Question

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

Answer 1

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.