```java
public class RdmaSecurityManager {
    public void validateMemoryAccess(RdmaMemoryRegion region, RdmaAccess requestedAccess) {
        if (!region.hasAccess(requestedAccess)) {
            throw new SecurityException("Insufficient RDMA memory access rights");
        }
    }
    
    public void validateRemoteAccess(long remoteAddress, int length, int remoteKey) {
        // Validate that remote memory access is authorized

sliceForAppend takes a slice and a requested number of bytes. It returns a // slice with the contents of the given slice followed by that many bytes and a // second slice that aliases into it and contains only the extra bytes. If the // original slice has sufficient capacity then no allocation is performed. func sliceForAppend(in []byte, n int) (head, tail []byte) { if total := len(in) + n; cap(in) >= total { head = in[:total] } else { head = make([]byte, total) copy(head, in) } tail = head[len(in):] return...

   * modeled by "storedPermits" variable. This variable is zero when there is no underutilization,
   * and it can grow up to maxStoredPermits, for sufficiently large underutilization. So, the
   * requested permits, by an invocation acquire(permits), are served from:
   *
   * - stored permits (if available)
   *
   * - fresh permits (for any remaining permits)
   *

### Reproduce el problema { #reproduce-the-problem }

En la mayoría de los casos y preguntas hay algo relacionado con el **código original** de la persona.

En muchos casos solo copiarán un fragmento del código, pero eso no es suficiente para **reproducir el problema**.

      super.addAll(elements);
      return this;
    }

    /**
     * If the specified collection is backed by an ObjectCountHashMap, it will be much more
     * efficient to iterate over it by index rather than an entry iterator, which will need to
     * allocate an object for each entry, so we check for that.
     */

 *   <li>(1) "yes" and (2) "yes", use {@link #forTree(SuccessorsFunction)}.
 *   <li>(1) "yes" and (2) "no", you can use either, but {@code forTree()} will be more efficient.
 *   <li>(1) "no" and (2) "yes", <b><i>neither will work</i></b>, but if you transform your node
 *       objects into a non-recursive form, you can use {@code forGraph()}.
 * </ul>
 *
 * @author Jens Nyman

  /**
   * Creates a new {@code ConcurrentHashMultiset} containing the specified elements, using the
   * default initial capacity, load factor, and concurrency settings.
   *
   * <p>This implementation is highly efficient when {@code elements} is itself a {@link Multiset}.
   *
   * @param elements the elements that the multiset should contain
   */
  public static <E> ConcurrentHashMultiset<E> create(Iterable<? extends E> elements) {

 *     .url("http://publicobject.com/helloworld.txt")
 *     .build();
 * ```
 *
 * If it is only necessary to force a cached response to be validated by the server, use the more
 * efficient `max-age=0` directive instead:
 *
 * ```java
 * Request request = new Request.Builder()
 *     .cacheControl(new CacheControl.Builder()
 *         .maxAge(0, TimeUnit.SECONDS)
 *         .build())

   * Reader#skip(long) skip} to the end of the stream, and return the total number of chars that
   * were skipped.
   *
   * <p>Note that for sources that implement {@link #lengthIfKnown} to provide a more efficient
   * implementation, it is <i>possible</i> that this method will return a different number of chars
   * than would be returned by reading all of the chars.
   *

    assertThat(is('a').retainFrom("")).isEqualTo("");
  }

  public void testPrecomputedOptimizations() {
    // These are testing behavior that's never promised by the API.
    // Some matchers are so efficient that it is a waste of effort to
    // build a precomputed version.
    CharMatcher m1 = is('x');
    assertThat(m1.precomputed()).isSameInstanceAs(m1);
    assertThat(m1.precomputed().toString()).isEqualTo(m1.toString());