CODEWITHSUNDEEP
scalatypes
Aslan986
Aslan986

Reputation: 10324

How to use function with Any input

I have to define a second order function that takes as a parameter a function.

In my application, the input function may have any input type, but the output type has to be a specified one (suppose Int, it does not matter).

I define the second order function has:

def sof(f : (Any => Int) ) = {...}

Now, if I have a function f : Int => Int, and I call:

sof(f)

I get:

 found   : Int => Int
 required: Any => Int

I guess I am misunderstanding the meaning of the Any type.

How can I make it work?

Upvotes: 3

Views: 145

Answers (2)

lambdista
lambdista

Reputation: 1905

You shouldn't use Any there, but a type parameter such as: 

def sof[A](f: A => Int) = {...}

However I don't think you can do much with that method. Probably you would want something like this: 

def sof[A](a: A)(f: A => Int) = f(a)

// usage example
val x = sof("hello")(_.size * 2) // the result is 10

// you can also partially apply it to obtain other functions
val sizeDoubler: String => Int = sof(_)(_.size * 2)

val helloDoubleSize = sizeDoubler("hello") // the result is 10

This way you can pass any type to sof plus you'll have the compiler by your side to signal any strange behaviour. Using Any you lose this ability.

Final Note: In case the syntax I used to pass two parameters (the A value and the A => Int function) to a method looks strange to you that's called currying. If you Google it you'll find many good articles about it.

Upvotes: 2

Clashsoft
Clashsoft

Reputation: 11882

The parameters of functions in Scala are contravariant. That means that Int => Int is not a subtype of Any => Int, but vice-versa. Imagine the following: You pass a String to the Any => Int function (that is actually implemented by a Int => Int function). How would the Int => Int handle the String argument?

Upvotes: 4

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