Пример #1
0
 public String toString() {
   StringBuffer buf = new StringBuffer("{RG ");
   int len = refs.size();
   if (len > 3) len = 3;
   for (int i = 0; i < len; i++) {
     if (i > 0) buf.append(',');
     buf.append(refs.elementAt(i));
   }
   if (len < refs.size()) buf.append(", ...");
   buf.append('}');
   return buf.toString();
 }
Пример #2
0
  private Cost computeRefCostSum(LoopHeaderChord loop, Vector<RefGroup> refGroups) {
    Cost lc = new Cost();
    int l = refGroups.size();
    for (int i = 0; i < l; i++) {
      RefGroup rg = refGroups.elementAt(i);
      SubscriptExpr se = rg.getRepresentative();
      Cost cost = computeRefCost(loop, se);

      if (cost == null) continue;

      lc.add(cost);
    }
    return lc;
  }
Пример #3
0
  public void perform() {
    if (trace) System.out.println("** LP " + scribble.getRoutineDecl().getName());

    LoopHeaderChord lt = scribble.getLoopTree();
    Vector<LoopHeaderChord> innerLoops = lt.getInnerLoops();

    for (int li = 0; li < innerLoops.size(); li++) {
      LoopHeaderChord loop = innerLoops.elementAt(li);
      InductionVar ivar = loop.getPrimaryInductionVar();

      if (trace) System.out.println("   lp " + loop.nestedLevel() + " " + loop);

      if ((ivar == null) || !loop.isPerfectlyNested()) {
        innerLoops.addVectors(loop.getInnerLoops());
        continue;
      }

      if (loop.nestedLevel() < 2) // no need to check permutation for a simple nest
      continue;

      tryPermute(loop);
    }
  }
Пример #4
0
  private void tryPermute(LoopHeaderChord topLoop) {
    Vector<LoopHeaderChord> loopNest = topLoop.getTightlyNestedLoops();
    if (loopNest == null) return;

    int loopDepth = loopNest.size();
    LoopHeaderChord bottom = loopNest.get(loopDepth - 1);
    if (!unsafe && !legalLoop(bottom)) return;

    // Set the loop costs for the loops in a loop nest. The cost for a loop is
    // the cost of executing the nest with that loop in the innermost nesting.

    Table<Declaration, SubscriptExpr> arrayRefs = new Table<Declaration, SubscriptExpr>();
    if (!topLoop.getSubscriptsRecursive(arrayRefs)) return;

    graph = topLoop.getDDGraph(false);
    if (graph == null) return;

    if (trace) System.out.println("     " + graph);

    int[] loopIndex = new int[loopDepth];
    Cost[] loopCostList = new Cost[loopDepth];
    Vector<RefGroup> refGroups = new Vector<RefGroup>(20);

    for (int i = 0; i < loopDepth; i++) {
      LoopHeaderChord loop = loopNest.elementAt(i);
      if (trace) System.out.println("   " + i + " " + loop);

      computeRefGroups(loop.getNestedLevel(), 2, 2, arrayRefs, refGroups);

      Cost lc = computeRefCostSum(loop, refGroups);
      if (trace) System.out.println("   " + i + " " + lc);

      loopCostList[i] = lc;
      loopIndex[i] = i; // the outtermost loop is at position 0

      Cost tp = tripProduct(loopNest, loop);
      if (trace) System.out.println("   " + i + " " + tp);
      lc.multiply(tp);
      if (trace) System.out.println("   " + i + " " + lc);
    }

    boolean permuted = sortByCost(loopCostList, loopIndex);

    if (!permuted) return;
    if (trace) {
      System.out.print("   permute " + loopDepth);
      System.out.print(":");
      for (int i = 0; i < loopIndex.length; i++) System.out.print(" " + loopIndex[i]);
      System.out.println("");
    }

    int[][] ddVec = getDDVec(arrayRefs, loopDepth);
    if (trace) printDDInfo(ddVec, loopDepth);

    if (!isLegal(loopIndex, ddVec)) return;

    if (trace) System.out.println("   permute " + loopDepth);

    int[] rank = new int[loopDepth];

    // We will do sorting on the rank vector, which corresponds to the interchange we need.

    for (int i = 0; i < loopDepth; i++) {
      int loopNum = loopIndex[i];
      rank[loopNum] = i;
    }

    if (trace) printOrder(rank);

    boolean changed = true;

    while (changed) {
      changed = false;

      for (int i = 0; i < loopDepth - 1; i++) {
        int j = i + 1;
        int outerRank = rank[i];
        int innerRank = rank[j];

        if (innerRank >= outerRank) continue;

        LoopHeaderChord innerLoop = loopNest.elementAt(j);
        LoopHeaderChord outerLoop = loopNest.elementAt(i);

        if (!outerLoop.isDDComplete() || outerLoop.inhibitLoopPermute()) continue;

        if (!innerLoop.isDDComplete() || innerLoop.inhibitLoopPermute()) continue;

        changed = true;

        rank[i] = innerRank;
        rank[j] = outerRank;

        loopNest.setElementAt(innerLoop, i);
        loopNest.setElementAt(outerLoop, j);

        performLoopInterchange(innerLoop, outerLoop);
      }
    }
  }
Пример #5
0
  /**
   * Compute the reference groups for this loop. Two array references are in the same group with
   * respect to this loop if - there is a loop-independent dependence between them, or - the
   * dependence distance(dependence vector entry dl) for this loop is less than some constant
   * *dist*, and all other dependence vector entries are 0. - the two array refer to the same array
   * and differ by at most *dist2* in the first subscript dimension, where d is less than or equal
   * to the cache line size in terms of array elements. All other subscripts must be identical.
   * Notes: Here we assume dist1 = dist2 = 2
   */
  private void computeRefGroups(
      int level,
      int dist1,
      int dist2,
      Table<Declaration, SubscriptExpr> arrayRefs,
      Vector<RefGroup> refGroups) {
    if (arrayRefs == null) return;

    Enumeration<Declaration> ek = arrayRefs.keys();
    while (ek.hasMoreElements()) {
      VariableDecl vd = (VariableDecl) ek.nextElement();
      String s = vd.getName(); // array name
      Object[] v = arrayRefs.getRowArray(vd); // vector of SubscriptExpr's
      int vi = v.length;

      for (int j = vi - 1; j >= 0; j--) {
        SubscriptExpr sr = (SubscriptExpr) v[j];
        Vector<LoopHeaderChord> allRelatedLoops =
            sr.allRelatedLoops(); // ** Incorrect when something like a[(j+i][j]
        int arls = allRelatedLoops.size();
        int firstsub = arls - 1; // ** Making an invalid assumption here

        // Process the list of references r' with which r has a data
        // dependence, and  r is the source(data flows from r to r').

        RefGroup rg = new RefGroup(s, sr);
        Object[] edges = graph.getEdges(sr);
        int len = edges.length;

        for (int i = 0; i < len; i++) {
          DDEdge edge = (DDEdge) edges[i];

          if (edge.isSpatial()) continue;

          // Condition(1)-(a) in McKinley's paper

          if (edge.isLoopIndependentDependency()) { // add rP to the RefGroup of r:
            rg.add(edge);
            continue;
          }

          // Condition(1)-(b) in McKinley's paper

          computeEdgeRefs(edge, rg, level, dist1);

          if (arls <= 0) continue;

          // Condition(2) in McKinley's paper
          // rlevel is the level of the loop related to the first subscript.

          int rlevel = allRelatedLoops.elementAt(firstsub).getNestedLevel();

          computeEdgeRefs(edge, rg, rlevel, dist2);
        }

        boolean isInExistingRefGroups = false;
        int rgl = refGroups.size();
        for (int i = 0; i < rgl; i++) {
          RefGroup rg2 = refGroups.elementAt(i);
          if (!rg2.getName().equals(s)) continue;

          isInExistingRefGroups = rg2.contains(rg);

          if (isInExistingRefGroups) {
            rg2.add(rg);
            break;
          }
        }

        if (!isInExistingRefGroups) refGroups.addElement(rg);
      }
    }
  }
Пример #6
0
 private void printRefGroups(Vector<RefGroup> refGroups) {
   for (int i = 0; i < refGroups.size(); i++) {
     RefGroup rg = refGroups.elementAt(i);
     System.out.println(rg);
   }
 }
Пример #7
0
    public SubscriptExpr getRepresentative() {
      if (refs.size() > 0) return (SubscriptExpr) refs.elementAt(0);

      return null;
    }