Placeholder Syntax in Scala Programming [duplicate] - scala

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What are all the uses of an underscore in Scala?
(7 answers)
Closed 2 years ago.
I came across the below placeholder syntax in scala. And could not able to under what '_' is doing in the code
def evenDef(n: Int): Int => Boolean = _ => n % 2 == 0
//evenDef: (n: Int)Int => Boolean
val result = evenDef(2)
//res: Int => Boolean = $$Lambda$5397/911325#3aa9352
result(9)
//res18: Boolean = true
I am little confused how the above is returning 'true' for the value 9, even though its a odd number.
Can anyone help me in understand the '_' in the code ?

The code is equivalent to
def evenDef(n: Int): Int => Boolean = (ignored: Int) => n % 2 == 0
Since ignored is not used anywhere it can be replaced with the underscore placeholder.
Admittedly the code is a bit useless the way it stands. For any integer n it returns a constant function that either returns true for even values of n and false otherwise. The argument of the returned function does not matter at all.

Related

In which scenario '=' used and '=>' used in scala? [duplicate]

This question already has an answer here:
Can anyone explain how the symbol "=>" is used in Scala
(1 answer)
Closed 4 years ago.
I am new to scala and I confused little bit with = and =>, I don't know in which scenario exactly = used and => used.
#RamanMishra's comment is correct, however I would like to expand it a bit hopping it becomes more clear to you.
a = b means a is now equals to b (in the imperative meaning of the word, which is better understood as assignment).
It is used in many places, including when defining a function\method, for example
def sum(a: Int, b: Int): Int = a + b
The above statement can be readed as:
"We defined a function with name sum which takes two parameters (a & b) both of type Int and returns an Int. And its body\implementation is 'equals' to a + b".
a => b means, there is a function which takes a parameters list a and has a body b, it is used a couple of places, including anonymous functions and higher-order functions.
val anonymous: Int => Int = x => x + 1
The above statement can be readed as:
"We defined a value called anonymous, whose type is a function with one parameter of type Int and a return type Int, which is 'equals' to the 'anonymous' function which takes one parameter x (its type is inferred in by the context, in this type the explicit type signature before) and its body is x + 1"
def higherOrder(x: Int, f: Int => Int): Int = f(x)
The above statement can be readed as:
"There is a function (sum), which takes another function as a parameter (f), the later is a function of type Int to Int".
Edit
As #Aki suggested, the => is also used in pattern matching to separate the cases and the block of code to execute for each one.
def fact(n: Long): Long = n match {
case 0L => 1L
case n => n * fact(n - 1)
}
"This is somewhat similar to a parameter list => function body, as the above examples".
PS: This is a 101 Scala question, SO is not the best place for such questions, because there is probably enough resources on the internet - like this cheatsheet, or better places to ask - like the scala gitter channel.

Understanding currying Scala [duplicate]

This question already has answers here:
Understanding Currying in Scala
(3 answers)
Closed 4 years ago.
def sum(f: Int => Int): (Int, Int) => Int = {
def sumF(a: Int, b: Int): Int =
if (a > b) 0
else f(a) + sumF(a + 1, b)
sumF
}
def sumCubes = sum(a => a * a * a)
sumCubes // output Function2
sumCubes(1,10) // output 3025.
sumCubes() // doesn't work
I feel I dont understand currying well enough. In the first statement , we are calling sumCubes without parameters , hence sum gets called with the anonymous function as parameter and returns a function2.
Whats really happening in 2nd and 3rd invocation ,
Why are we able to do
sum(a => a * a * a)(1,10)
but not
sumCubes()(1,10)
My understanding is that in sum(a => a * a * a)(1,10) , we are partially applying sum to the anonymous function, which returns a Function2 ,which is applied to the second pair of parameters (1,10) and hence we are getting 3025,
However the same should happen in case of sumCubes()(1,10) , invoking sumCubes without parameters first , would inturn invoke sum with the anonymous function and the Function2 returned would be applied to (1,10)
Why does sumCubes(1,10) work but not sumCubes()(1,10) , shouldn't sumCubes and sumCubes() mean the same thing , invocation of function sumCubes. Also if a simple reference sumCubes is invoking it , how can I pass it around. I feel like I am not understanding something fundamental about Scala.
Scala's methods can have multiple argument lists.
For example, here is a method foo that has ten argument lists, of which the first seven are empty:
def foo()()()()()()()(a: Int)(b: Int)(c: Int): Int = a + b + c
You can invoke it as follows:
println(foo()()()()()()()(1)(20)(300))
and it will print 321.
Note that when you invoke a method, the number of argument lists, as well as the number of arguments in each list (and also their types) have to match the declaration.
Another example. The following method has two argument lists:
def bar(i: Int)(f: Int => Int) = f(i)
you can invoke it as follows:
bar(42)(x => x * x)
but not as
bar()(x => x * x)
bar()(42)(x => x * x)
bar(42)()
or anything like it.
Completely analogously, if you have defined a method with zero argument lists
def baz = (x: Int) => x * x
then you must invoke it with zero argument lists too:
baz
Since it returns an Int => Int function, you can of course apply the result to an Int:
baz(42)
which is the same as (baz)(42), but you cannot do the following:
baz()(42)
because baz itself has no argument lists, and () does not contain a single integer argument.
Note that all of the above is actually a simplification: under certain circumstances, methods that have an empty argument list can be called without parentheses, i.e. def foo(): Unit = ... can be invoked as foo, without (). This is a somewhat strange feature, and I can't say exactly why it's there. My best guess would be: it has something to do with java-interop, where you really want to omit parentheses on zero-ary getters.

Functional Sets / Apply method

I have just started learning scala from the Coursera course by Martin Odesky. I am really struggling at the functional sets problem. The thing I am not able to catch is the union method-
object Functionalset {
type Set = Int => Boolean
def singletonSet(element: Int): Set = i => element == i
def union(a: Set, b: Set): Set = i => a(i) || b(i)
def main(args:Array[String]): Unit = {
print(singletonSet(2)(4));
}
}
In the union method I am essentially returning a function that takes an Int i and calls a(i) || b(i) to check whether i is in the set or not. The apply method is not defined anywhere. What does the a(i) do? Does it call the singletonSet(element:Int):Set? If so how?
What does the a(i) do? Does it call the singletonSet(element: Int)
No, there is no relationship between the two methods. a(i) in this particular case is simply the invocation of the function a, which is a function from Int to Boolean.
a is a of type Set, which is a type alias for a function from Int to Boolean (Int => Boolean with Scala syntax sugar).
Think of it like this:
def union(a: Int => Boolean, b: Int => Boolean): Int => Boolean = i => a(i) || b(i)
Since a is a function of type Function1[Int, Boolean], and so is b, calling a(i) is simply invoking the function with the passed value of i. Now, what may be confusing is the notation of:
i => a(i) || b(i)
Since union itself returns a function, it needs a placeholder for any future Int which may be passed in, this happens when you call singletonSet(2)(4), if we desugarize it a little, you have:
val intermidiateFunction: Int => Boolean = singletonSet(2)
val result: Boolean = intermidiateFunction(4)
Think of it as a two step invocation of a method, when you do the first invocation, you get back another function, and only when you call it the second time, you get back the result of the generated function.
Let's invoke union and see what happens:
scala> union(i => i < 3, i => i > 1)
res6: Set = <function1>
scala> val result = union(i => i < 3, i => i > 1)
result: Set = <function1>
scala> result(1)
res7: Boolean = true
union requires us to pass in two functions, so we create these simplified functions that check whether an Int is smaller than 3 or bigger than 1. The result of the first invocation is itself a function. When we'll now pass in an Int to the result, for example: result(1), we will check 1 < 3 || 1 > 1, which is equivalent to a(i) || b(i), where a is the first function, and b is the second.
In this lab, a set is defined as its characteristic function.
This translates into:
type Set = Int => Boolean
Which means that a Set is simply a function taking an Int and returning a Boolean (this is the characteristic function of the Set, an oracle which tells whether a given integer is in the set or not).
Your variable a, being a Set, is a function Int => Boolean, so a(i) is well-defined if i is an Int.
The characteristic function of a union of two sets is a function which returns true if its argument is in one of the two sets, false otherwise.

Why does not ; equivalent to the eol? [duplicate]

This question already has answers here:
toList on Range with suffix notation causes type mismatch
(2 answers)
Closed 6 years ago.
Here is the code that compiles as intended
def coarse_grained: Int = {
def fib: Int = List(1,2) sum ;
fib
}
and one which does not
def coarse_grained: Int = {
def fib: Int = List(1,2) sum
fib
}
The only difference is ; after the sum.
As you know, List(2,6,9).drop(1) can also be written as List(2,6,9) drop 1. In fact, it can also be written like this.
List(2,6,9) drop
1
The compiler keeps looking for the final argument, even past a newline. So if you want to do this List(1,2).sum like this List(1,2) sum, you'll need to use the semicolon ; to tell the compiler to stop looking for the final argument. It's not coming.

Union between sets

I am taking a Coursera Functional Programming in Scala class. This is the second week and I hit a wall. In the assignment we are working with Sets, but not the kind of Set we all meet in Java, for example. It is a Set that returns true if the value is in there and false otherwise. They say it's not a container, it's just a function.
To get it clear, I need your help. I don't want you to solve my assignment, it's just an example that I want to get the idea of what I should do.
/**
* We represent a set by its characteristic function, i.e.
* its `contains` predicate.
*/
type Set = Int => Boolean
/**
* Indicates whether a set contains a given element.
*/
def contains(s: Set, elem: Int): Boolean = s(elem)
/**
* Returns the set of the one given element.
*/
def singletonSet(elem: Int): Set = Set(elem)
/**
* Returns the union of the two given sets,
* the sets of all elements that are in either `s` or `t`.
*/
def union(s: Set, t: Set): Set = ???
This is the code. In the singletonSet I guess the way to solve it is to return the Set(elem), right?
If that is good, how am I supposed to make the union between the two? I am not new to programming but I can't see any way to do it. Since I shouldn't return a "set" of numbers.
This is what another student told me about sets: "But all a "Set" is is a function that takes an Int and returns a Boolean (Int => Boolean). Any function that takes an Int and returns a Boolean fits the type 'Set'."
What I tried in the union function is to have something like:
def union(s: Set, t: Set): Set = (s | t) //value | not a member of Int => Boolean
Any help would be appreciated :)
It seems the wall you are hitting is that you are unfamiliar with defining functions in Scala. In this particular case you need to define functions of type Int => Boolean, they take an Int and return a Boolean.
Here are some examples of function literals of type Int => Boolean. Try them in the Scala console or the Scala IDE worksheet:
(x: Int) => true
(x: Int) => false
(x: Int) => x == 2
(x: Int) => x == 10
(x: Int) => x == 2 || x == 10
(x: Int) => x % 2 == 0
Then all you have to do for the assignment is to use the same syntax, starting with (x: Int) => and then translate the meaning of union, intersect, ... into the right hand side of the expression.
Part of learning is giving it a genuine effort. I believe you can resubmit the solution multiple times, so don't hesitate to submit and iterate if you don't get 10/10 on the first try. All you need is compiling code. Good luck!
A possible hint is to look at the types. Look at the Set type. It is actually a type alias to a function from Int into Boolean.
Thus, when you have two sets, you actually have two functions. How can you use them to provide a function that represent the union of these Set? It must be your starting point.