Compiler fails when applying default argument with the same name as variable being bound #6867
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Imported From: https://issues.scala-lang.org/browse/SI-6867?orig=1 |
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Rob Norris (rnorris) said: |
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@adriaanm said: scala> case class Foo[A,B](f: (A, B) => A)
defined class Foo
scala> case class Person(name:String)
defined class Person
scala> val name = Foo[Person,String]((a, b) => a.copy(name = b))
<console>:11: error: not found: value b
val name = Foo[Person,String]((a, b) => a.copy(name = b))
^
<console>:11: warning: type-checking the invocation of method copy checks if the named argument expression 'name = ...' is a valid assignment
in the current scope. The resulting type inference error (see above) can be fixed by providing an explicit type in the local definition for name.
val name = Foo[Person,String]((a, b) => a.copy(name = b))
^
scala> val wibble = Foo[Person,String]((a, b) => a.copy(name = b))
wibble: Foo[Person,String] = Foo(<function2>)
scala>
scala> val wibble = Foo[Person,String]((a, b) => a.copy(name = b))
wibble: Foo[Person,String] = Foo(<function2>) |
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In some cases the compiler becomes confused if you are binding a variable whose name is the same as a default argument in the expression being bound, as in (1) below.
// Setup
case class Foo[A,B](f: (A, B) => A)
case class Person(name:String)
// (1) Doesn't compile:
// REPL says value 'b' not found.
// Scala IDE says "error applying default arguments"
val name = Foo[Person,String]((a, b) => a.copy(name = b))
// (2) Ok -- val is not called 'name'
val wibble = Foo[Person,String]((a, b) => a.copy(name = b))
// (3) Ok -- type ascription on LHS
val name:Foo[Person,String] = Foo((a, b) => a.copy(name = b))
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