I have a definition of next methods:
def add1(x: Int, y: Int) = x + y
def add2(x: Int)(y: Int) = x + y
the second one is curried version of first one. Then if I want to partially apply second function I have to write val res2 = add2(2) _. Everything is fine. Next I want add1 function to be curried. I write
val curriedAdd = (add1 _).curried
Am I right that curriedAdd is similiar to add2?
But when I try to partially apply curriedAdd in a such way val resCurried = curriedAdd(4) _ I get a compilation error. Then I fix it to
val resCurried = curriedAdd(4)
Why the result of a Functions.curried differs from curried version of add function(from add2)?
Firstly curriedAdd is same as add2 _ and not add2. add2 is just a method.
scala> curriedAdd
res52: Int => (Int => Int) = <function1>
scala> add2 _
res53: Int => (Int => Int) = <function1>
About the second question. I think the below is the reason. Doing
scala> val i = curriedAdd(23)
i: Int => Int = <function1>
scala> i _
res54: () => Int => Int = <function0>
scala> curriedAdd(23) _
<console>:10: error: _ must follow method; cannot follow Int => Int
curriedAdd(23) _
curriedAdd(23) _ does not work. Lets look at scala manual (§6.7)-
The expression e _ is well-formed if e is of method type or if e is a
call-by-name parameter. If e is a method with parameters, e _
represents e converted to a function type by eta expansion (§6.26.5).
If e is a parameterless method or call-by-name parameter of type =>T ,
e _ represents the function of type () => T , which evaluates e when
it is applied to the empty parameterlist ().
Remember it only evaluates if it is a method or call-by-name parameter. In curriedAdd(23) _, it does not evaluate curriedAdd(23) but checks if it is a method or call-by-name. It is not a method nor a call-by-name parameter.
It is not by-name because by-name is the property of variable. Above you get a by-name parameter after evaluating curriedAdd(23) but curriedAdd(23) in itself is not a by-name variable. Hence the error (Ideally the compiler should have coverted it). Note that the below works:
scala> curriedAdd(23)
res80: Int => Int = <function1>
scala> res80 _
res81: () => Int => Int = <function0>
The above works because res80 _, here you are applying _ to a call-by-name parameter and hence does the conversion.
To answer this question, let's take a look at the REPL.
First we define the two functions as you did.
scala> def add1(x: Int, y: Int) = x + y
add1: (x: Int, y: Int)Int
scala> def add2(x: Int)(y: Int) = x + y
add2: (x: Int)(y: Int)Int
We have defined two functions. The first one expects two parameters in one parameterlist. The second one expects two parameters, each one in an own parameterlist. The result type is the same.
Let's move on.
scala> val curriedAdd = (add1 _).curried
curriedAdd: Int => (Int => Int) = <function1>
You just created a partial applied function, that expects one parameter and returns a partial applied function of type Int => Int. This is not as similar to add2 as you expect it to be.
To achieve the same for add2, you would need to call
scala> val curriedAdd2 = add2 _
curriedAdd2: Int => (Int => Int) = <function1>
Related
Given
scala> def method(x: Int) = x
method: (x: Int)Int
scala> val func = (x: Int) => x
func: Int => Int = <function1>
Consider the following code:
scala> method _
res0: Int => Int = <function1>
scala> func(_)
res1: Int => Int = <function1>
scala> func _
res2: () => Int => Int = <function0>
I can understand that res0 is eta expansion and res1 is equivalent to lambda function (x) => func(x). But I cannot figure out the output of res2. Could anyone please explain that for me?
This is actually a bit tricky. First, let's see what happens outside REPL:
It doesn't work when func is a local variable:
object Main extends App {
def foo() = {
val f = (_: Int) + 1
f _
}
println(foo())
}
[error] /tmp/rendereraEZGWf9f1Q/src/main/scala/test.scala:8: _ must follow method; cannot follow Int => Int
[error] f _
[error] ^
But if you put it outside def foo, it compiles:
object Main extends App {
val f = (_: Int) + 1
val f1 = f _
println(f1)
}
because f is both a field of Main and a method without arguments which returns the value of this field.
The final piece is that REPL wraps each line into an object (because Scala doesn't allow code to appear outside a trait/class/object), so
scala> val func = (x: Int) => x
func: Int => Int = <function1>
is really something like
object Line1 {
val func = (x: Int) => x
}
import Line1._
// print result
So func on the next line refers to Line1.func which is a method and so can be eta-expanded.
You've used the eta Expansion to turn res2 into a function that takes 0 parameters and returns a function that takes a single parameter.
res2: () => Int => Int = <function0>
So you can now do this:
val zero = func _
val f: Int => Int = zero()
To answer your question about res2 - the appended underscore _ syntax is used to partially apply a function.
So.
scala> func _
Means you have partially applied your <function1>.
This results in a new curried form, the first function of which takes zero arguments (hence <function0>)
() =>
Which returns your original <function1> which takes 1 argument.
Int => Int = <function1>
The complete result being the chain of functions.
res2: () => Int => Int = <function0>
The rule that might be useful for you to remember is that functions associate to the right so the following are equivalent.
() => Int => Int
() => (Int => Int)
This other post might be useful reading for you.
val func1 = func _
This returns a function0 which takes no arguments and returns the func function.
You can use this like:
func1()(2) // outputs 2
You can continue doing this kind of expansion ad infinitum:
val func2 = func1 _
func2()()(2) // outputs 2
When I create a partial function, why can't I immediately invoke it?
Both res6 and res8 are the same type (function1) so I'm not sure how come res7 works (immediately invoking it) and what would be res9 fails
scala> ((x: Int) => x + 1)
res6: Int => Int = <function1>
scala> ((x: Int) => x + 1)(1)
res7: Int = 2
scala> def adder(a: Int, b: Int) = a + b
adder: (a: Int, b: Int)Int
scala> adder(1, _: Int)
res8: Int => Int = <function1>
scala> adder(1, _: Int)(1)
<console>:12: error: Int does not take parameters
adder(1, _: Int)(1)
^
scala> (adder(1, _: Int))(1)
res10: Int = 2
I think you've just found one of the little Scala compiler quirks.
I don't know how this case is implemented exactly in the parser, but from the looks of it Scala thinks you are invoking a function with multiple parameter lists (of the form f(...)(...)). Thus you need to explicitly surround the partially applied function with brackets here, so that compiler could disambiguate between f()() and f(_)() forms. In res8 you don't have any parameters following the function, so there is no ambiguity. Otherwise, Scala would require you to follow the function with underscore if you ommit parameter list: f _ or f() _.
You can still immediately apply that function as you show in res10.
adder(1, _: Int)
This is caused by scala compiler will expand it to:
((x: Int) => adder(1, x)(1))
and this is caused by the compiler can't infer the wildcard's context, like:
_ + _
scala> _ + _
<console>:17: error: missing parameter type for expanded function ((x$1, x$2) => x$1.$plus(x$2))
_ + _
^
<console>:17: error: missing parameter type for expanded function ((x$1: <error>, x$2) => x$1.$plus(x$2))
_ + _
so you can do it with context bounds, as your way:
(adder(1, _: Int))(1)
it will be expanded like:
((x: Int) => adder(1, x))(1)
I am trying to declare a simple method in scala with multiple parameter lists.
These two work.
scala> def add(a:Int, b:Int) = a+ b
add: (a: Int, b: Int)Int
scala> def add(a:Int)(b:Int) = a + b
add: (a: Int)(b: Int)Int
This does not...
scala> val add = (a:Int)(b:Int)=>a + b
<console>:1: error: not a legal formal parameter.
Note: Tuples cannot be directly destructured in method or function parameters.
Either create a single parameter accepting the Tuple1,
or consider a pattern matching anonymous function: `{ case (param1, param1) => ... }
val add = (a:Int)(b:Int)=>a + b
But why ... all I am trying to do is to assign a anonymous function which takes multiple parameter lists to a value. that works with a single parameter list but not with multiple parameter lists.
It's just a matter of syntax when declaring the curried arguments.
scala> val add = (a:Int) => (b:Int) => a + b
add: Int => (Int => Int) = <function1>
scala> add(4)
res5: Int => Int = <function1>
scala> res5(9)
res6: Int = 13
Example of anonymous usage:
scala> def doit(f: Int => Int => String): String = f(2)(5)
doit: (f: Int => (Int => String))String
scala> doit(a => b => (a+b).toString)
res8: String = 7
Why I can't invoke g2 with the second parameter in the line g2(1, _: Int)(9)?
g2(1, _: Int) returns a function <function1> and then I invoke that <function1> with parameter. At the same time in the example above I can invoke f3(9) however it is the same as g2(1, _: Int)(9) IMHO.
// OK
def f1: (Int, Int) => Int = _ + _
val f2: (Int, Int) => Int = f1
val f3: (Int) => Int = f2(1, _: Int)
println {
f3(9)
}
// FAIL
def g1: (Int, Int) => Int = _ + _
val g2: (Int, Int) => Int = g1
println {
g2(1, _: Int)(9) // Error: 'g2' does not take parameters
}
It's the issue with the scope of underscore in anonymous functions. See here for the rules: What are the rules to govern underscore to define anonymous function?
According to the rule 1 from the link above g2(1, _: Int)(9) is interpreted as (x: Int) => g2(1, x)(9). So you need to add parentheses, for the rule 2 to start working: (g2(1, _: Int))(9)
As you add in a comment, (g2(1, _: Int))(9) works.
Scala has multiple parameter lists, so f(x)(y) parses differently from (f(x))(y). As g takes only one parameter list, it can't take the second parameter list, hence the explicit grouping is necessary.
g2(1, _: Int)(9) means: take the function g2(x, y)(z), and apply parameters x = 1, z = 9.
(g2(1, _: Int))(9) means: take the function g2(x, y), and apply parameter x = 1. Then apply y = 9 to the resulting function.
Scala allows functions with no parameter lists to be invoked without parentheses:
scala> def theAnswer() = 42
theAnswer: ()Int
scala> theAnswer
res5: Int = 42
How would I construct a scala expression that evaluates to the function theAnswer itself, rather than the result of theAnswer? Or in other words, how would I modify the expression theAnswer, so that the result was of type () => Int, and not type Int?
It can be done with:
scala> theAnswer _
res0: () => Int = <function0>
From the answer to the similar question:
The rule is actually simple: you have to write the _ whenever the
compiler is not explicitly expecting a Function object.
This call will create the new instance every time, because you're "converting" method to function (what is called "ETA expansion").
Simply:
scala> val f = () => theAnswer
f: () => Int = <function0>
scala> val g = theAnswer _
g: () => Int = <function0>