Scala string pattern matching for mathematical symbols - scala

Question

I have the following code:
val z: String = tree.symbol.toString
z match {
case "method +" | "method -" | "method *" | "method ==" =>
println("no special op")
false
case "method /" | "method %" =>
println("we have the special div operation")
true
case _ =>
false
}
Is it possible to create a match for the primitive operations in Scala:
"method *".matches("(method) (+-*==)")
I know that the (+-*) signs are used as quantifiers. Is there a way to match them anyway?
Thanks from a avidly Scala scholar!

Answer 1

Sure.
val z: String = tree.symbol.toString
val noSpecialOp = "method (?:[-+*]|==)".r
val divOp = "method [/%]".r
z match {
case noSpecialOp() =>
println("no special op")
false
case divOp() =>
println("we have the special div operation")
true
case _ =>
false
}
Things to consider:
I choose to match against single characters using [abc] instead of (?:a|b|c).
Note that - has to be the first character when using [], or it will be interpreted as a range. Likewise, ^ cannot be the first character inside [], or it will be interpreted as negation.
I'm using (?:...) instead of (...) because I don't want to extract the contents. If I did want to extract the contents -- so I'd know what was the operator, for instance, then I'd use (...). However, I'd also have to change the matching to receive the extracted content, or it would fail the match.
It is important not to forget () on the matches -- like divOp(). If you forget them, a simple assignment is made (and Scala will complain about unreachable code).
And, as I said, if you are extracting something, then you need something inside those parenthesis. For instance, "method ([%/])".r would match divOp(op), but not divOp().

Answer 2

Much the same as in Java. To escape a character in a regular expression, you prefix the character with \. However, backslash is also the escape character in standard Java/Scala strings, so to pass it through to the regular expression processing you must again prefix it with a backslash. You end up with something like:
scala> "+".matches("\\+")
res1 : Boolean = true
As James Iry points out in the comment below, Scala also has support for 'raw strings', enclosed in three quotation marks: """Raw string in which I don't need to escape things like \!""" This allows you to avoid the second level of escaping, that imposed by Java/Scala strings. Note that you still need to escape any characters that are treated as special by the regular expression parser:
scala> "+".matches("""\+""")
res1 : Boolean = true

Answer 3

Escaping characters in Strings works like in Java.
If you have larger Strings which need a lot of escaping, consider Scala's """.
E. g. """String without needing to escape anything \n \d"""
If you put three """ around your regular expression you don't need to escape anything anymore.