Пример #1
0
 private Bound(Expression min, Expression max) {
   this.flags = NORMAL;
   setMin(min);
   setMax(max);
   if (bounds == null) bounds = new Vector<Bound>(10);
   bounds.addElement(this);
 }
Пример #2
0
 public void add(RefGroup r) {
   int l = r.refs.size();
   for (int i = 0; i < l; i++) {
     Object ref = r.refs.elementAt(i);
     if (!refs.contains(ref)) refs.addElement(ref);
   }
 }
Пример #3
0
 /** Add the flags necessary for the C preprocessor. */
 public void addCPPFlags(Vector<String> v) {
   v.addElement("__i386__=1");
 }
Пример #4
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  /**
   * 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 #5
0
    public void add(DDEdge edge) {
      if (refs.contains(edge)) return;

      refs.addElement(edge);
    }
Пример #6
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    public RefGroup(String name, SubscriptExpr r) {
      this.refs = new Vector<Object>();
      this.aname = name;

      refs.addElement(r);
    }