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;

      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;
  }

        for (String result : results) {
            assertEquals(result, result.toLowerCase());
        }
    }

    public void test_convertHalfWidth() {
        // Test that full-width characters are converted to half-width
        List<String> results = converter.convert("アイウ", null);
        for (String result : results) {
            assertFalse(result.contains("Ａ")); // Full-width A

 * possible readings.
 * </p>
 *
 * <p>
 * The class also utilizes ICU4J's {@link com.ibm.icu.text.Transliterator} to convert full-width characters
 * to half-width and to convert any characters to lowercase, ensuring consistency in the output.
 * </p>
 *
 * <p>
 * Example:
 * For the input "キャ", the converter might return a list containing "kya".

      }
      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;

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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.

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    </div>
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     * @param failureThreshold number of failures before opening
     * @param successThreshold number of successes in half-open before closing
     * @param resetTimeoutMillis time to wait before attempting reset (ms)
     * @param halfOpenMaxAttempts max attempts in half-open state
     */