What is the proper way to explicitly return the Unit type from a method, using () or Unit? It appears to me that both work in all cases I've tried myself.
For context, this often occurs if I'm writing a method with side effects and returns Unit that calls another method, which also performs side effects but returns some value instead of Unit. e.g.
def effectAndReturn(): String = {
val msg = "Hello, SO"
println(msg)
msg
}
def doEffect(): Unit = {
val _ = effectAndReturn()
() // `Unit` also works here
}
From my understanding () is the only value of type Unit that exists. Returning the token Unit in doEffect() is referencing the Unit companion object; I'm confused how this would return a value as there's not even an apply method defined on it. Returning the companion object for a given abstract class's type isn't valid as a return value as far as I know.
Plugging these into a Scala REPL is also interesting
scala> val parenUnit = ()
parenUnit: Unit = ()
scala> parenUnit
// Returns blank line
scala> val wordUnit = Unit
wordUnit: Unit.type = object scala.Unit
scala> wordUnit
res1: Unit.type = object scala.Unit
scala> res1
res2: Unit.type = object scala.Unit
() is simply a Unit value, whereas Unit gives back a type, which doesn't make sense to me as no other companion objects do this as far as I can tell. My guess is that the compiler handles Unit in a particular and unique way compared to any other type, but how exactly?
So, we have:
The Unit type
The unit value, i.e. () which is of type Unit
The Unit companion object which causes the confusion. It is not of type Unit. It is of type Unit.type (its own singleton type). However... Scala automatically implicitly converts everything to Unit and that's why you can use it where Unit type is expected.
TLDR: Use ()
Suppose we have two methods in a class called TestUnit:
class TestUnit {
def foo(): Unit = 2217
def bar(): Int = 1478
}
Let's look at its bytecode:
// access flags 0x1
public foo()V
L0
LINENUMBER 4 L0
SIPUSH 2217
POP
RETURN // returns void
L1
LOCALVARIABLE this Lunit/TestUnit; L0 L1 0
MAXSTACK = 1
MAXLOCALS = 1
// access flags 0x1
public bar()I
L0
LINENUMBER 5 L0
SIPUSH 1478
IRETURN // returns integer because it is declared in method
L1
LOCALVARIABLE this Lunit/TestUnit; L0 L1 0
MAXSTACK = 1
MAXLOCALS = 1
My assumption is - the Scala compiler just put RETURN instruction (which returns void) in every method where Unit is declared as returning type. (you can look listings here)
So you can return any type in doEffect(). But as #ghik said, it's better to use ().
Related
What is the proper way to explicitly return the Unit type from a method, using () or Unit? It appears to me that both work in all cases I've tried myself.
For context, this often occurs if I'm writing a method with side effects and returns Unit that calls another method, which also performs side effects but returns some value instead of Unit. e.g.
def effectAndReturn(): String = {
val msg = "Hello, SO"
println(msg)
msg
}
def doEffect(): Unit = {
val _ = effectAndReturn()
() // `Unit` also works here
}
From my understanding () is the only value of type Unit that exists. Returning the token Unit in doEffect() is referencing the Unit companion object; I'm confused how this would return a value as there's not even an apply method defined on it. Returning the companion object for a given abstract class's type isn't valid as a return value as far as I know.
Plugging these into a Scala REPL is also interesting
scala> val parenUnit = ()
parenUnit: Unit = ()
scala> parenUnit
// Returns blank line
scala> val wordUnit = Unit
wordUnit: Unit.type = object scala.Unit
scala> wordUnit
res1: Unit.type = object scala.Unit
scala> res1
res2: Unit.type = object scala.Unit
() is simply a Unit value, whereas Unit gives back a type, which doesn't make sense to me as no other companion objects do this as far as I can tell. My guess is that the compiler handles Unit in a particular and unique way compared to any other type, but how exactly?
So, we have:
The Unit type
The unit value, i.e. () which is of type Unit
The Unit companion object which causes the confusion. It is not of type Unit. It is of type Unit.type (its own singleton type). However... Scala automatically implicitly converts everything to Unit and that's why you can use it where Unit type is expected.
TLDR: Use ()
Suppose we have two methods in a class called TestUnit:
class TestUnit {
def foo(): Unit = 2217
def bar(): Int = 1478
}
Let's look at its bytecode:
// access flags 0x1
public foo()V
L0
LINENUMBER 4 L0
SIPUSH 2217
POP
RETURN // returns void
L1
LOCALVARIABLE this Lunit/TestUnit; L0 L1 0
MAXSTACK = 1
MAXLOCALS = 1
// access flags 0x1
public bar()I
L0
LINENUMBER 5 L0
SIPUSH 1478
IRETURN // returns integer because it is declared in method
L1
LOCALVARIABLE this Lunit/TestUnit; L0 L1 0
MAXSTACK = 1
MAXLOCALS = 1
My assumption is - the Scala compiler just put RETURN instruction (which returns void) in every method where Unit is declared as returning type. (you can look listings here)
So you can return any type in doEffect(). But as #ghik said, it's better to use ().
How possible that the first is correct Scala code but the second won't even compile?
The one that does compile
object First {
class ABC(body: => Unit) {
val a = 1
val b = 2
println(body)
}
def main(args: Array[String]): Unit = {
val x = new ABC {
a + b
}
}
}
This one doesn't compile on Scala 2.11 and 2.12
object Second {
class ABC(body: => Int) {
val a = 1
val b = 2
println(body)
}
def main(args: Array[String]): Unit = {
val x = new ABC {
a + b
}
}
}
It's not strange at all. Let's look at the first example:
You declare your class ABC to receive a pass by name parameter that returns Unit and you think this snippet:
val x = new ABC {
a + b
}
is passing that body parameter, it isn't.What's really happening is:
val x = new ABC(()) { a + b }
If you run that code you will see that println(body) prints () because you're not passing a value for your body parameter, the compiler allows it to compile because as the scaladoc states there is only 1 value of type Unit:
Unit is a subtype of scala.AnyVal. There is only one value of type Unit, (), and it is not represented by any object in the underlying runtime system. A method with return type Unit is analogous to a Java method which is declared void.
Since there is only one value the compiler allows you to omit it and it will fill in the gap. This doesn't happen with singleton objects because they don't extend AnyVal. Just has the default value for Int is 0 the default value for Unit is () and because there is only this value available the compiler accepts it.
From documentation:
If ee has some value type and the expected type is Unit, ee is converted to the expected type by embedding it in the term { ee; () }.
Singleton objects don't extend AnyVal so they don't get treated the same.
When you use syntax like:
new ABC {
// Here comes code that gets executed after the constructor code.
// Code here can returns Unit by default because a constructor always
// returns the type it is constructing.
}
You're merely adding things to the constructor body, you are not passing parameters.
The second example doesn't compile because the compiler cannot infer a default value for body: => Int thus you have to explicitly pass it.
Conclusion
Code inside brackets to a constructor is not the same as passing a parameter. It might look the same in same cases, but that's due to "magic".
You cannot pass a single argument to a constructor in curly braces, because this would be parsed as defining an anonymous class. If you want to do this, you need to enclose the curly braces in normal braces as well, like this:
new ABC({
a + b
})
As for why does compiler accept new ABC {a + b}, the explanation is a bit intricate and unexpected:
new ABC {...} is equivalent to new ABC() {...}
new ABC() can be parsed as new ABC(()) because of automatic tupling, which is a feature of the parser not mentioned in the specs, see SI-3583 Spec doesn't mention automatic tupling. The same feature casues the following code to compile without an error:
def f(a: Unit) = {}
f()
def g(a: (Int, Int)) = {}
g(0,1)
Note the call produces a warning (even your original example does):
Adaptation of argument list by inserting () has been deprecated: this is unlikely to be what you want.
The warning is produced since 2.11, see issue SI-8035 Deprecate automatic () insertion.
I was coding in Scala and doing some quick refactoring in Intellij, when I stumbled upon the following piece of weirdness...
package misc
/**
* Created by abimbola on 05/10/15.
*/
object WTF extends App {
val name: String = name
println(s"Value is: $name")
}
I then noticed that the compiler didn't complain, so I decided to attempt to run this and I got a very interesting output
Value is: null
Process finished with exit code 0
Can anyone tell me why this works?
EDIT:
First problem, the value name is assigned a reference to itself even though it does not exist yet; why exactly does the Scala compiler not explode with errors???
Why is the value of the assignment null?
1.) Why does the compiler not explode
Here is a reduced example. This compiles because through given type a default value can be inferred:
class Example { val x: Int = x }
scalac Example.scala
Example.scala:1: warning: value x in class Example does nothing other than call itself recursively
class Example { val x: Int = x }
This does not compile because no default value can be inferred:
class ExampleDoesNotCompile { def x = x }
scalac ExampleDoesNotCompile.scala
ExampleDoesNotCompile.scala:1: error: recursive method x needs result type
class ExampleDoesNotCompile { def x = x }
1.1 What happens here
My interpretation. So beware: The uniform access principle kicks in.
The assignment to the val x calls the accessor x() which returns the unitialized value of x.
So x is set to the default value.
class Example { val x: Int = x }
^
[[syntax trees at end of cleanup]] // Example.scala
package <empty> {
class Example extends Object {
private[this] val x: Int = _;
<stable> <accessor> def x(): Int = Example.this.x;
def <init>(): Example = {
Example.super.<init>();
Example.this.x = Example.this.x();
()
}
}
} ^
2.) Why the value is null
The default values are determined by the environment Scala is compiled to.
In the example you have given it looks like you run on the JVM. The default value for Object here is null.
So when you do not provide a value the default value is used as a fallback.
Default values JVM:
byte 0
short 0
int 0
long 0L
float 0.0f
double 0.0d
char '\u0000'
boolean false
Object null // String are objects.
Also the default value is a valid value for given type:
Here is an example in the REPL:
scala> val x : Int = 0
x: Int = 0
scala> val x : Int = null
<console>:10: error: an expression of type Null is ineligible for implicit conversion
val x : Int = null
^
scala> val x : String = null
x: String = null
why exactly does the Scala compiler not explode with errors?
Because this problem can't be solved in the general case. Do you know the halting problem? The halting problem says that it is not possible to write an algorithm that finds out if a program ever halts. Since the problem of finding out if a recursive definition would result in a null assignment can be reduced to the halting problem, it is also not possible to solve it.
Well, now it is quite easy to forbid recursive definitions at all, this is for example done for values that are no class values:
scala> def f = { val k: String = k+"abc" }
<console>:11: error: forward reference extends over definition of value k
def f = { val k: String = k+"abc" }
^
For class values this feature is not forbidden for a few reasons:
Their scope is not limited
The JVM initializes them with a default value (which is null for reference types).
Recursive values are useful
Your use case is trivial, as is this:
scala> val k: String = k+"abc"
k: String = nullabc
But what about this:
scala> object X { val x: Int = Y.y+1 }; object Y { val y: Int = X.x+1 }
defined object X
defined object Y
scala> X.x
res2: Int = 2
scala> Y.y
res3: Int = 1
scala> object X { val x: Int = Y.y+1 }; object Y { val y: Int = X.x+1 }
defined object X
defined object Y
scala> Y.y
res4: Int = 2
scala> X.x
res5: Int = 1
Or this:
scala> val f: Stream[BigInt] = 1 #:: 1 #:: f.zip(f.tail).map { case (a,b) => a+b }
f: Stream[BigInt] = Stream(1, ?)
scala> f.take(10).toList
res7: List[BigInt] = List(1, 1, 2, 3, 5, 8, 13, 21, 34, 55)
As you can see it is quite easy to write programs where it is not obvious anymore to which value they will result. And since the halting problem is not solvable we can not let the compiler do the work for us in non trivial cases.
This also means that trivial cases, as the one shown in your question, could be hardcoded in the compiler. But since there can't exist a algorithm that can detect all possible trivial cases, all cases that are ever found need to be hardcoded in the compiler (not to mention that a definition of a trivial case does not exist). Therefore it wouldn't be wise to even start hardcoding some of these cases. It would ultimately result in a slower compiler and a compiler that is more difficult to maintain.
One could argue that for an use case that burns every second user it would be wise to at least hardcode such an extreme scenario. On the other hand, some people just need to be burned in order to learn something new. ;)
I think #Andreas' answer already has the necessary info. I'll just try to provide additional explanation:
When you write val name: String = name at the class level, this does a few different things at the same time:
create the field name
create the getter name()
create code for the assignment name = name, which becomes part of the primary constructor
This is what's made explicit by Andreas' 1.1
package <empty> {
class Example extends Object {
private[this] val x: Int = _;
<stable> <accessor> def x(): Int = Example.this.x;
def <init>(): Example = {
Example.super.<init>();
Example.this.x = Example.this.x();
()
}
}
}
The syntax is not Scala, it is (as suggested by [[syntax trees at end of cleanup]]) a textual representation of what the compiler will later convert into bytecode. Some unfamiliar syntax aside, we can interpret this, like the JVM would:
the JVM creates an object. At this point, all fields have default values. val x: Int = _; is like int x; in Java, i.e. the JVM's default value is used, which is 0 for I (i.e. int in Java, or Int in Scala)
the constructor is called for the object
(the super constructor is called)
the constructor calls x()
x() returns x, which is 0
x is assigned 0
the constructor returns
as you can see, after the initial parsing step, there is nothing in the syntax tree that seems immediately wrong, even though the original source code looks wrong. I wouldn't say that this is the behavior I expect, so I would imagine one of three things:
Either, the Scala devs saw it as too intricate to recognize and forbid
or, it's a regression and simply wasn't found as a bug
or, it's a "feature" and there is legitimate need for this behavior
(ordering reflects my opinion of likeliness, in decreasing order)
Why does this compile
scala> import scala.concurrent.Future
import scala.concurrent.Future
scala> val f: Unit = Future.successful(())
f: Unit = ()
I expected the compiler to complain about the assignment.
This is called "Value Discarding". Citing the scala specification (6.26.1):
Value Discarding
If e has some value type and the expected type is Unit, e is converted to the expected type by embedding it in the term { e; () }.
In other words, any value, whatever its type, is implicitly converted to Unit, effectively discarding it.
It you want to be warned about such discarding (which can in some cases hide a bug), you can pass the -Ywarn-value-discard option to the compiler. You'll then have to explicitly return () every time you call a method only for its side effect, but that method does return a non-Unit value.
The compiler is fine since applying f will only execute the call
val f: Unit = Future.successful(())
and the return value will go into the nirvana.
Basically this is the same as:
val f: Unit = {
Future.successful(())
()
}
If the compiler don't find the Unit it expects in the last value of the method it will put it there.
I was making my way through the Scala playframework tutorial and I came across this snippet of code which had me puzzled:
def newTask = Action { implicit request =>
taskForm.bindFromRequest.fold(
errors => BadRequest(views.html.index(Task.all(), errors)),
label => {
Task.create(label)
Redirect(routes.Application.tasks())
}
)
}
So I decided to investigate and came across this post.
I still don't get it.
What is the difference between this:
implicit def double2Int(d : Double) : Int = d.toInt
and
def double2IntNonImplicit(d : Double) : Int = d.toInt
other than the obvious fact they have different method names.
When should I use implicit and why?
I'll explain the main use cases of implicits below, but for more detail see the relevant chapter of Programming in Scala.
Implicit parameters
The final parameter list on a method can be marked implicit, which means the values will be taken from the context in which they are called. If there is no implicit value of the right type in scope, it will not compile. Since the implicit value must resolve to a single value and to avoid clashes, it's a good idea to make the type specific to its purpose, e.g. don't require your methods to find an implicit Int!
example:
// probably in a library
class Prefixer(val prefix: String)
def addPrefix(s: String)(implicit p: Prefixer) = p.prefix + s
// then probably in your application
implicit val myImplicitPrefixer = new Prefixer("***")
addPrefix("abc") // returns "***abc"
Implicit conversions
When the compiler finds an expression of the wrong type for the context, it will look for an implicit Function value of a type that will allow it to typecheck. So if an A is required and it finds a B, it will look for an implicit value of type B => A in scope (it also checks some other places like in the B and A companion objects, if they exist). Since defs can be "eta-expanded" into Function objects, an implicit def xyz(arg: B): A will do as well.
So the difference between your methods is that the one marked implicit will be inserted for you by the compiler when a Double is found but an Int is required.
implicit def doubleToInt(d: Double) = d.toInt
val x: Int = 42.0
will work the same as
def doubleToInt(d: Double) = d.toInt
val x: Int = doubleToInt(42.0)
In the second we've inserted the conversion manually; in the first the compiler did the same automatically. The conversion is required because of the type annotation on the left hand side.
Regarding your first snippet from Play:
Actions are explained on this page from the Play documentation (see also API docs). You are using
apply(block: (Request[AnyContent]) ⇒ Result): Action[AnyContent]
on the Action object (which is the companion to the trait of the same name).
So we need to supply a Function as the argument, which can be written as a literal in the form
request => ...
In a function literal, the part before the => is a value declaration, and can be marked implicit if you want, just like in any other val declaration. Here, request doesn't have to be marked implicit for this to type check, but by doing so it will be available as an implicit value for any methods that might need it within the function (and of course, it can be used explicitly as well). In this particular case, this has been done because the bindFromRequest method on the Form class requires an implicit Request argument.
WARNING: contains sarcasm judiciously! YMMV...
Luigi's answer is complete and correct. This one is only to extend it a bit with an example of how you can gloriously overuse implicits, as it happens quite often in Scala projects. Actually so often, you can probably even find it in one of the "Best Practice" guides.
object HelloWorld {
case class Text(content: String)
case class Prefix(text: String)
implicit def String2Text(content: String)(implicit prefix: Prefix) = {
Text(prefix.text + " " + content)
}
def printText(text: Text): Unit = {
println(text.content)
}
def main(args: Array[String]): Unit = {
printText("World!")
}
// Best to hide this line somewhere below a pile of completely unrelated code.
// Better yet, import its package from another distant place.
implicit val prefixLOL = Prefix("Hello")
}
In scala implicit works as:
Converter
Parameter value injector
Extension method
There are some uses of Implicit
Implicitly type conversion : It converts the error producing assignment into intended type
val x :String = "1"
val y:Int = x
String is not the sub type of Int , so error happens in line 2. To resolve the error the compiler will look for such a method in the scope which has implicit keyword and takes a String as argument and returns an Int .
so
implicit def z(a:String):Int = 2
val x :String = "1"
val y:Int = x // compiler will use z here like val y:Int=z(x)
println(y) // result 2 & no error!
Implicitly receiver conversion: We generally by receiver call object's properties, eg. methods or variables . So to call any property by a receiver the property must be the member of that receiver's class/object.
class Mahadi{
val haveCar:String ="BMW"
}
class Johnny{
val haveTv:String = "Sony"
}
val mahadi = new Mahadi
mahadi.haveTv // Error happening
Here mahadi.haveTv will produce an error. Because scala compiler will first look for the haveTv property to mahadi receiver. It will not find. Second it will look for a method in scope having implicit keyword which take Mahadi object as argument and returns Johnny object. But it does not have here. So it will create error. But the following is okay.
class Mahadi{
val haveCar:String ="BMW"
}
class Johnny{
val haveTv:String = "Sony"
}
val mahadi = new Mahadi
implicit def z(a:Mahadi):Johnny = new Johnny
mahadi.haveTv // compiler will use z here like new Johnny().haveTv
println(mahadi.haveTv)// result Sony & no error
Implicitly parameter injection: If we call a method and do not pass its parameter value, it will cause an error. The scala compiler works like this - first will try to pass value, but it will get no direct value for the parameter.
def x(a:Int)= a
x // ERROR happening
Second if the parameter has any implicit keyword it will look for any val in the scope which have the same type of value. If not get it will cause error.
def x(implicit a:Int)= a
x // error happening here
To slove this problem compiler will look for a implicit val having the type of Int because the parameter a has implicit keyword.
def x(implicit a:Int)=a
implicit val z:Int =10
x // compiler will use implicit like this x(z)
println(x) // will result 10 & no error.
Another example:
def l(implicit b:Int)
def x(implicit a:Int)= l(a)
we can also write it like-
def x(implicit a:Int)= l
Because l has a implicit parameter and in scope of method x's body, there is an implicit local variable(parameters are local variables) a which is the parameter of x, so in the body of x method the method-signature l's implicit argument value is filed by the x method's local implicit variable(parameter) a implicitly.
So
def x(implicit a:Int)= l
will be in compiler like this
def x(implicit a:Int)= l(a)
Another example:
def c(implicit k:Int):String = k.toString
def x(a:Int => String):String =a
x{
x => c
}
it will cause error, because c in x{x=>c} needs explicitly-value-passing in argument or implicit val in scope.
So we can make the function literal's parameter explicitly implicit when we call the method x
x{
implicit x => c // the compiler will set the parameter of c like this c(x)
}
This has been used in action method of Play-Framework
in view folder of app the template is declared like
#()(implicit requestHreader:RequestHeader)
in controller action is like
def index = Action{
implicit request =>
Ok(views.html.formpage())
}
if you do not mention request parameter as implicit explicitly then you must have been written-
def index = Action{
request =>
Ok(views.html.formpage()(request))
}
Extension Method
Think, we want to add new method with Integer object. The name of the method will be meterToCm,
> 1 .meterToCm
res0 100
to do this we need to create an implicit class within a object/class/trait . This class can not be a case class.
object Extensions{
implicit class MeterToCm(meter:Int){
def meterToCm={
meter*100
}
}
}
Note the implicit class will only take one constructor parameter.
Now import the implicit class in the scope you are wanting to use
import Extensions._
2.meterToCm // result 200
Why and when you should mark the request parameter as implicit:
Some methods that you will make use of in the body of your action have an implicit parameter list like, for example, Form.scala defines a method:
def bindFromRequest()(implicit request: play.api.mvc.Request[_]): Form[T] = { ... }
You don't necessarily notice this as you would just call myForm.bindFromRequest() You don't have to provide the implicit arguments explicitly. No, you leave the compiler to look for any valid candidate object to pass in every time it comes across a method call that requires an instance of the request. Since you do have a request available, all you need to do is to mark it as implicit.
You explicitly mark it as available for implicit use.
You hint the compiler that it's "OK" to use the request object sent in by the Play framework (that we gave the name "request" but could have used just "r" or "req") wherever required, "on the sly".
myForm.bindFromRequest()
see it? it's not there, but it is there!
It just happens without your having to slot it in manually in every place it's needed (but you can pass it explicitly, if you so wish, no matter if it's marked implicit or not):
myForm.bindFromRequest()(request)
Without marking it as implicit, you would have to do the above. Marking it as implicit you don't have to.
When should you mark the request as implicit? You only really need to if you are making use of methods that declare an implicit parameter list expecting an instance of the Request. But to keep it simple, you could just get into the habit of marking the request implicit always. That way you can just write beautiful terse code.
Also, in the above case there should be only one implicit function whose type is double => Int. Otherwise, the compiler gets confused and won't compile properly.
//this won't compile
implicit def doubleToInt(d: Double) = d.toInt
implicit def doubleToIntSecond(d: Double) = d.toInt
val x: Int = 42.0
I had the exact same question as you had and I think I should share how I started to understand it by a few really simple examples (note that it only covers the common use cases).
There are two common use cases in Scala using implicit.
Using it on a variable
Using it on a function
Examples are as follows
Using it on a variable. As you can see, if the implicit keyword is used in the last parameter list, then the closest variable will be used.
// Here I define a class and initiated an instance of this class
case class Person(val name: String)
val charles: Person = Person("Charles")
// Here I define a function
def greeting(words: String)(implicit person: Person) = person match {
case Person(name: String) if name != "" => s"$name, $words"
case _ => "$words"
}
greeting("Good morning") // Charles, Good moring
val charles: Person = Person("")
greeting("Good morning") // Good moring
Using it on a function. As you can see, if the implicit is used on the function, then the closest type conversion method will be used.
val num = 10 // num: Int (of course)
// Here I define a implicit function
implicit def intToString(num: Int) = s"$num -- I am a String now!"
val num = 10 // num: Int (of course). Nothing happens yet.. Compiler believes you want 10 to be an Int
// Util...
val num: String = 10 // Compiler trust you first, and it thinks you have `implicitly` told it that you had a way to covert the type from Int to String, which the function `intToString` can do!
// So num is now actually "10 -- I am a String now!"
// console will print this -> val num: String = 10 -- I am a String now!
Hope this can help.
A very basic example of Implicits in scala.
Implicit parameters:
val value = 10
implicit val multiplier = 3
def multiply(implicit by: Int) = value * by
val result = multiply // implicit parameter wiil be passed here
println(result) // It will print 30 as a result
Note: Here multiplier will be implicitly passed into the function multiply. Missing parameters to the function call are looked up by type in the current scope meaning that code will not compile if there is no implicit variable of type Int in the scope.
Implicit conversions:
implicit def convert(a: Double): Int = a.toInt
val res = multiply(2.0) // Type conversions with implicit functions
println(res) // It will print 20 as a result
Note: When we call multiply function passing a double value, the compiler will try to find the conversion implicit function in the current scope, which converts Int to Double (As function multiply accept Int parameter). If there is no implicit convert function then the compiler will not compile the code.