com.google.re2j.Simplify Maven / Gradle / Ivy

Go to download
Show more of this group Show more artifacts with this name
Show all versions of hudi-cli-bundle_2.13 Show documentation
The newest version!
/*
 * Copyright (c) 2020 The Go Authors. All rights reserved.
 *
 * Use of this source code is governed by a BSD-style
 * license that can be found in the LICENSE file.
 */
// Original Go source here:
// http://code.google.com/p/go/source/browse/src/pkg/regexp/syntax/simplify.go

package com.google.re2j;

import java.util.ArrayList;

class Simplify {

  // Simplify returns a regexp equivalent to re but without counted
  // repetitions and with various other simplifications, such as
  // rewriting /(?:a+)+/ to /a+/.  The resulting regexp will execute
  // correctly but its string representation will not produce the same
  // parse tree, because capturing parentheses may have been duplicated
  // or removed.  For example, the simplified form for /(x){1,2}/ is
  // /(x)(x)?/ but both parentheses capture as $1.  The returned regexp
  // may share structure with or be the original.
  static Regexp simplify(Regexp re) {
    if (re == null) {
      return null;
    }
    switch (re.op) {
      case CAPTURE:
      case CONCAT:
      case ALTERNATE:
        {
          // Simplify children, building new Regexp if children change.
          Regexp nre = re;
          for (int i = 0; i < re.subs.length; ++i) {
            Regexp sub = re.subs[i];
            Regexp nsub = simplify(sub);
            if (nre == re && nsub != sub) {
              // Start a copy.
              nre = new Regexp(re); // shallow copy
              nre.runes = null;
              nre.subs = Parser.subarray(re.subs, 0, re.subs.length); // clone
            }
            if (nre != re) {
              nre.subs[i] = nsub;
            }
          }
          return nre;
        }
      case STAR:
      case PLUS:
      case QUEST:
        {
          Regexp sub = simplify(re.subs[0]);
          return simplify1(re.op, re.flags, sub, re);
        }
      case REPEAT:
        {
          // Special special case: x{0} matches the empty string
          // and doesn't even need to consider x.
          if (re.min == 0 && re.max == 0) {
            return new Regexp(Regexp.Op.EMPTY_MATCH);
          }

          // The fun begins.
          Regexp sub = simplify(re.subs[0]);

          // x{n,} means at least n matches of x.
          if (re.max == -1) {
            // Special case: x{0,} is x*.
            if (re.min == 0) {
              return simplify1(Regexp.Op.STAR, re.flags, sub, null);
            }

            // Special case: x{1,} is x+.
            if (re.min == 1) {
              return simplify1(Regexp.Op.PLUS, re.flags, sub, null);
            }

            // General case: x{4,} is xxxx+.
            Regexp nre = new Regexp(Regexp.Op.CONCAT);
            ArrayList subs = new ArrayList();
            for (int i = 0; i < re.min - 1; i++) {
              subs.add(sub);
            }
            subs.add(simplify1(Regexp.Op.PLUS, re.flags, sub, null));
            nre.subs = subs.toArray(new Regexp[subs.size()]);
            return nre;
          }

          // Special case x{0} handled above.

          // Special case: x{1} is just x.
          if (re.min == 1 && re.max == 1) {
            return sub;
          }

          // General case: x{n,m} means n copies of x and m copies of x?
          // The machine will do less work if we nest the final m copies,
          // so that x{2,5} = xx(x(x(x)?)?)?

          // Build leading prefix: xx.
          ArrayList prefixSubs = null;
          if (re.min > 0) {
            prefixSubs = new ArrayList();
            for (int i = 0; i < re.min; i++) {
              prefixSubs.add(sub);
            }
          }

          // Build and attach suffix: (x(x(x)?)?)?
          if (re.max > re.min) {
            Regexp suffix = simplify1(Regexp.Op.QUEST, re.flags, sub, null);
            for (int i = re.min + 1; i < re.max; i++) {
              Regexp nre2 = new Regexp(Regexp.Op.CONCAT);
              nre2.subs = new Regexp[] {sub, suffix};
              suffix = simplify1(Regexp.Op.QUEST, re.flags, nre2, null);
            }
            if (prefixSubs == null) {
              return suffix;
            }
            prefixSubs.add(suffix);
          }
          if (prefixSubs != null) {
            Regexp prefix = new Regexp(Regexp.Op.CONCAT);
            prefix.subs = prefixSubs.toArray(new Regexp[prefixSubs.size()]);
            return prefix;
          }

          // Some degenerate case like min > max or min < max < 0.
          // Handle as impossible match.
          return new Regexp(Regexp.Op.NO_MATCH);
        }
    }

    return re;
  }

  // simplify1 implements Simplify for the unary OpStar,
  // OpPlus, and OpQuest operators.  It returns the simple regexp
  // equivalent to
  //
  //      Regexp{Op: op, Flags: flags, Sub: {sub}}
  //
  // under the assumption that sub is already simple, and
  // without first allocating that structure.  If the regexp
  // to be returned turns out to be equivalent to re, simplify1
  // returns re instead.
  //
  // simplify1 is factored out of Simplify because the implementation
  // for other operators generates these unary expressions.
  // Letting them call simplify1 makes sure the expressions they
  // generate are simple.
  private static Regexp simplify1(Regexp.Op op, int flags, Regexp sub, Regexp re) {
    // Special case: repeat the empty string as much as
    // you want, but it's still the empty string.
    if (sub.op == Regexp.Op.EMPTY_MATCH) {
      return sub;
    }
    // The operators are idempotent if the flags match.
    if (op == sub.op && (flags & RE2.NON_GREEDY) == (sub.flags & RE2.NON_GREEDY)) {
      return sub;
    }
    if (re != null
        && re.op == op
        && (re.flags & RE2.NON_GREEDY) == (flags & RE2.NON_GREEDY)
        && sub == re.subs[0]) {
      return re;
    }

    re = new Regexp(op);
    re.flags = flags;
    re.subs = new Regexp[] {sub};
    return re;
  }

  private Simplify() {} // uninstantiable
}