示例#1
0
  private int[][] getDDVec(Table<Declaration, SubscriptExpr> arrayRefs, int loopDepth) {
    if (arrayRefs == null) return new int[0][0];

    int numEdges = 0;
    Stack<DDEdge> wl = WorkArea.<DDEdge>getStack("g<SubscriptExpr>etDDVec");

    Enumeration<DDEdge> k = graph.getEdges();
    while (k.hasMoreElements()) { // Check out every array variable in the loop nest.
      DDEdge edge = k.nextElement();
      if (edge.isSpatial()) continue;

      if (edge.isLoopIndependentDependency()) continue;

      if (edge.representsAllInput()) continue;

      wl.push(edge);
      numEdges++;
    }

    int[][] DDVector = new int[numEdges][loopDepth];

    int i = 0;
    while (!wl.empty()) {
      DDEdge edge = wl.pop();
      if (trace) System.out.println("   edge " + edge);

      // Find dependence distance.

      for (int level = 1; level <= loopDepth; level++)
        DDVector[i][level - 1] = edge.getDistance(level);

      if (trace) {
        System.out.print(i);
        System.out.print(" ");
        System.out.println(edge);
      }

      i++;
    }

    WorkArea.<DDEdge>returnStack(wl);

    return DDVector;
  }
示例#2
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);
      }
    }
  }