intersectionsubtree.java

一个完整的XACML工程,学习XACML技术的好例子!

/*
* Copyright (c) 2006, University of Kent
* All rights reserved.
*
* Redistribution and use in source and binary forms, with or without 
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* list of conditions and the following disclaimer.
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* Redistributions in binary form must reproduce the above copyright notice, 
* this list of conditions and the following disclaimer in the documentation 
* and/or other materials provided with the distribution. 
*
* 1. Neither the name of the University of Kent nor the names of its 
* contributors may be used to endorse or promote products derived from this 
* software without specific prior written permission. 
*
* 2. THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS  
* IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO,
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* 4. YOU AGREE THAT THE EXCLUSIONS IN PARAGRAPHS 2 AND 3 ABOVE ARE REASONABLE
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* 5. This license is governed, except to the extent that local laws
* necessarily apply, by the laws of England and Wales.
*/
package issrg.pba.rbac.policies;

import issrg.utils.repository.Entry;

/**
 * This class is an intersection of two subtrees. An entry is contained
 * in the IntersectionSubtree, if it is contained in all the subtrees that
 * it is an intersection of.
 */
public class IntersectionSubtree implements Subtree {
  protected Subtree tree1;
  protected Subtree tree2;

  protected IntersectionSubtree(){}

  /**
   * This constructor builds an IntersectionSubtree, given two Subtrees. Both
   * of the subtrees are treated equally.
   *
   * @param t1 - one Subtree
   * @param t2 - another Subtree
   */
  public IntersectionSubtree(Subtree t1, Subtree t2){
    tree1=t1;
    tree2=t2;
  }

  /**
   * This method tells whether the given Entry is contained in the Intersection
   * of the two Subtrees passed to the constructor.
   *
   * @param e - the Entry to test
   *
   * @return true, if and only if the Entry is contained in both Subtrees; 
   *    otherwise false
   */
  public boolean contains(Entry e){
    return tree1.contains(e) && tree2.contains(e);
  }
}