What is the following methods' difference?
def sum1() = 1+2
def sum2(a:Unit) = 1+2
I think they are semantically identical, is it right?
With sum1, you can call it with or without parentheses:
val x = sum1 // x: Int = 3
val y = sum1() // y: Int = 3
But with sum2 you are forced to provide parentheses.. I think that if you call sum2(), you are actually calling sum2 with () as the argument a.
val x2 = sum2 // error
val y2 = sum2() // y2: Int = 3
Note that passing unit as an argument to an expression lets you simulate lazy evaluation in a strict language. By "moving evaluation under a lambda" you ensure that the expression isn't eval'd until the () gets passed in. This can be useful for e.g. auto-memoizing data structures, which collapse from a function to a value the first time they're inspected.
These methods are not identical. Once receives a parameter, the other does not. See here:
scala> sum1(println("Hi, there!"))
<console>:9: error: too many arguments for method sum1: ()Int
sum1(println("Hi, there!"))
^
scala> sum2(println("Hi, there!"))
Hi, there!
res1: Int = 3
Related
I’m learning scala and read this concept called ‘currying’. I know it is used to divide the parameters in a function and pass them one by one. So from what I understood I tried creating a simple basic currying function below:
def add_num(a:Int,b:Int)(c:Int):Int={a+b+c}
add_num(10,20)
add_num(10)
But it gives me an error when I call the function with values 10 and 20. What i understood is it will remember the values and then I can pass the third value as 10 separately. I’m unable to understand this concept. Can someone help me in understanding this concept in most basic terms.
Error is missing argument list for method add_num. Unsupplied methods are only converted to functions when a function type is executed.
A curried function is applied to multiple argument lists, instead of just
one. An example of a non-curried function, which adds two Int parameters, x and y:
scala> def plainOldSum(x: Int, y: Int) = x + y
plainOldSum: (x: Int, y: Int)Int
scala> plainOldSum(1, 2)
res4: Int = 3
A similar function that’s curried:
scala> def curriedSum(x: Int)(y: Int) = x + y
curriedSum: (x: Int)(y: Int)Int
scala> curriedSum(1)(2)
res5: Int = 3
What's happening here is that when you invoke curriedSum, you actually
get two traditional function invocations back-to-back. The first function
invocation takes a single Int parameter named x, and returns a function
value for the second function. This second function takes the Int parameter
y.
Here's a function named first that does in spirit what the first traditional
function invocation of curriedSum would do:
scala> def first(x: Int) = (y: Int) => x + y
first: (x: Int)Int => Int
Applying the first function to 1—in other words, invoking the first function
and passing in 1—yields the second function:
scala> val second = first(1)
second: Int => Int = <function1>
Applying the second function to 2 yields the result:
scala> second(2)
res6: Int = 3
These first and second functions are just an illustration of the currying
process. They are not directly connected to the curriedSum function. Nevertheless,
there is a way to get an actual reference to curriedSum's "second"
function. You can use the placeholder notation to use curriedSum in a partially
applied function expression, like this:
scala> val onePlus = curriedSum(1)_
onePlus: Int => Int = <function1>
The underscore in curriedSum(1)_ is a placeholder for the second parameter
list.2 The result is a reference to a function which, when invoked, adds one
to its sole Int argument and returns the result:
scala> onePlus(2)
res7: Int = 3
And here's how you'd get a function that adds two to its sole Int argument:
scala> val twoPlus = curriedSum(2)_
twoPlus: Int => Int = <function1>
scala> twoPlus(2)
res8: Int = 4
I am writing a DSL in Scala where I'd like to achieve a chain of method calls as follows:
def x(i:Int) = i
x 1 equals 1 //doesn't compile
I am not sure why the compiler is happy if I leave out the second parentheses but not the first one:
x(1) equals 1 //works fine
Is there a way to achieve the first version?
You can invoke methods without parenthesis, but not functions.
So the following works:
scala> object Foo {
| def x(i:Int) = i
| }
defined object Foo
scala> Foo x 1
res9: Int = 1
scala> Foo x 1 equals 1
res10: Boolean = true
Take this code:
var x = 10
val y = () => x + 1
I then want to treat y as if its a variable that holds an Int and changes anytime x changes. Essentially I want to bind y to x.
Problem is, if I just type y then I get the result res0: () => Int = <function0>
I was under the impression that you could invoke 0-arity functions without any parens, but I am required to use y() to get the behavior I am trying to achieve.
Is there a better way to define the anonymous function or do I need to use a different approach?
Why do you need to do it like that? Just do def y = x + 1
scala> var x = 1
x: Int = 1
scala> def y = x + 1
y: Int
scala> y
res0: Int = 2
scala> x = 3
x: Int = 3
scala> y
res1: Int = 4
EDIT to address some comments:
If you define val y = () => x + 1, you are defining y as a value that holds a function that takes no argument and returns an Int. To call the function that is held by this variable, you will need to call it with (), to tell the compiler that you don't want to pass the value, but to execute (evaluate) the function that is within it.
If you define val y = x + 1, you are defining a value (constant), that is assigned in the moment it is executed, you could postpone evaluation using lazy, but once it is assigned, it will not be evaluated again.
If you define def y = x + 1, you are defining a function that returns x+1, which is what you want.
You have not defined a 0-arity function. You have actually defined a Unit => Int function. That is why you can not invoke it like you would like to invoke it. I've rarely seen 0-arity functions outside the context of come contained function scope:
def something[V](value: Int)(f: => V) =
if(value %2 == 0) f
else throw new Exception("not even (and also not evaluated f)")
where it is used as a lazily deferred execution body.
Edit: I would use the other person's answer.
I'm having trouble understanding underscores in function literals.
val l = List(1,2,3,4,5)
l.filter(_ > 0)
works fine
l.filter({_ > 0})
works fine
l.filter({val x=1; 1+_+3 > 0}) // ie you can have multiple statements in your function literal and use the underscore not just in the first statement.
works fine
And yet:
l.filter({val x=_; x > 0})
e>:1: error: unbound placeholder parameter
l.filter({val x=_; x > 0})
I can't assign the _ to a variable, even though the following is legal function literal:
l.filter(y => {val x=y; x > 0})
works fine.
What gives? Is my 'val x=_' getting interpreted as something else? Thanks!
Actually, you have to back up a step.
You are misunderstanding how the braces work.
scala> val is = (1 to 5).toList
is: List[Int] = List(1, 2, 3, 4, 5)
scala> is map ({ println("hi") ; 2 * _ })
hi
res2: List[Int] = List(2, 4, 6, 8, 10)
If the println were part of the function passed to map, you'd see more greetings.
scala> is map (i => { println("hi") ; 2 * i })
hi
hi
hi
hi
hi
res3: List[Int] = List(2, 4, 6, 8, 10)
Your extra braces are a block, which is some statements followed by a result expression. The result expr is the function.
Once you realize that only the result expr has an expected type that is the function expected by map, you wouldn't think to use underscore in the preceding statements, since a bare underscore needs the expected type to nail down what the underscore means.
That's the type system telling you that your underscore isn't in the right place.
Appendix: in comments you ask:
how can I use the underscore syntax to bind the parameter of a
function literal to a variable
Is this a "dumb" question, pardon the expression?
The underscore is so you don't have to name the parameter, then you say you want to name it.
One use case might be: there are few incoming parameters, but I'm interested in naming only one of them.
scala> (0 /: is)(_ + _)
res10: Int = 15
scala> (0 /: is) { case (acc, i) => acc + 2 * i }
res11: Int = 30
This doesn't work, but one may wonder why. That is, we know what the fold expects, we want to apply something with an arg. Which arg? Whatever is left over after the partially applied partial function.
scala> (0 /: is) (({ case (_, i) => _ + 2 * i })(_))
or
scala> (0 /: is) (({ case (_, i) => val d = 2 * i; _ + 2 * d })(_))
SLS 6.23 "placeholder syntax for anonymous functions" mentions the "expr" boundary for when you must know what the underscore represents -- it's not a scope per se. If you supply type ascriptions for the underscores, it will still complain about the expected type, presumably because type inference goes left to right.
The underscore syntax is mainly user for the following replacement:
coll.filter(x => { x % 2 == 0});
coll.filter(_ % 2 == 0);
This can only replace a single parameter. This is the placeholder syntax.
Simple syntactic sugar for a lambda.
In the breaking case you are attempting null initialization/defaulting.
For primitive types with init conventions:
var x: Int = _; // x will be 0
The general case:
var y: List[String] = _; // y is null
var z: Any = _; // z = null;
To get pedantic, it works because null is a ref to the only instance of scala.Null, a sub-type of any type, which will always satisfy the type bound because of covariance. Look HERE.
A very common usage scenario, in ScalaTest:
class myTest extends FeatureTest with GivenWhenThen with BeforeAndAfter {
var x: OAuthToken = _;
before {
x = someFunctionThatReturnsAToken;
}
}
You can also see why you shouldn't use it with val, since the whole point is to update the value after initialization.
The compiler won't even let you, failing with: error: unbound placeholder parameter.
This is your exact case, the compiler thinks you are defaulting, a behaviour undefined for vals.
Various constraints, such as timing or scope make this useful.
This is different from lazy, where you predefine the expression that will be evaluated when needed.
For more usages of _ in Scala, look HERE.
Because in this two cases underscore (_) means two different things. In case of a function it's a syntactic sugar for lambda function, your l.filter(_ > 0) later desugares into l.filter(x => x > 0). But in case of a var it has another meaning, not a lambda function, but a default value and this behavior is defined only for var's:
class Test {
var num: Int = _
}
Here num gonna be initialized to its default value determined by its type Int. You can't do this with val cause vals are final and if in case of vars you can later assign them some different values, with vals this has no point.
Update
Consider this example:
l filter {
val x = // compute something
val z = _
x == z
}
According to your idea, z should be bound to the first argument, but how scala should understand this, or you you have more code in this computation and then underscore.
Update 2
There is a grate option in scala repl: scala -Xprint:type. If you turn it on and print your code in (l.filter({val x=1; 1+_+3 > 0})), this what you'll see:
private[this] val res1: List[Int] = l.filter({
val x: Int = 1;
((x$1: Int) => 1.+(x$1).+(3).>(0))
});
1+_+3 > 0 desugares into a function: ((x$1: Int) => 1.+(x$1).+(3).>(0)), what filter actually expects from you, a function from Int to Boolean. The following also works:
l.filter({val x=1; val f = 1+(_: Int)+3 > 0; f})
cause f here is a partially applied function from Int to Boolean, but underscore isn't assigned to the first argument, it's desugares to the closes scope:
private[this] val res3: List[Int] = l.filter({
val x: Int = 1;
val f: Int => Boolean = ((x$1: Int) => 1.+((x$1: Int)).+(3).>(0));
f
});
Why can't i define a variable recursively in a code block?
scala> {
| val test: Stream[Int] = 1 #:: test
| }
<console>:9: error: forward reference extends over definition of value test
val test: Stream[Int] = 1 #:: test
^
scala> val test: Stream[Int] = 1 #:: test
test: Stream[Int] = Stream(1, ?)
lazy keyword solves this problem, but i can't understand why it works without a code block but throws a compilation error in a code block.
Note that in the REPL
scala> val something = "a value"
is evaluated more or less as follows:
object REPL$1 {
val something = "a value"
}
import REPL$1._
So, any val(or def, etc) is a member of an internal REPL helper object.
Now the point is that classes (and objects) allow forward references on their members:
object ForwardTest {
def x = y // val x would also compile but with a more confusing result
val y = 2
}
ForwardTest.x == 2
This is not true for vals inside a block. In a block everything must be defined in linear order. Thus vals are no members anymore but plain variables (or values, resp.). The following does not compile either:
def plainMethod = { // could as well be a simple block
def x = y
val y = 2
x
}
<console>: error: forward reference extends over definition of value y
def x = y
^
It is not recursion which makes the difference. The difference is that classes and objects allow forward references, whereas blocks do not.
I'll add that when you write:
object O {
val x = y
val y = 0
}
You are actually writing this:
object O {
val x = this.y
val y = 0
}
That little this is what is missing when you declare this stuff inside a definition.
The reason for this behavior depends on different val initialization times. If you type val x = 5 directly to the REPL, x becomes a member of an object, which values can be initialized with a default value (null, 0, 0.0, false). In contrast, values in a block can not initialized by default values.
This tends to different behavior:
scala> class X { val x = y+1; val y = 10 }
defined class X
scala> (new X).x
res17: Int = 1
scala> { val x = y+1; val y = 10; x } // compiles only with 2.9.0
res20: Int = 11
In Scala 2.10 the last example does not compile anymore. In 2.9.0 the values are reordered by the compiler to get it to compile. There is a bug report which describes the different initialization times.
I'd like to add that a Scala Worksheet in the Eclipse-based Scala-IDE (v4.0.0) does not behave like the REPL as one might expect (e.g. https://github.com/scala-ide/scala-worksheet/wiki/Getting-Started says "Worksheets are like a REPL session on steroids") in this respect, but rather like the definition of one long method: That is, forward referencing val definitions (including recursive val definitions) in a worksheet must be made members of some object or class.