CODEWITHSUNDEEP
scalatype-conversionscala-implicits
Buck Shlegeris
Buck Shlegeris

Reputation: 517

Scala not letting me add integers

Here's some code which looks reasonable enough to me:

val myMap: Map[Int, Int] = ((x: Int) => Map[Int, Int](1 -> x + 1, 2 -> x + 2))(4)

When I try to compile it, I get two errors like this:

Error:(20, 68) type mismatch;
 found   : Int(1)
 required: String
    val myMap: Map[Int, Int] = ((x: Int) => Map[Int, Int](1 -> x + 1, 2 -> x + 2))(4)
                                                                   ^

I understand that my compiler is trying to use the string addition implementation of +. But why is it doing that? How do I ask the compiler to use integer addition here?

(Changing Int to Integer doesn't help.)

Upvotes: 0

Views: 72

Answers (3)

Bakkot
Bakkot

Reputation: 1245

'->' has the same precedence as '+', so 1 -> x + 1 parses as (1 -> x) + 1. Adding parentheses as 1 -> (x + 1) fixes this error.

Upvotes: 1

Buck Shlegeris
Buck Shlegeris

Reputation: 517

The problem is that according to Scala's precedence rules, -> binds as tightly as +, because it starts with -.

It's fixed by adding parentheses: 

val myMap: Map[Int, Int] = ((x: Int) => Map[Int, Int](1 -> (x + 1), 2 -> (x + 2)))(4)

Upvotes: 0

Michael Zajac
Michael Zajac

Reputation: 55569

Here is a simpler example that reproduces the error:

scala> def fail(x: Int) = 1 -> x + 1
<console>:10: error: type mismatch;
 found   : Int(1)
 required: String
       def fail(x: Int) = 1 -> x + 1
                                   ^

All of the operators in play are left-associative, so everything reads left-to-right. That is, the call chain looks like this:

1.->(x).+(1)

or using infix notation:

(1 -> x) + 1

1 -> x returns a tuple (1, x) and a tuple does not have a + method, so the compiler attempts to implicitly convert it to a String via toString, because String does have a + method, which eventually fails because Int is required.

Just use parentheses to group things appropriately:

scala> def foo(x: Int) = 1 -> (x + 1)
foo: (x: Int)(Int, Int)

Upvotes: 5

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