*/
  object DisconnectAfterRequest : SocketPolicy

  /**
   * Half close connection (InputStream for client) after reading the request but before
   * writing the response. Use this to simulate late connection pool failures.
   */
  object HalfCloseAfterRequest : SocketPolicy

  /** Close connection after reading half of the request body (if present). */
  object DisconnectDuringRequestBody : SocketPolicy

  private final long k1;

  /**
   * @param c the number of compression rounds (must be positive)
   * @param d the number of finalization rounds (must be positive)
   * @param k0 the first half of the key
   * @param k1 the second half of the key
   */
  SipHashFunction(int c, int d, long k0, long k1) {
    checkArgument(
        c > 0, "The number of SipRound iterations (c=%s) during Compression must be positive.", c);

	// Tolerance is not set, defaults to half of the locker clients.
	tolerance := len(restClnts) / 2

	// Quorum is effectively = total clients subtracted with tolerance limit
	quorum := len(restClnts) - tolerance
	if !isReadLock {
		// In situations for write locks, as a special case
		// to avoid split brains we make sure to acquire
		// quorum + 1 when tolerance is exactly half of the
		// total locker clients.

MinIO protects data against hardware failures and silent data corruption using erasure code and checksums. With the highest level of redundancy, you may lose up to half (N/2) of the total drives and still be able to recover the data.

## What is Erasure Code?

  private final long k1;

  /**
   * @param c the number of compression rounds (must be positive)
   * @param d the number of finalization rounds (must be positive)
   * @param k0 the first half of the key
   * @param k1 the second half of the key
   */
  SipHashFunction(int c, int d, long k0, long k1) {
    checkArgument(
        c > 0, "The number of SipRound iterations (c=%s) during Compression must be positive.", c);

   *   <li>awaits executor service termination for half of the specified timeout.
   *   <li>if the timeout expires, it calls {@link ExecutorService#shutdownNow()}, cancelling
   *       pending tasks and interrupting running tasks.
   *   <li>awaits executor service termination for the other half of the specified timeout.
   * </ol>
   *

      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;
    return oldSlowFactorial(n1, mid).multiply(oldSlowFactorial(mid, n2));
  }

  @Benchmark
  int slowFactorial(int reps) {
    int tmp = 0;

      return entry.value;
    }
    entry = cacheEntry2;
    if (entry != null && entry.key == key) {
      // Promote second cache entry to first so the access pattern
      // [K1, K2, K1, K3, K1, K4...] still hits the cache half the time.
      addToCache(entry);
      return entry.value;
    }
    return null;
  }

  @Override
  void clearCache() {
    super.clearCache();
    cacheEntry1 = null;
    cacheEntry2 = null;
  }

          access.registerAsUsed(testClass)
          registerTest(access, testClass)
        }
      } catch (e: Exception) {
        // If you throw an exception here then native image building fails half way through
        // silently without rewriting the binary. So we report noisily, but keep going and prefer
        // running most tests still.
        e.printStackTrace()
      }
    }
  }

  }

  private static long random() {
    return RANDOM_SOURCE.nextLong();
  }

  // A random value that cannot be 0 and that is unsigned-less-than or equal
  // to the given dividend, so that we don't have half of our divisions being
  // trivial because the divisor is bigger than the dividend.
  // Using remainder here does not give us a uniform distribution but it should
  // not have a big impact on the measurement.