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There is a program where I would like to limit the range on a set of ints from 5 to 15.
Is there a way to define a type which allows this?
An example of how would like to use this:
// Define type Good X as range from 5 to 15
class Foo(val x: GoodX)
{
//blah blah
}
I would also like to preserve the "Int-iness" of GoodX.
val base:GoodX=5
val f=Foo(base+4)
Take a look at https://github.com/fthomas/refined . It allows you to refine (constrain) existing types at type level. E.g. positive integers, which still have a subtype relationship with integers.
The syntax is a bit verbose, and it will box primitives (see below for details). But other than that it does exactly what you want.
Here is a short demo. Define a refinement and a method using a refined type:
import eu.timepit.refined._
import eu.timepit.refined.api.Refined
import eu.timepit.refined.auto._
import eu.timepit.refined.numeric._
type FiveToFifteen = GreaterEqual[W.`5`.T] And Less[W.`15`.T]
type IntFiveToFifteen = Int Refined FiveToFifteen
def sum(a: IntFiveToFifteen, b: IntFiveToFifteen): Int = a + b
Use it with constants (note the good compile error messages):
scala> sum(5,5)
res6: Int = 10
scala> sum(0,10)
<console>:60: error: Left predicate of (!(0 < 5) && (0 < 15)) failed: Predicate (0 < 5) did not fail.
sum(0,10)
^
scala> sum(5,20)
<console>:60: error: Right predicate of (!(20 < 5) && (20 < 15)) failed: Predicate failed: (20 < 15).
sum(5,20)
^
When you have variables, you do not know at compile time whether they are in range or not. So downcasting from Int to a refined int can fail. Throwing exceptions is not considered good style in functional libraries. So the refineV method returns an Either:
val x = 20
val y = 5
scala> refineV[FiveToFifteen](x)
res14: Either[String,eu.timepit.refined.api.Refined[Int,FiveToFifteen]] = Left(Right predicate of (!(20 < 5) && (20 < 15)) failed: Predicate failed: (20 < 15).)
scala> refineV[FiveToFifteen](y)
res16: Either[String,eu.timepit.refined.api.Refined[Int,FiveToFifteen]] = Right(5)
I think Partial Function would help.
case class GoodX(x: Int)
object GoodX {
def apply: PartialFunction[Int, GoodX] =
{ case i if i > 5 && i < 15 => new GoodX(i) }
}
// implicits to remain int-fulness
implicit def goodXToInt(goodX: GoodX): Int = goodX.x
GoodX(5) // throw Match Error
GoodX(10) // GoodX(10)
This solution requires no library.
Hope this help.
As shown in examples section of refined library we can define a custom refined type whose value is between 7 and 77
// Here we define a refined type "Int with the predicate (7 <= value < 77)".
scala> type Age = Int Refined Interval.ClosedOpen[W.`7`.T, W.`77`.T]
Furthermore, if on scala 2.13.x, one can also use literal based singleton types as shown below there by not needing Witness from shapeless ;)
import eu.timepit.refined.numeric.Interval.Closed
type AgeOfChild = Int Refined Closed[2, 12]
case class Child(name: NonEmptyString, age:AgeOfChild = 2)
Please refer to SIP and official documentation for more details.
Sure ...
object FiveToFifteen extends Enumeration {
val _5 = Value(5)
val _6,_7,_8,_9,_10,_11,_12,_13,_14,_15 = Value
}
Edit if you want to "preserve int-ness", you could also add conversions like this:
implicit def toInt(v: Value) = v.id
implicit def fromInt(i: Int) = apply(i)
But this, obviously, won't make your type much more "int-ful" then it already is (which is, pretty much none), because things like
val v: Value = _15 - _10 or val v: Value = _5 * 3 or even val v = _15 * _5 will work, but others, like val v: Value = _5 - 1 will crash
This question already has answers here:
Equivalent to Ruby's #tap method in Scala [duplicate]
(1 answer)
how to keep return value when logging in scala
(6 answers)
Closed 9 years ago.
Often I have functions like this:
{
val x = foo;
bar(x);
x
}
For example, bar is often something like Log.debug.
Is there a shorter, idiomatic way how to run it? For example, a built-in function like
def act[A](value: A, f: A => Any): A = { f(value); value }
so that I could write just act(foo, bar _).
I'm not sure if i understood the question correctly, but if i do, then i often use this method taken from the Spray toolkit:
def make[A, B](obj: A)(f: A => B): A = { f(obj); obj }
then you can write the following things:
utils.make(new JobDataMap()) { map =>
map.put("phone", m.phone)
map.put("medicine", m.medicine.name)
map.put("code", utils.genCode)
}
Using your act function as written seems perfectly idiomatic to me. I don't know of a built-in way to do it, but I'd just throw this kind of thing in a "commons" or "utils" project that I use everywhere.
If the bar function is usually the same (e.g. Log.debug) then you could also make a specific wrapper function for that. For instance:
def withDebug[A](prefix: String)(value: A)(implicit logger: Logger): A = {
logger.debug(prefix + value)
value
}
which you can then use as follows:
implicit val loggerI = logger
def actExample() {
// original method
val c = act(2 + 2, logger.debug)
// a little cleaner?
val d = withDebug("The sum is: ") {
2 + 2
}
}
Or for even more syntactic sugar:
object Tap {
implicit def toTap[A](value: A): Tap[A] = new Tap(value)
}
class Tap[A](value: A) {
def tap(f: A => Any): A = {
f(value)
value
}
def report(prefix: String)(implicit logger: Logger): A = {
logger.debug(prefix + value)
value
}
}
object TapExample extends Logging {
import Tap._
implicit val loggerI = logger
val c = 2 + 2 tap { x => logger.debug("The sum is: " + x) }
val d = 2 + 2 report "The sum is: "
assert(d == 4)
}
Where tap takes an arbitrary function, and report just wraps a logger. Of course you could add whatever other commonly used taps you like to the Tap class.
Note that Scala already includes a syntactically heavyweight version:
foo match { case x => bar(x); x }
but creating the shorter version (tap in Ruby--I'd suggest using the same name) can have advantages.
In short: I try to write something like A <N B for a DSL in Scala, for an integer N and A,B of Type T. Is there a nice possibility to do so?
Longer: I try to write a DSL for TGrep2 in Scala. I'm currently interested to write
A <N B B is the Nth child of A (the rst child is <1).
in a nice way and as close as possible to the original definition in Scala. Is there a way to overload the < Operator that it can take a N and a B as a argument.
What I tried: I tried two different possibilities which did not make me very happy:
scala> val N = 10
N: Int = 10
scala> case class T(n:String) {def <(i:Int,j:T) = println("huray!")}
defined class T
scala> T("foo").<(N,T("bar"))
huray!
and
scala> case class T(n:String) {def <(i:Int) = new {def apply(j:T) = println("huray!")}}
defined class T
scala> (T("foo")<N)(T("bar"))
warning: there were 1 feature warnings; re-run with -feature for details
huray!
Id suggest you use something like nth instead of the < symbol which makes the semantics clear. A nth N is B would make a lot of sense to me at least. It would translate to something like
case class T (label:String){
def is(j:T) = {
label equals j.label
}
}
case class J(i:List[T]){
def nth(index:Int) :T = {
i(index)
}
}
You can easily do:
val t = T("Mice")
val t1 = T("Rats")
val j = J(List(t1,t))
j nth 1 is t //res = true
The problem is that apply doesn't work as a postfix operator, so you can't write it without the parantheses, you could write this:
case class T(n: String) {
def <(in: (Int, T)) = {
in match {
case (i, t) =>
println(s"${t.n} is the ${i} child of ${n}")
}
}
}
implicit class Param(lower: Int) {
def apply(t: T) = (lower, t)
}
but then,
T("foo") < 10 T("bar")
would still fail, but you could work it out with:
T("foo") < 10 (T("bar"))
there isn't a good way of doing what you want without adding parenthesis somewhere.
I think that you might want to go for a combinational parser instead if you really want to stick with this syntax. Or as #korefn proposed, you break the compatibility and do it with new operators.
I worked my way implementing a recursive version of selection and quick sort,i am trying to modify the code in a way that it can sort a list of any generic type , i want to assume that the generic type supplied can be converted to Comparable at runtime.
Does anyone have a link ,code or tutorial on how to do this please
I am trying to modify this particular code
'def main (args:Array[String]){
val l = List(2,4,5,6,8)
print(quickSort(l))
}
def quickSort(x:List[Int]):List[Int]={
x match{
case xh::xt =>
{
val (first,pivot,second) = partition(x)
quickSort (first):::(pivot :: quickSort(second))
}
case Nil => {x}
}
}
def partition (x:List[Int])=
{
val pivot =x.head
var first:List[Int]=List ()
var second : List[Int]=List ()
val fun=(i:Int)=> {
if (i<pivot)
first=i::first
else
second=i::second
}
x.tail.foreach(fun)
(first,pivot,second)
}
enter code here
def main (args:Array[String]){
val l = List(2,4,5,6,8)
print(quickSort(l))
}
def quickSort(x:List[Int]):List[Int]={
x match{
case xh::xt =>
{
val (first,pivot,second) = partition(x)
quickSort (first):::(pivot :: quickSort(second))
}
case Nil => {x}
}
}
def partition (x:List[Int])=
{
val pivot =x.head
var first:List[Int]=List ()
var second : List[Int]=List ()
val fun=(i:Int)=> {
if (i<pivot)
first=i::first
else
second=i::second
}
x.tail.foreach(fun)
(first,pivot,second)
} '
Language: SCALA
In Scala, Java Comparator is replaced by Ordering (quite similar but comes with more useful methods). They are implemented for several types (primitives, strings, bigDecimals, etc.) and you can provide your own implementations.
You can then use scala implicit to ask the compiler to pick the correct one for you:
def sort[A]( lst: List[A] )( implicit ord: Ordering[A] ) = {
...
}
If you are using a predefined ordering, just call:
sort( myLst )
and the compiler will infer the second argument. If you want to declare your own ordering, use the keyword implicit in the declaration. For instance:
implicit val fooOrdering = new Ordering[Foo] {
def compare( f1: Foo, f2: Foo ) = {...}
}
and it will be implicitly use if you try to sort a List of Foo.
If you have several implementations for the same type, you can also explicitly pass the correct ordering object:
sort( myFooLst )( fooOrdering )
More info in this post.
For Quicksort, I'll modify an example from the "Scala By Example" book to make it more generic.
class Quicksort[A <% Ordered[A]] {
def sort(a:ArraySeq[A]): ArraySeq[A] =
if (a.length < 2) a
else {
val pivot = a(a.length / 2)
sort (a filter (pivot >)) ++ (a filter (pivot == )) ++
sort (a filter(pivot <))
}
}
Test with Int
scala> val quicksort = new Quicksort[Int]
quicksort: Quicksort[Int] = Quicksort#38ceb62f
scala> val a = ArraySeq(5, 3, 2, 2, 1, 1, 9, 39 ,219)
a: scala.collection.mutable.ArraySeq[Int] = ArraySeq(5, 3, 2, 2, 1, 1, 9, 39, 21
9)
scala> quicksort.sort(a).foreach(n=> (print(n), print (" " )))
1 1 2 2 3 5 9 39 219
Test with a custom class implementing Ordered
scala> case class Meh(x: Int, y:Int) extends Ordered[Meh] {
| def compare(that: Meh) = (x + y).compare(that.x + that.y)
| }
defined class Meh
scala> val q2 = new Quicksort[Meh]
q2: Quicksort[Meh] = Quicksort#7677ce29
scala> val a3 = ArraySeq(Meh(1,1), Meh(12,1), Meh(0,1), Meh(2,2))
a3: scala.collection.mutable.ArraySeq[Meh] = ArraySeq(Meh(1,1), Meh(12,1), Meh(0
,1), Meh(2,2))
scala> q2.sort(a3)
res7: scala.collection.mutable.ArraySeq[Meh] = ArraySeq(Meh(0,1), Meh(1,1), Meh(
2,2), Meh(12,1))
Even though, when coding Scala, I'm used to prefer functional programming style (via combinators or recursion) over imperative style (via variables and iterations), THIS TIME, for this specific problem, old school imperative nested loops result in simpler code for the reader. I don't think falling back to imperative style is a mistake for certain classes of problems (such as sorting algorithms which usually transform the input buffer (like a procedure) rather than resulting to a new sorted one
Here it is my solution:
package bitspoke.algo
import scala.math.Ordered
import scala.collection.mutable.Buffer
abstract class Sorter[T <% Ordered[T]] {
// algorithm provided by subclasses
def sort(buffer : Buffer[T]) : Unit
// check if the buffer is sorted
def sorted(buffer : Buffer[T]) = buffer.isEmpty || buffer.view.zip(buffer.tail).forall { t => t._2 > t._1 }
// swap elements in buffer
def swap(buffer : Buffer[T], i:Int, j:Int) {
val temp = buffer(i)
buffer(i) = buffer(j)
buffer(j) = temp
}
}
class SelectionSorter[T <% Ordered[T]] extends Sorter[T] {
def sort(buffer : Buffer[T]) : Unit = {
for (i <- 0 until buffer.length) {
var min = i
for (j <- i until buffer.length) {
if (buffer(j) < buffer(min))
min = j
}
swap(buffer, i, min)
}
}
}
As you can see, rather than using java.lang.Comparable, I preferred scala.math.Ordered and Scala View Bounds rather than Upper Bounds. That's certainly works thanks to many Scala Implicit Conversions of primitive types to Rich Wrappers.
You can write a client program as follows:
import bitspoke.algo._
import scala.collection.mutable._
val sorter = new SelectionSorter[Int]
val buffer = ArrayBuffer(3, 0, 4, 2, 1)
sorter.sort(buffer)
assert(sorter.sorted(buffer))
Is there any reason for Scala not support the ++ operator to increment primitive types by default?
For example, you can not write:
var i=0
i++
Thanks
My guess is this was omitted because it would only work for mutable variables, and it would not make sense for immutable values. Perhaps it was decided that the ++ operator doesn't scream assignment, so including it may lead to mistakes with regard to whether or not you are mutating the variable.
I feel that something like this is safe to do (on one line):
i++
but this would be a bad practice (in any language):
var x = i++
You don't want to mix assignment statements and side effects/mutation.
I like Craig's answer, but I think the point has to be more strongly made.
There are no "primitives" -- if Int can do it, then so can a user-made Complex (for example).
Basic usage of ++ would be like this:
var x = 1 // or Complex(1, 0)
x++
How do you implement ++ in class Complex? Assuming that, like Int, the object is immutable, then the ++ method needs to return a new object, but that new object has to be assigned.
It would require a new language feature. For instance, let's say we create an assign keyword. The type signature would need to be changed as well, to indicate that ++ is not returning a Complex, but assigning it to whatever field is holding the present object. In Scala spirit of not intruding in the programmers namespace, let's say we do that by prefixing the type with #.
Then it could be like this:
case class Complex(real: Double = 0, imaginary: Double = 0) {
def ++: #Complex = {
assign copy(real = real + 1)
// instead of return copy(real = real + 1)
}
The next problem is that postfix operators suck with Scala rules. For instance:
def inc(x: Int) = {
x++
x
}
Because of Scala rules, that is the same thing as:
def inc(x: Int) = { x ++ x }
Which wasn't the intent. Now, Scala privileges a flowing style: obj method param method param method param .... That mixes well C++/Java traditional syntax of object method parameter with functional programming concept of pipelining an input through multiple functions to get the end result. This style has been recently called "fluent interfaces" as well.
The problem is that, by privileging that style, it cripples postfix operators (and prefix ones, but Scala barely has them anyway). So, in the end, Scala would have to make big changes, and it would be able to measure up to the elegance of C/Java's increment and decrement operators anyway -- unless it really departed from the kind of thing it does support.
In Scala, ++ is a valid method, and no method implies assignment. Only = can do that.
A longer answer is that languages like C++ and Java treat ++ specially, and Scala treats = specially, and in an inconsistent way.
In Scala when you write i += 1 the compiler first looks for a method called += on the Int. It's not there so next it does it's magic on = and tries to compile the line as if it read i = i + 1. If you write i++ then Scala will call the method ++ on i and assign the result to... nothing. Because only = means assignment. You could write i ++= 1 but that kind of defeats the purpose.
The fact that Scala supports method names like += is already controversial and some people think it's operator overloading. They could have added special behavior for ++ but then it would no longer be a valid method name (like =) and it would be one more thing to remember.
I think the reasoning in part is that +=1 is only one more character, and ++ is used pretty heavily in the collections code for concatenation. So it keeps the code cleaner.
Also, Scala encourages immutable variables, and ++ is intrinsically a mutating operation. If you require +=, at least you can force all your mutations to go through a common assignment procedure (e.g. def a_=).
The primary reason is that there is not the need in Scala, as in C. In C you are constantly:
for(i = 0, i < 10; i++)
{
//Do stuff
}
C++ has added higher level methods for avoiding for explicit loops, but Scala has much gone further providing foreach, map, flatMap foldLeft etc. Even if you actually want to operate on a sequence of Integers rather than just cycling though a collection of non integer objects, you can use Scala range.
(1 to 5) map (_ * 3) //Vector(3, 6, 9, 12, 15)
(1 to 10 by 3) map (_ + 5)//Vector(6, 9, 12, 15)
Because the ++ operator is used by the collection library, I feel its better to avoid its use in non collection classes. I used to use ++ as a value returning method in my Util package package object as so:
implicit class RichInt2(n: Int)
{
def isOdd: Boolean = if (n % 2 == 1) true else false
def isEven: Boolean = if (n % 2 == 0) true else false
def ++ : Int = n + 1
def -- : Int = n - 1
}
But I removed it. Most of the times when I have used ++ or + 1 on an integer, I have later found a better way, which doesn't require it.
It is possible if you define you own class which can simulate the desired output however it may be a pain if you want to use normal "Int" methods as well since you would have to always use *()
import scala.language.postfixOps //otherwise it will throw warning when trying to do num++
/*
* my custom int class which can do ++ and --
*/
class int(value: Int) {
var mValue = value
//Post-increment
def ++(): int = {
val toReturn = new int(mValue)
mValue += 1
return toReturn
}
//Post-decrement
def --(): int = {
val toReturn = new int(mValue)
mValue -= 1
return toReturn
}
//a readable toString
override def toString(): String = {
return mValue.toString
}
}
//Pre-increment
def ++(n: int): int = {
n.mValue += 1
return n;
}
//Pre-decrement
def --(n: int): int = {
n.mValue -= 1
return n;
}
//Something to get normal Int
def *(n: int): Int = {
return n.mValue
}
Some possible test cases
scala>var num = new int(4)
num: int = 4
scala>num++
res0: int = 4
scala>num
res1: int = 5 // it works although scala always makes new resources
scala>++(num) //parentheses are required
res2: int = 6
scala>num
res3: int = 6
scala>++(num)++ //complex function
res4: int = 7
scala>num
res5: int = 8
scala>*(num) + *(num) //testing operator_*
res6: Int = 16
Of course you can have that in Scala, if you really want:
import scalaz._
import Scalaz._
case class IncLens[S,N](lens: Lens[S,N], num : Numeric[N]) {
def ++ = lens.mods(num.plus(_, num.one))
}
implicit def incLens[S,N:Numeric](lens: Lens[S,N]) =
IncLens[S,N](lens, implicitly[Numeric[N]])
val i = Lens[Int,Int](identity, (x, y) => y)
val imperativeProgram = for {
_ <- i := 0;
_ <- i++;
_ <- i++;
x <- i++
} yield x
def runProgram = imperativeProgram ! 0
And here you go:
scala> runProgram
runProgram: Int = 3
It isn't included because Scala developers thought it make the specification more complex while achieving only negligible benefits and because Scala doesn't have operators at all.
You could write your own one like this:
class PlusPlusInt(i: Int){
def ++ = i+1
}
implicit def int2PlusPlusInt(i: Int) = new PlusPlusInt(i)
val a = 5++
// a is 6
But I'm sure you will get into some trouble with precedence not working as you expect. Additionally if i++ would be added, people would ask for ++i too, which doesn't really fit into Scala's syntax.
Lets define a var:
var i = 0
++i is already short enough:
{i+=1;i}
Now i++ can look like this:
i(i+=1)
To use above syntax, define somewhere inside a package object, and then import:
class IntPostOp(val i: Int) { def apply(op: Unit) = { op; i } }
implicit def int2IntPostOp(i: Int): IntPostOp = new IntPostOp(i)
Operators chaining is also possible:
i(i+=1)(i%=array.size)(i&=3)
The above example is similar to this Java (C++?) code:
i=(i=i++ %array.length)&3;
The style could depend, of course.