CODEWITHSUNDEEP
scala
Rogach
Rogach

Reputation: 27200

Scala method and values names

Why this fails to compile:

scala> val a? = true
<console>:1: error: illegal start of simple pattern
   val a? = true
          ^

and this works?

scala>  val a_? = true
a_?: Boolean = true

Upvotes: 6

Views: 1921

Answers (2)

S. Kucherenko
S. Kucherenko

Reputation: 585

Scala's grammar for identifiers is defined in such a way. ? is defined to be an operator character. And an identifier must obey the following rules: it must be a lower-case letter which may be followed by an element of an 'idrest' syntactic category, which is defined as 'letters or digits, possibly followed by _ and an op char.' See Scala Language Specification for more details.

Upvotes: 2

S.R.I
S.R.I

Reputation: 1870

According to the Scala language specification (looking at 2.8, doubt things have changed much since):

idrest ::= {letter | digit} [`_' op]

That is, an identifier can start with a letter or a digit followed by an underscore character, and further operator characters. That makes identifiers such as foo_!@! valid identifiers. Also, note that identifiers may also contain a string of operator characters alone. Consider the following REPL session:

Welcome to Scala version 2.9.1.final (Java HotSpot(TM) Client VM, Java 1.6.0_16).

scala> val +aff = true
<console>:1: error: illegal start of simple pattern
val +aff = true
^

scala> val ??? = true
???: Boolean = true

scala> val foo_!@! = true
foo_!@!: Boolean = true

scala> val %^@%@ = true
%^@%@: Boolean = true

scala> val ^&*!%@ = 42
^&*!%@: Int = 42

Hope this answers your question.

Upvotes: 6

Related Questions