Пример #1
0
 public boolean handlesTime() {
   for (BArm arm : arms) {
     if (arm.patterns.length == 1) {
       Pattern p = arm.patterns[0];
       if (p instanceof PTerm) {
         PTerm term = (PTerm) p;
         if (term.getName().equals("Time")) return true;
       }
     }
   }
   return false;
 }
Пример #2
0
 private boolean isDisjointArms() {
   // Returns true when the arms of the case are all headed by unique term patterns...
   Set<String> terms = new HashSet<String>();
   for (BArm arm : arms) {
     Pattern[] patterns = arm.patterns;
     Pattern pattern = patterns[0];
     if (pattern instanceof PNil) {
       if (terms.contains("[]")) return false;
       else terms.add("[]");
     } else if (pattern instanceof PCons) {
       if (terms.contains(":")) return false;
       else terms.add(":");
     } else if (pattern instanceof ast.patterns.PTerm) {
       ast.patterns.PTerm term = (ast.patterns.PTerm) pattern;
       String functor = term.getName();
       if (terms.contains(functor)) return false;
       else terms.add(functor);
     } else return false;
   }
   return true;
 }