Scala Parser Combinators :Extract List of a type

Question

I am writing parser and grammer using scala.util.parsing.combinator . My input is ":za >= 1 alok && :ft == 9"

case class Expression(op1:Operand,operator:Operator,op2:Operand)

def word: Parser[String] = """[a-z\\\s]+""".r ^^ { _.toString }

def colonWord:Parser[Operand]=":"~> word ^^{case variable => Operand(variable)}

def constant:Parser[Operand]="""^[a-zA-Z0-9\\\s-]*""".r ^^ {case constant => Operand(constant)}

def expression:Parser[Expression] = colonWord ~ operator ~ constant  ^^{ case op1~operator~op2 => Expression(op1, operator, op2)}

def expressions = expression ~ opt(" && " ~> expression)*

but when I parse sample String , the result is not expected. The second expression after && is not parsed. Please note there can be multiple expression joined using &&.

When i execute:

val expr= ":za >= 1 alok && :ft == 9"
    parse(expressions, expr) match {
      case Success(matched, _) => println(matched)
      case ..}

Output is :

List((Expression(za ,>= ,1 alok )~None))

I dont see the second expression being parsed. Can anyone help what have i missed here?

EDIT -----------------------------------

The requirement is to get List[Expression]. when I say, incorporting the changes mentioned in Ans :

def expressions = expression ~ ("&&" ~> expression)*

The return type of expressions is not List[Expression]. For eg: If I write another def :

case class Condition(exprs: List[Expression], st:Statement)
def condition = expressions ~","~statement ^^{
    case exprs~commaa~statement => Condition(exprs,statement) //This is giving error.

the error is: type mismatch; found : ~[Expression,Expression]] required: Expressions.

So how do i convert [Expression, Expression] to List[Expressions]?

Thanks

In your output List((Expression(za ,>= ,1 alok )~None)). Try to notice that extra space after 1 alok. And extra space is the culprit, as your next expressions parser expects two expressions to be delimited by a <space>&&<space> — sarveshseri
– sarveshseri, Commented Sep 19, 2016 at 8:46
That is why, it is a good practice to have <space> as a first class member of your grammar. Don't ever belittle space. — sarveshseri
– sarveshseri, Commented Sep 19, 2016 at 8:49
Thanks. Yes , that is the problem . I tried removing spaces from <space>&&<space>. It worked in this case. But i gave another expression => ":za >= 1&&:ft == 9&&:wq == 12" , In this case it took 2 expression , rather than 3 . The output i get is List((Expression(za ,>= ,1)~Some(Expression(ft ,== ,9)))) — Alok
– Alok, Commented Sep 19, 2016 at 10:02

Jatin · Accepted Answer · 2016-09-20 08:44:21Z

2

The correction needed was:

expression ~ opt("&&" ~> expression)*

Remove space across && and it should work. This is because you are already covering space in your constant parser.

Edit: Based on edited question, is this what you want:

 def expressions = expression ~ (("&&" ~> expression)*) ^^{
    case x ~ y => x :: y
  }

Now the return type of expressions is List[Expression]. Your condition will now compile

edited Sep 20, 2016 at 8:44

answered Sep 19, 2016 at 8:44

Jatin

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4 Comments

Alok Over a year ago

Thanks. Yes , that is the problem . I tried removing spaces from <space>&&<space>. It worked in this case. But i gave another expression => ":za >= 1&&:ft == 9&&:wq == 12" , In this case it took 2 expression , rather than 3 . The output i get is List((Expression(za ,>= ,1)~Some(Expression(ft ,== ,9))))

Alok Over a year ago

I believe the problem is also with def expressions = expression ~ opt(" && " ~> expression)* . I was expecting this to return me List[Expression] . But it returns [Expression, Option[Expression]]

sarveshseri Over a year ago

because you have opt in opt(" && " ~> expression). Now since you are using * which means 0 or more you do not need that opt.

Alok Over a year ago

But this would not solve the problem . To make the problem more clearer , I have edited it.

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