fn1 : (Int => Int) = {println("calling fn1 "); _ + 100}

println(fn1(1))
println(fn1(2))
println(fn1(3))

Output:

calling fn1
fn1: Int => Int = <function>da62626

101
102
103

In this last case "body of the function executed 3 times", but only last part "_ + 100" and the first part only 1 time "println("calling fn1 ");"

Huh.

Upvotes: 2

Views: 53

Answers (1)

sarveshseri
sarveshseri

Reputation: 13985

Well... So your first case is bit weird and causing some confusion to you.

So, your first case,

val fn1 : (Int => Int) = {
  println("calling fn1 ")
  _ + 100
}

is actually,

val fn1 : (Int => Int) = {
  println("calling fn1 ")
  val xx: Int => Int = _ + 100
  xx
}

It is similar to,

val i: Int = {
  println("creating int")
  5
}

So, your fn1 is just that xx. It has no relation with println("calling fn1 ") line.

Where as in your second case println("calling fn2 ") is also part of fn2.

Upvotes: 1

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