scala.util.regexp

WordExp

class WordExp extends Base

The class WordExp provides regular word expressions. Users have to instantiate type member _regexpT <: RegExp (from class Base) and a type member _labelT <: Label. Here is a short example: 

 import scala.util.regexp._
 import scala.util.automata._
 object MyLang extends WordExp {
   type _regexpT = RegExp
   type _labelT = MyChar
   case class MyChar(c:Char) extends Label
 }
 import MyLang._
 // (a* | b)*
 val rex = Star(Alt(Star(Letter(MyChar('a'))),Letter(MyChar('b'))))
 object MyBerriSethi extends WordBerrySethi {
   override val lang = MyLang
 }
 val nfa = MyBerriSethi.automatonFrom(Sequ(rex), 1)

attributes: abstract
known subclasses: ContentModel

Inherits

  1. Base
  2. AnyRef
  3. Any

Type Members

  1. class Alt extends RegExp

  2. class Label extends AnyRef

  3. class Letter(a: _labelT) extends RegExp with Product

  4. class Meta extends RegExp

    this class can be used to add meta information to regexps

  5. class RegExp extends AnyRef

  6. class Sequ extends RegExp

  7. class Star(r: _regexpT) extends RegExp with Product

  8. class Wildcard() extends RegExp with Product

  9. type _labelT <: Label

  10. type _regexpT <: RegExp

Value Members

  1. object Alt extends AnyRef

  2. object Eps extends RegExp with Product

  3. object Sequ extends AnyRef

  4. def equals(arg0: Any): Boolean

    This method is used to compare the receiver object (this) with the argument object (arg0) for equivalence

    This method is used to compare the receiver object (this) with the argument object (arg0) for equivalence.

    The default implementations of this method is an equivalence relation:

    • It is reflexive: for any instance x of type Any, x.equals(x) should return true.
    • It is symmetric: for any instances x and y of type Any, x.equals(y) should return true if and only if y.equals(x) returns true.
    • It is transitive: for any instances x, y, and z of type AnyRef if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.

    If you override this method, you should verify that your implementation remains an equivalence relation. Additionally, when overriding this method it is often necessary to override hashCode to ensure that objects that are "equal" (o1.equals(o2) returns true) hash to the same Int (o1.hashCode.equals(o2.hashCode)).

    arg0

    the object to compare against this object for equality.

    returns

    true if the receiver object is equivalent to the argument; false otherwise.

    definition classes: AnyRef ⇐ Any

  5. def hashCode(): Int

    Returns a hash code value for the object

    Returns a hash code value for the object.

    The default hashing algorithm is platform dependent.

    Note that it is allowed for two objects to have identical hash codes (o1.hashCode.equals(o2.hashCode)) yet not be equal (o1.equals(o2) returns false). A degenerate implementation could always return 0. However, it is required that if two objects are equal (o1.equals(o2) returns true) that they have identical hash codes (o1.hashCode.equals(o2.hashCode)). Therefore, when overriding this method, be sure to verify that the behavior is consistent with the equals method.

    definition classes: AnyRef ⇐ Any

  6. def toString(): String

    Returns a string representation of the object

    Returns a string representation of the object.

    The default representation is platform dependent.

    definition classes: AnyRef ⇐ Any

Instance constructors

  1. new WordExp()