Scala compilation error- found : Int required: Int => Int

Question

I am new to scala. I don't understand the compilation error for the below code:

def delayed( t:(Int)=> Int):Unit={
    println("In delayed method")
    var y=t; 
    println(y)
}

def time(x:Int):Int={
    x*2
}

and when I call

delayed(time(8))

I get the following error:

scala> delayed(time(8))
<console>:15: error: type mismatch;
found   : Int
required: Int => Int
delayed(time(8))
        ^

Please explain what is the issue? Please also suggest a good link to understand functions and function literals in scala. I am not able to understand fully.

Thanks so much

Edit:

Please tell the difference between

        def delayed( t: Int => Int):Unit = {

and

        def delayed(t: =>Int):Unit {

and

        def delayed(t:=>Int):Unit { (without space b/w ":" and "=>"))

Answer 1

Your function delayed expects function as an argument, however, you passed Int. That's why you get the error.

The type of the argument of delayed is Int=>Int, which means it is a function accept one Int as an argument and returns Int.

Your function time is Int=>Int function, however, when you pass time(8) to the delayed function, time(8) will be evaluated before it is passed to delay, and the evaluation result is just an Int.

scala> delayed(time(8))
<console>:14: error: type mismatch;
 found   : Int
 required: Int => Int
       delayed(time(8))
                   ^

If you pass the time function only, it will work.

scala> delayed(time)
In delayed method
<function1>

If you want to pass time(8) as a function argument, you should change time function to return function:

scala> def time(x:Int) = () => x*2

You also need to modify delayed function like the below:

def delayed(t:()=>Int) {
    println("In delayed method")
    var y=t();
    println(y)
}

Then you can pass time(8) to delayed.

scala> delayed(time(8))
In delayed method
16

Or you can use call by name as @Tzach mentioned in the comment.

scala> def delayed(t: =>Int) {
     |   println("In delayed method")
     |   var y = t
     |   println(y)
     | }
delayed: (t: => Int)Unit

scala> def time(t:Int) = t*2
time: (t: Int)Int

scala> delayed(time(8))
In delayed method
16

Answer 2

Method delayed expects a function with Input param Int and return type Int But in your example you are passing result of time function.

This will solve your issue.

delayed(time)