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);

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?

            circuitBreaker.executeWithCircuitBreaker(() -> {
                assertEquals(State.HALF_OPEN, circuitBreaker.getState(), "Should be in HALF_OPEN state");
                throw new RuntimeException("Failure in half-open");
            });
            fail("Should have thrown exception");
        } catch (CIFSException e) {
            // Expected
        }

      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;
  }

     * Creates a simple circuit breaker
     *
     * @param name the circuit breaker name
     * @param failureThreshold number of consecutive failures to open circuit
     * @param successThreshold number of successes in half-open to close circuit
     * @param timeoutMillis timeout before attempting to close open circuit
     */
    public SimpleCircuitBreaker(String name, int failureThreshold, int successThreshold, long timeoutMillis) {

clock in a twinkling!  Half-past one, time for dinner!'

  (`I only wish it was,' the March Hare said to itself in a
whisper.)

  `That would be grand, certainly,' said Alice thoughtfully:
`but then--I shouldn't be hungry for it, you know.'

  `Not at first, perhaps,' said the Hatter:  `but you could keep
it to half-past one as long as you liked.'

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    </div>
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  }

  private static long random() {
    return randomSource.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.

            return roundFloorAsDouble;
          } else if (diff > 0) { // closer to ceiling
            return roundCeilingAsDouble;
          }
          // halfway between the representable values; do the half-whatever logic
          switch (mode) {
            case HALF_EVEN:
              // roundFloorAsDouble and roundCeilingAsDouble are neighbors, so precisely
              // one of them should have an even long representation

      }
      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;