What's meaning of "(=>A-Certain-Type)" in scala?

Question

Here is a piece of scala codes:

var hof: (=> Int) => Int = {r=>r+1}

Can anybody explain to me what's meaning of (=> Int)? before, i think it's the same to (() => Int) => Int, but it's not. I tested it as following:

scala> var hof: (=> Int) => Int = {r=>r+1}
hof: (=> Int) => Int = <function1>

scala> hof({()=>1})
<console>:13: error: type mismatch;
 found   : () => Int
 required: Int
       hof({()=>1})

you can see it's wrong, if i change it to hof(1), if works. Can anybody tell me what's meaning of (=>A-Certain-Type)?

Nagarjuna Pamu · Accepted Answer · 2016-10-31T07:56:14.403

You can call this ( => Int) a code block which evaluates to Int only when called by name.

code: => T is a call by name parameter

def repeat[T](code: => T)(times: Int): List[T] = {
  (1 to times).toList.map {_ =>  code }
}

Consider the above code snippet. code in the above function can accept any expression or group of expressions which will evaluate to T. The code block gets evaluated only when it is called by name.

Scala REPL

scala> repeat { println("hello") }(3)
hello
hello
hello
res2: List[Unit] = List((), (), ())

scala> repeat { println("hello"); 1 }(3)
hello
hello
hello
res3: List[Int] = List(1, 1, 1)

scala> repeat(println("foo"))(1)
foo
res4: List[Unit] = List(())

Call by name lets you build interesting things like loops

def loop(cond: => Boolean)(codeBlock: => Unit): Unit = {
 if (cond) {
   codeBlock
   loop(cond)(codeBlock)
 } else ()
}

Magic happens because of the lazy evaluation of the code block every time it is called by name

Scala REPL

scala>     def loop(cond: => Boolean)(codeBlock: => Unit): Unit = {
     |      if (cond) {
     |        codeBlock
     |        loop(cond)(codeBlock)
     |      } else ()
     |     }
loop: (cond: => Boolean)(codeBlock: => Unit)Unit

scala> var i = 0
i: Int = 0

scala> loop(i < 10) { println("foobar"); i += 1}
foobar
foobar
foobar
foobar
foobar
foobar
foobar
foobar
foobar
foobar

What's meaning of "(=>A-Certain-Type)" in scala?

1 Answers1