Execute functions in a List - scala

I have Object in scala in which I have defined some functions.
Object Sample {
val listFunction = Seq(func1(a,b),func2(p,q))
def func1(a: Int,b : Int) : Int ={
val c = a+b
c
}
def func2(p: Int,q : Int) : Int ={
val d = p+q
}
}
def main(args: Array[String]): Unit = {
//Want to call the list and execute the functions
ListFunction
}
How to call the list in main method and execute it?

Given
def func1(a: Int, b: Int): Int = a + b
def func2(p: Int, q: Int): Int = p + q
Consider the difference between
val x: Int = func1(2, 3) // applied function
val f: (Int, Int) => Int = func1 // function as value
So you have to use functions as values as you pass them to the sequence like so
val listFunction: Seq[(Int, Int) => Int] = Seq(func1, func2)
and then map over the list to apply the functions
listFunction.map(f => f.apply(2, 3))
listFunction.map(f => f(2, 3))
listFunction.map(_(2, 3))
scastie

Related

Invoke method based on list

I have object in scala, where I have defined functions. I have Seq where all the functions are listed as the elements as below:
object ABC {
def abc(a: Int, b: Int): Int = a+b
def pqr(p: Int, q: Int): Int = p - q
val listFunction: Seq[(Int, Int) => Int] = Seq(abc, pqr)
}
Now, I want to call the list which I defined as string and execute all the functions inside it, by taking the parameters. Similar to the below:
val listName = "listFunction"
val method = example.ABC.getClass.getMethod(listName,1,2)
Can someone help me, how to invoke the listFunction list and execute all the functions inside it.
To call the methods do this:
val result: Seq[Int] = ABC.listFunction.map(_(1, 2))
This calls each function in turn and puts the result in a new Seq.
If you want multiple lists indexed by name, use a Map like this:
object ABC {
def abc(a: Int, b: Int): Int = a+b
def pqr(p: Int, q: Int): Int = p - q
private val functions = Map(
"listFunction" -> Seq(abc _, pqr _),
"listReverse" -> Seq(pqr _, abc _),
)
def apply(name: String) = functions(name)
}
val listName = "listFunction"
val result: Seq[Int] = ABC(listName).map(_(1, 2))
Try
val listName = "listFunction"
example.ABC.getClass.getMethod(listName).invoke(example.ABC).asInstanceOf[Seq[(Int, Int) => Int]].map(_.apply(1, 2)) // List(3, -1)

Why scala can't pass a function name directly to a val?

I have this scala code:
object C extends App{
def sum(a:Int,b:Int,c:Int)=a+b+c
//val d=sum
val d=sum _
println(d(1,2,3))
}
I have to write sum _,not sum
but In f#,I can write this:
let sum a b c=a+b+c
let d=sum
[<EntryPoint>]
let main argv =
printfn "%A" (d 1 2 3)
0 // return an integer exit code
why in scala I can't just write
val d=sum
Scala has
methods (by default def is a method) and
functions
And they are not same.
Taking java methods as example, you can not assign methods to a variable without evaluating the method. But you can do that with function and to achieve that in scala define sum as function.
scala> def sum: (Int, Int, Int) => Int = (a, b, c) => a + b + c
sum: (Int, Int, Int) => Int
scala> val sumVal = sum
sumVal: (Int, Int, Int) => Int = $$Lambda$912/0x0000000801667840#716f94c1
scala> sumVal(1, 2, 3)
res1: Int = 6
Longer version of defining a function is,
scala> def sum = new Function3[Int, Int, Int, Int] {
| def apply(a: Int, b: Int, c: Int): Int = a +b + c
| }
sum: (Int, Int, Int) => Int
scala> val sumVal = sum
sumVal: (Int, Int, Int) => Int = <function3>
scala> sumVal(1, 2, 3)
res2: Int = 6

Partial Functions as an input parameter in scala

I am trying to pass the partial function as input to square function.
What is worng with the following code?
package PartialFunction
object PartialFunctionSum {
val yourConstant = 10
val pf: PartialFunction[(Int, Int), Int] = {
case (x, y) => x + y + yourConstant
}
def square(cb:(Int, Int) => Int): Unit = {
println(cb(5,10))
}
def main(args: Array[String]): Unit= {
square(pf)
}
}
A PartialFunction is a Function1 (a => b) and square needs a Function2 ((a,b) => c)
One way to accomplish what you want to do is changing cb type to Function1.
object PartialFunctionSum extends App {
val yourConstant = 10
val pf: PartialFunction[(Int, Int), Int] = {
case (x, y) => x + y + yourConstant
}
def square(cb:((Int, Int)) => Int): Unit = {
println(cb(5,10))
}
square(pf)
}
Also, anywhere where you need a FunctionX, you can define a PartialFunction, so
square {
case (x, y) => x + y + yourConstant
}
Or transform your PartialFunction in a Function2 with a lambda
square( pf.apply(_, _) )
Anywhere you need a function you can a pass a PartialFunction,
Your problem is with square definition:
def square(cb:(Int, Int) => Int): Unit = {
println(cb(5,10))
}
That means that cb is a function that receives two Ints and returns another Int.
Just change square signature as follows:
def square(cb: ((Int, Int)) => Int): Unit = {
println(cb(5 -> 10))
}
And now your types are compatible and the rest of your code works as you expected.
(Int, Int) => Int is an instance of Function2[Int, Int, Int](a Function that takes 2 Int parameters and returns an Int), not PartialFunction[(Int, Int), Int] (a partial function that takes an input of type (Int, Int) and returns an Int)
If you want to pass your parameters as a Tuple then:
def square(cb:PartialFunction[(Int, Int),Int]): Unit = {
println(cb((5,10)))
}
Example:
def square(cb:PartialFunction[(Int, Int),Int]): Unit = {
println(cb((5,10)))
}
square: (cb: PartialFunction[(Int, Int),Int])Unit
scala> square{case (i, j) => i * j}
50

Why does this partial application not compile?

The following:
val add = (a: Int, b: Int) => a + b
gets converted to:
object add extends Function2[Int, Int, Int] {
def apply(a: Int, b: Int) = a + b
}
while
val a1 = add(_: Int, 3)
gets converted to:
object a1 extends Function1[Int, Int] {
def apply(x: Int): Int = {
add(x, 3)
}
}
But when I do:
scala> val a2 = add _
a2: () => (Int, Int) => Int = <function0>
And then call a2, it throws an error:
scala> a2(1, 2)
<console>:11: error: too many arguments for method apply: ()(Int, Int) => Int in trait Function0
a2(1, 2)
^
Why is this? Why does the following work?
a2()(1, 2)
add is already a Function2[Int, Int, Int]. If you want a2 to have the same type, then a simple assignment will suffice.
scala> val a2 = add
a2: (Int, Int) => Int = <function2>
scala> a2(1, 2)
res3: Int = 3
What you're thinking of is eta-expansion of a method into a function. If we had:
def add(a: Int, b: Int): Int = a + b
Then, we would use add _ to get the eta-expansion to assign to a value.
scala> def a2 = add _
a2: (Int, Int) => Int
scala> a2(1, 2)
res4: Int = 3
But add is already a function, so the underscore has a different meaning. add is now a value and not a method. Since add is a value, it is like a parameter-less method that returns a Function2[Int, Int, Int]. And if we try to get the eta-expansion of that, we get () => Function2[Int, Int, Int].
Consider a simpler example where we have a simple val a = 1. a is essentially the same as a parameter-less method that returns 1 (def a = 1). If I try to obtain the eta-expansion, I will get () => Int.
scala> val a = 1
a: Int = 1
scala> val a2 = a _
a2: () => Int = <function0>

Implicit conversions for defs/lambdas in Scala?

I just ran into a strange disparity between functions and objects (scala 2.10):
implicit def conv(c: Int => String) : (PrintStream => Int => Unit) = p => v => p.println(c(v))
def f(h: PrintStream => Int => Unit) : Unit = h(System.out)(1)
def a(x: Int) = x.toString
val b = (x: Int) => x.toString
// def main(args: Array[String]) = f(a) // fail
// def main(args: Array[String]) = f((x: Int) => x.toString) // fail
def main(args: Array[String]) = f(b) // ok
Why is there a difference between defs/lambda literals and lambda vals?
Update: apparently, the Problem does not occur for binary functions: Implicit conversion of a function to a second-order-function only works if the function to convert has at least two parameters
I checked this, and indeed the following code works:
implicit def conv(c: (Int,Unit) => String) : (PrintStream => Int => Unit) = p => v => p.println(c(v,()))
def f(h: PrintStream => Int => Unit) : Unit = h(System.out)(1)
def a(x: Int, y : Unit) = x.toString
val b = (x: Int, y : Unit) => x.toString
def main(args: Array[String]) = f(a) // ok
def main(args: Array[String]) = f((x: Int, y: Unit) => x.toString) // ok
def main(args: Array[String]) = f(b) // ok
Likewise, Nullary functions don't pose a problem, either:
implicit def conv(c: () => String) : (PrintStream => Int => Unit) = p => v => p.println(c())
def f(h: PrintStream => Int => Unit) : Unit = h(System.out)(1)
def a() = "1"
val b = () => "1"
def main(args: Array[String]) = f(a) // ok
def main(args: Array[String]) = f(() => "1") // ok
def main(args: Array[String]) = f(b) // ok
So, rephrasing the question: why does this not work for UNARY methods and functions?
Update: the problem also seems to be related to the target type (the type of f's argument h). The following also works (this time, in favour of "eta-expansion counts as hop", because we need to create a method value from a using _)
implicit def conv(c: Int => String) : Unit = ()
def f(h: Unit) : Unit = System.out.print("?")
def a(x: Int) = x.toString
val b = (x: Int) => x.toString
def main(args: Array[String]) = f(a _) // ok
def main(args: Array[String]) = f((x: Int) => x.toString) // ok
def main(args: Array[String]) = f(b) // ok
In scala defs are methods and are diffrent from functions.
scala> def a( x: Int, y: Int ): Int = x + y
a: (x: Int, y:Int)Int
scala> (x: Int, y: Int) => x + y
res0: (Int, Int) => Int = <function2>
You can convert a method to function by partially applying it.
scala> b _
res1: (Int, Int) => Int = <function2>
So.. you can do,
implicit def conv(c: Int => String) : (PrintStream => Int => Unit) = p => v => p.println(c(v))
def f(h: PrintStream => Int => Unit) : Unit = h(System.out)(1)
def a(x: Int) = x.toString
val af = a _
def main( args: Array[ String ] ) = f( af )
Alse, as #srgfed01 mentioned in his comment... for the second case the problem is that... they types are not explicitly specified, if you specify the type correctly... the second case will work.
scala> f( ( a => a.toString ): (Int => String) )
1
or
scala> f( ( _.toString ): (Int => String) )
1
Now, about differences between methods and functions...
You can call a method taking no arguments without parenthesis ()... but you can not call a function without ().
scala> def g() = 5
g: ()Int
scala> g
res15: Int = 5
scala> () => 5
res13: () => Int = <function0>
scala> res13
res14: () => Int = <function0>
scala> res13()
res15: 5
One of the most important reasons for methods being different from functions is because creators of Scala wanted seamless inter-interoperability with Java without being stuck with Java's limitations.
So methods (def) are very much similar to Java methods and keeping functions different from methods enabled them with limitless freedom to create Scala, the way they wanted.
Also... Another major difference is that methods can accept Type-classes where as functions can not. Basically you can have generic methods like
scala> :paste
trait Behave {
def behave
}
class A( elem: String ) extends Behave {
def behave() {
println( elem )
}
}
// Exiting paste mode, now interpreting.
defined trait Behave
defined class A
Now you can define a generic method,
scala> def check[ T <: Behave ]( t: T ): Unit = t.behave()
check: [T <: Behave](t: T)Unit
But you can not define a function like this,
scala> ( t: T ) => t.behave()
<console>:8: error: not found: type T
( t: T ) => t.behave()
or like this
scala> ( t: (T <: Behave) ) => t.behave()
<console>:1: error: ')' expected but '<:' found.
( t: (T <: A) ) => t.behave()