package soot.jimple.spark.sets;

/*-
 * #%L
 * Soot - a J*va Optimization Framework
 * %%
 * Copyright (C) 1997 - 2018 Raja Vallée-Rai and others
 * %%
 * This program is free software: you can redistribute it and/or modify
 * it under the terms of the GNU Lesser General Public License as
 * published by the Free Software Foundation, either version 2.1 of the
 * License, or (at your option) any later version.
 * 
 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Lesser Public License for more details.
 * 
 * You should have received a copy of the GNU General Lesser Public
 * License along with this program.  If not, see
 * <http://www.gnu.org/licenses/lgpl-2.1.html>.
 * #L%
 */

import java.util.List;

import soot.Type;
import soot.jimple.spark.pag.Node;
import soot.jimple.spark.pag.PAG;
import soot.util.BitSetIterator;
import soot.util.BitVector;

/*
 * Possible sources of inefficiency:
 * -It seems like there must be a way to immediately know which subset bitvector
 * will exist, since sets are shared when a node with a smaller points-to set is
 * pointing to another node.  Why not just take that node's bitvector as a base?
 * -addAll could probably use many improvements.
 * -Cast masking - calling typeManager.get
 * -An interesting problem is that when merging a bitvector into an overflow list, if
 * the one being merged
 * in has a bitvector, the mask or exclude might mask it down to a bitvector with very
 * few ones.  (In fact, this might even result in one with 0 ones!)
 * Should that new bitvector stay as a bitvector, or be converted to an
 * overflow list?  And how can we tell when it will have few ones?  (Modify BitVector?)
 *
 */
/**
 * A shared representation of a points-to set which uses a bit vector + a list of extra elements, an "overflow list", to make
 * adding single elements fast in most cases.
 *
 * The bit vector may be shared by multiple points-to sets, while the overflow list is specific to each points-to set.
 *
 * To facilitate sharing of the bitvectors, there is a "hash table" of all existing bitvectors kept, called
 * BitVectorLookupMap, where the ith element contains a list of all existing bitvectors of cardinality i (i.e. has i one
 * bits).
 *
 * @author Adam Richard
 *
 */

public class SharedHybridSet extends PointsToSetInternal {

  public SharedHybridSet(Type type, PAG pag) {
    // I'm not sure what "type" is for, but this is the way the other set
    // representations
    // did it
    super(type);
    this.pag = pag;
    // System.out.println("Using new heintze set");
  }

  // The following 2 constants should be tweaked for efficiency
  public final static int OVERFLOW_SIZE = 16;

  /**
   * The max number of elements allowed in the set before creating a new bitvector for it.
   */
  public final static int OVERFLOW_THRESHOLD = 5;

  /**
   * When the overflow list overflows, the maximum number of elements that may remain in the overflow list (the rest are
   * moved into the base bit vector)
   */

  public boolean contains(Node n) {
    // Which should be checked first, bitVector or overflow? (for
    // performance)
    // I think the bit vector, since it only takes O(1) to check many
    // elements

    // Check the bit vector
    if (bitVector != null && bitVector.contains(n)) {
      return true;
    }

    // Check overflow
    if (overflow.contains(n)) {
      return true;
    }

    return false;
  }

  public boolean isEmpty() {
    return numElements == 0;
  }

  /**
   * @return an overflow list of all elements in a that aren't in b (b is assumed to be a subset of a)
   */
  private OverflowList remainder(PointsToBitVector a, PointsToBitVector b) {
    // Since a contains everything b contains, doing an XOR will give
    // everything
    // in a not in b
    PointsToBitVector xorResult = new PointsToBitVector(a);
    xorResult.xor(b);
    // xorResult must now contain <= 20 elements, assuming
    // OVERFLOW_THRESHOLD <= OVERFLOW_SIZE
    return new OverflowList(xorResult);
  }

  // Look for an existing bitvector in the lookupMap which is a subset of the
  // newBitVector (the bitVector to set as the new points-to set), with only a
  // few
  // elements missing. If we find one, make that set the new `bitVector', and
  // the leftovers the new `overflow'
  // szBitVector is the size of the ORIGINAL bit vector, NOT the size of newBitVector
  private void findAppropriateBitVector(PointsToBitVector newBitVector, PointsToBitVector otherBitVector, int otherSize,
      int szBitvector) {
    // First check "other" and "this"'s bitvector, to maximize sharing and
    // minimize searching for a new bitvector
    if (otherBitVector != null && otherSize <= numElements && otherSize + OVERFLOW_THRESHOLD >= numElements
        && otherBitVector.isSubsetOf(newBitVector)) {
      setNewBitVector(szBitvector, otherBitVector);
      overflow = remainder(newBitVector, otherBitVector);
    } else if (bitVector != null && szBitvector <= numElements && szBitvector + OVERFLOW_THRESHOLD >= numElements
        && bitVector.isSubsetOf(newBitVector)) {
      overflow = remainder(newBitVector, bitVector);
    } else {
      for (int overFlowSize = 0; overFlowSize < OVERFLOW_THRESHOLD; ++overFlowSize) {
        int bitVectorCardinality = numElements - overFlowSize;
        if (bitVectorCardinality < 0) {
          break; // We might be trying to add a bitvector
        }
        // with <OVERFLOW_THRESHOLD ones (in fact, there might be bitvectors with 0
        // ones). This results from merging bitvectors and masking out certain values.
        if (bitVectorCardinality < AllSharedHybridNodes.v().lookupMap.map.length
            && AllSharedHybridNodes.v().lookupMap.map[bitVectorCardinality] != null) {
          List<PointsToBitVector> lst = AllSharedHybridNodes.v().lookupMap.map[bitVectorCardinality];
          for (PointsToBitVector candidate : lst) {
            // for each existing bit vector with bitVectorCardinality
            // ones
            if (candidate.isSubsetOf(newBitVector)) {
              setNewBitVector(szBitvector, candidate);
              overflow = remainder(newBitVector, candidate);
              return;
            }
          }
        }
      }
      // Didn't find an appropriate bit vector to use as a base; add the new
      // bit vector to the map of all bit vectors and set it as the new base
      // bit vector
      setNewBitVector(szBitvector, newBitVector);
      overflow.removeAll();
      AllSharedHybridNodes.v().lookupMap.add(numElements, newBitVector);
    }
  }

  // Allows for reference counting and deleting the old bit vector if it
  // isn't being shared
  private void setNewBitVector(int size, PointsToBitVector newBitVector) {
    newBitVector.incRefCount();
    if (bitVector != null) {
      bitVector.decRefCount();

      if (bitVector.unused()) {
        // delete bitVector from lookupMap
        AllSharedHybridNodes.v().lookupMap.remove(size, bitVector);
      }
    }
    bitVector = newBitVector;
  }

  public boolean add(Node n) {
    /*
     * This algorithm is described in the paper "IBM Research Report: Fast Pointer Analysis" by Hirzel, Dincklage, Diwan, and
     * Hind, pg. 11
     */
    if (contains(n)) {
      return false;
    }
    ++numElements;

    if (!overflow.full()) {
      overflow.add(n);
    } else {

      // Put everything in the bitvector
      PointsToBitVector newBitVector;
      if (bitVector == null) {
        newBitVector = new PointsToBitVector(pag.getAllocNodeNumberer().size());
      } else {
        newBitVector = new PointsToBitVector(bitVector);
      }
      newBitVector.add(n); // add n to it
      add(newBitVector, overflow);

      // Now everything is in newBitVector, and it must have numElements
      // ones

      // The algorithm would still work without this step, but wouldn't be
      // a
      // shared implmentation at all.
      findAppropriateBitVector(newBitVector, null, 0, numElements - overflow.size() - 1);

    }
    return true;
  }

  private boolean nativeAddAll(SharedHybridSet other, SharedHybridSet exclude) {
    /*
     * If one of the shared hybrid sets has a bitvector but the other doesn't, set that bitvector as the base bitvector and
     * add the stuff from the other overflow list. If they both have a bitvector, AND them together, then add it to the
     * lookupMap. If neither of them has a bitvector, just combine the overflow lists.
     */

    BitVector mask = getBitMask(other, pag);

    if (exclude != null) {
      if (exclude.overflow.size() > 0) {
        // Make exclude only a bitvector, for simplicity
        PointsToBitVector newBitVector;
        if (exclude.bitVector == null) {
          newBitVector = new PointsToBitVector(pag.getAllocNodeNumberer().size());
        } else {
          newBitVector = new PointsToBitVector(exclude.bitVector);
        }
        add(newBitVector, exclude.overflow);
        exclude = new SharedHybridSet(type, pag);
        exclude.bitVector = newBitVector;
      }

      // It's possible at this point that exclude could have been passed in non-null,
      // but with no elements. Simplify the rest of the algorithm by setting it to null
      // in that case.
      else if (exclude.bitVector == null) {
        exclude = null;
      }
    }

    int originalSize = size(), originalOnes = originalSize - overflow.size(),
        otherBitVectorSize = other.size() - other.overflow.size();

    // Decide on the base bitvector
    if (bitVector == null) {
      bitVector = other.bitVector;
      if (bitVector != null) { // Maybe both bitvectors were null; in
        // that case, no need to do this
        bitVector.incRefCount();

        // Since merging in new bits might add elements that
        // were
        // already in the overflow list, we have to remove and re-add
        // them all.
        // TODO: Can this be avoided somehow?
        // Maybe by allowing an element to be both in the overflow set
        // and
        // the bitvector?
        // Or could it be better done by checking just the bitvector and
        // removing elements that are there?
        OverflowList toReAdd = overflow;
        overflow = new OverflowList();

        boolean newBitVectorCreated = false; // whether a new bit vector
        // was created, which is used to decide whether to re-add the
        // overflow list as an overflow list again or merge it into the
        // new bit vector.

        numElements = otherBitVectorSize;
        if (exclude != null || mask != null) {
          PointsToBitVector result = new PointsToBitVector(bitVector);
          if (exclude != null) {
            result.andNot(exclude.bitVector);
          }
          if (mask != null) {
            result.and(mask);
          }
          if (!result.equals(bitVector)) {
            add(result, toReAdd);
            int newBitVectorSize = result.cardinality();
            numElements = newBitVectorSize;
            findAppropriateBitVector(result, other.bitVector, otherBitVectorSize, otherBitVectorSize);
            newBitVectorCreated = true;
          }
        }

        if (!newBitVectorCreated) // if it was, then toReAdd has
        // already been re-added
        {
          for (OverflowList.ListNode i = toReAdd.overflow; i != null; i = i.next) {
            add(i.elem);
          }
        }
      }
    } else if (other.bitVector != null) {
      // Now both bitvectors are non-null; merge them
      PointsToBitVector newBitVector = new PointsToBitVector(other.bitVector);
      if (exclude != null) {
        newBitVector.andNot(exclude.bitVector);
      }
      if (mask != null) {
        newBitVector.and(mask);
      }

      newBitVector.or(bitVector);

      if (!newBitVector.equals(bitVector)) // if some elements were
      // actually added
      {

        // At this point newBitVector is bitVector + some new bits

        // Have to make a tough choice - is it better at this point to
        // put both overflow lists into this bitvector (which involves
        // recalculating bitVector.cardinality() again since there might
        // have been overlap), or is it better to re-add both the
        // overflow lists to the set?
        // I suspect the former, so I'll do that.

        // Basically we now want to merge both overflow lists into this
        // new
        // bitvector (if it is indeed a new bitvector), then add that
        // resulting
        // huge bitvector to the lookupMap, unless a subset of it is
        // already there.

        if (other.overflow.size() != 0) {
          PointsToBitVector toAdd = new PointsToBitVector(newBitVector.size());
          add(toAdd, other.overflow);
          if (mask != null) {
            toAdd.and(mask);
          }
          if (exclude != null) {
            toAdd.andNot(exclude.bitVector);
          }
          newBitVector.or(toAdd);
        }
        // At this point newBitVector is still bitVector + some new bits

        int numOnes = newBitVector.cardinality(); // # of bits in the
        // new bitvector
        int numAdded = add(newBitVector, overflow);
        numElements += numOnes - originalOnes // number of new bits
            + numAdded - overflow.size(); // might be negative due to
        // elements in overflow already being in the new bits

        if (size() > originalSize) {
          findAppropriateBitVector(newBitVector, other.bitVector, otherBitVectorSize, originalOnes);
          // checkSize();
          return true;
        } else {
          // checkSize();
          return false; // It might happen that the bitvector being merged in adds some bits
          // to the existing bitvector, but that those new bits are all elements that were already
          // in the overflow list. In that case, the set might not change, and if not we return false.
          // We also leave the set the way it was by not calling findAppropriateBitvector,
          // which maximizes sharing and is fastest in the short term. I'm not sure whether it
          // would be faster overall to keep the already calculated bitvector anyway.
        }
      }
    }
    // Add all the elements in the overflow list of other, unless they're in
    // exclude
    OverflowList overflow = other.overflow;
    for (OverflowList.ListNode i = overflow.overflow; i != null; i = i.next) {
      // for (int i = 0; i < overflow.size(); ++i) {
      Node nodeToMaybeAdd = i.elem;
      if ((exclude == null) || !exclude.contains(nodeToMaybeAdd)) {
        if (mask == null || mask.get(nodeToMaybeAdd.getNumber())) {
          add(nodeToMaybeAdd);
        }
      }
    }

    // checkSize();
    return size() > originalSize;
  }

  /**
   * @ Adds the Nodes in arr to this bitvector.
   *
   * @return The number of new nodes actually added.
   */
  private int add(PointsToBitVector p, OverflowList arr) {
    // assert size <= arr.length;
    int retVal = 0;
    for (OverflowList.ListNode i = arr.overflow; i != null; i = i.next) {
      if (p.add(i.elem)) {
        ++retVal;
        /*
         * int num = arr[i].getNumber(); if (!get(num)) { set(num); ++retVal; }
         */
      }
    }
    return retVal;
  }

  /*
   * //A class invariant - numElements correctly holds the size //Only used for testing private void checkSize() { int
   * realSize = overflow.size(); if (bitVector != null) realSize += bitVector.cardinality(); if (numElements != realSize) {
   * throw new RuntimeException("Assertion failed."); } }
   */

  public boolean addAll(PointsToSetInternal other, final PointsToSetInternal exclude) {
    // Look at the sort of craziness we have to do just because of a lack of
    // multimethods
    if (other == null) {
      return false;
    }
    if ((!(other instanceof SharedHybridSet)) || (exclude != null && !(exclude instanceof SharedHybridSet))) {
      return super.addAll(other, exclude);
    } else {
      return nativeAddAll((SharedHybridSet) other, (SharedHybridSet) exclude);
    }
  }

  public boolean forall(P2SetVisitor v) {
    // Iterate through the bit vector. Ripped from BitPointsToSet again.
    // It seems there should be a way to share code between BitPointsToSet
    // and
    // SharedHybridSet, but I don't know how at the moment.
    if (bitVector != null) {
      for (BitSetIterator it = bitVector.iterator(); it.hasNext();) {
        v.visit((Node) pag.getAllocNodeNumberer().get(it.next()));
      }
    }
    // Iterate through the overflow list
    for (OverflowList.ListNode i = overflow.overflow; i != null; i = i.next) {
      v.visit(i.elem);
    }
    return v.getReturnValue();
  }

  // Ripped from the other points-to sets - returns a factory that can be
  // used to construct SharedHybridSets
  public final static P2SetFactory getFactory() {
    return new P2SetFactory() {
      public final PointsToSetInternal newSet(Type type, PAG pag) {
        return new SharedHybridSet(type, pag);
      }
    };
  }

  private PointsToBitVector bitVector = null; // Shared with other points-to
  // sets

  private OverflowList overflow = new OverflowList();

  private PAG pag; // I think this is needed to get the size of the bit
  // vector and the mask for casting

  private int numElements = 0; // # of elements in the set

  public int size() {
    return numElements;
  }

  private class OverflowList {
    public class ListNode {
      public Node elem;
      public ListNode next;

      public ListNode(Node elem, ListNode next) {
        this.elem = elem;
        this.next = next;
      }
    }

    public OverflowList() {
    }

    public OverflowList(PointsToBitVector bv) {
      BitSetIterator it = bv.iterator(); // Iterates over only the 1 bits
      while (it.hasNext()) {
        // Get the next node in the bitset by looking it up in the
        // pointer assignment graph.
        // Ripped from BitPointsToSet.
        Node n = (Node) (pag.getAllocNodeNumberer().get(it.next()));
        add(n);
      }

    }

    public void add(Node n) {
      if (full()) {
        throw new RuntimeException("Can't add an element to a full overflow list.");
      }
      overflow = new ListNode(n, overflow);
      ++overflowElements;
    }

    public int size() {
      return overflowElements;
    }

    public boolean full() {
      return overflowElements == OVERFLOW_SIZE;
    }

    public boolean contains(Node n) {
      for (ListNode l = overflow; l != null; l = l.next) {
        if (n == l.elem) {
          return true;
        }
      }
      return false;
    }

    public void removeAll() {
      overflow = null;
      overflowElements = 0;
    }

    /*
     * public ListNode next() { return overflow.next; } public Node elem() { return overflow.elem; }
     */
    public ListNode overflow = null; // Not shared with
    // other points-to sets - the extra elements besides the ones in bitVector
    private int overflowElements = 0; // # of elements actually in the
    // array `overflow'

  }

}
