Пример #1
0
 /**
  * Gives string representation of a sequence, parameterized by how to represent each position
  * (i.e., constructs a string by iterating over all positions of the sequence, concatenating the
  * specified string representation of each position). Iteration is in the natural order for a
  * sequence.
  *
  * @param s sequence to stringify
  * @param pts how to stringify each position
  * @return the string representation of s
  */
 public static String stringfor(InspectableSequence s, PositionToString pts) {
   String toReturn = "[";
   PositionIterator pp = s.positions();
   // first element should have no comma
   if (pp.hasNext()) toReturn += " " + pts.stringfor(pp.nextPosition());
   while (pp.hasNext()) {
     toReturn += ", " + pts.stringfor(pp.nextPosition());
   }
   toReturn += " ]";
   return toReturn;
 }
Пример #2
0
 /**
  * Recursively dumps a byte representation of a subtree onto a stream. Used by stringfor( InspTree
  * ). Publicly available in case you want to stringify a subtree and really think this way is less
  * work than doing it yourself. You'll want to make sure the "spaces" static variable is
  * initialized first :)
  *
  * @param subtreeRoot root of subtree to represent
  * @param pts way to stringify each position
  * @param ostream stream on which to represent the subtree
  * @param t tree, so children(.) can be called
  * @param indentation number of spaces to indent this level
  * @param indentation_increment amount by which indentation will be increased before writing
  *     children
  */
 public static void writeNodeAndChildren(
     Position subtreeRoot,
     PositionToString pts,
     DataOutputStream ostream,
     InspectableTree t,
     int indentation,
     int indentation_increment) {
   try {
     // indent, then write the subtreeRoot
     ostream.write(spaces, 0, indentation);
     ostream.writeBytes(pts.stringfor(subtreeRoot));
     ostream.writeBytes("\n");
   } catch (java.io.IOException e) {
     System.err.println("\nAn I/O error occurred: " + e);
     return;
   }
   PositionIterator pp = t.children(subtreeRoot);
   // recur on all children
   while (pp.hasNext())
     writeNodeAndChildren(
         pp.nextPosition(),
         pts,
         ostream,
         t,
         indentation + indentation_increment,
         indentation_increment);
 }