Scala parser combinator all but last

Question

What is the correct way to parse a list consuming all but the last element using scala parser-combinators? For example, if I want to parse "spam spam spam spam spam and eggs" into List("spam", "spam", "spam", "spam") ~ "spam and eggs".

If I just use "spam".+, then all the spam is consumed before I get to match "spam and eggs". Naturally, I could match on "and eggs" but in practice I might already have a complicated "spam and eggs" parser that I'd rather not alter.

It isn't a complete answer but ("spam and eggs" | "spam").+ would work for this case, if accepting "spam and eggs spam" etc. is acceptable. If not and nobody does earlier, I'll try to answer tomorrow. — Alexey Romanov
– Alexey Romanov, Commented Feb 14, 2020 at 21:03
The best I came up with was def allButOne[T](p:Parser[T]):Parser[List[T]] = ((p ~ guard(p) ~ allButOne(p)) ^^ { case a ~ b ~ c => List(a) ++ c })|(p <~ guard(p) ^^ { a => List(a) }) but wondered if there was a more natural pattern. — James Riordan
– James Riordan, Commented Feb 14, 2020 at 21:18

Alexey Romanov · Accepted Answer · 2020-02-14 21:27:18Z

3

Here is a simple but inefficient version:

def allButOne[T](p: Parser[T]): Parser[List[T]] = (p <~ guard(p)).+

answered Feb 14, 2020 at 21:27

Alexey Romanov

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1 Comment

James Riordan Over a year ago

That's far more elegant. Thanks.

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