doubles[i] = randomDouble(Long.SIZE);
      factorials[i] = RANDOM_SOURCE.nextInt(100);
    }
  }

  @Benchmark
  long log2(int reps) {
    long tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += Double.doubleToRawLongBits(DoubleMath.log2(positiveDoubles[j]));
    }
    return tmp;
  }

  @Benchmark
  long factorial(int reps) {
    long tmp = 0;

    }
  }

  /** Returns the product of {@code n1} exclusive through {@code n2} inclusive. */
  private static BigInteger oldSlowFactorial(int n1, int n2) {
    assert n1 <= n2;
    if (IntMath.log2(n2, CEILING) * (n2 - n1) < Long.SIZE - 1) {
      // the result will definitely fit into a long
      long result = 1;
      for (int i = n1 + 1; i <= n2; i++) {
        result *= i;
      }

     *
     * The key idea is that based on the number of leading zeros (equivalently, floor(log2(x))), we
     * can narrow the possible floor(log10(x)) values to two. For example, if floor(log2(x)) is 6,
     * then 64 <= x < 128, so floor(log10(x)) is either 1 or 2.
     */
    int y = maxLog10ForLeadingZeros[Long.numberOfLeadingZeros(x)];
    /*

      nonzero1[i] = randomNonZeroBigInteger(1024);
      nonzero2[i] = randomNonZeroBigInteger(1024);
    }
  }

  @Benchmark
  int log2(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += BigIntegerMath.log2(positive[j], mode);
    }
    return tmp;
  }

  @Benchmark
  int log10(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {

        }
      }
    }
  }

  // Relies on the correctness of BigIntegerMath.log2 for all modes except UNNECESSARY.
  public void testLog2MatchesBigInteger() {
    for (int x : POSITIVE_INTEGER_CANDIDATES) {
      for (RoundingMode mode : ALL_SAFE_ROUNDING_MODES) {
        assertEquals(BigIntegerMath.log2(valueOf(x), mode), IntMath.log2(x, mode));
      }
    }
  }

  // Relies on the correctness of isPowerOfTwo(int).

    }
    return tmp;
  }

  @Benchmark
  int log2Round(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += DoubleMath.log2(positiveDoubles[j], mode);
    }
    return tmp;
  }

    top.offer(1);
    for (int i = 1; i < n; i++) {
      top.offer(0);
    }
    assertThat(top.topK()).containsExactlyElementsIn(Collections.nCopies(k, 0));
    assertThat(compareCalls[0]).isAtMost(10L * n * IntMath.log2(k, RoundingMode.CEILING));
  }

  public void testExceedMaxIteration() {
    /*
     * Bug #5692 occurred when TopKSelector called Arrays.sort incorrectly.
     */

    }
  }

  /** Returns the product of {@code n1} exclusive through {@code n2} inclusive. */
  private static BigInteger oldSlowFactorial(int n1, int n2) {
    assert n1 <= n2;
    if (IntMath.log2(n2, CEILING) * (n2 - n1) < Long.SIZE - 1) {
      // the result will definitely fit into a long
      long result = 1;
      for (int i = n1 + 1; i <= n2; i++) {
        result *= i;
      }

    top.offer(1);
    for (int i = 1; i < n; i++) {
      top.offer(0);
    }
    assertThat(top.topK()).containsExactlyElementsIn(Collections.nCopies(k, 0));
    assertThat(compareCalls[0]).isAtMost(10L * n * IntMath.log2(k, RoundingMode.CEILING));
  }

  public void testExceedMaxIteration() {
    /*
     * Bug #5692 occurred when TopKSelector called Arrays.sort incorrectly.
     */

    // This is an extremely fast implementation of BigInteger.doubleValue(). JDK patch pending.
    BigInteger absX = x.abs();
    int exponent = absX.bitLength() - 1;
    // exponent == floor(log2(abs(x)))
    if (exponent < Long.SIZE - 1) {
      return x.longValue();
    } else if (exponent > MAX_EXPONENT) {
      return x.signum() * POSITIVE_INFINITY;
    }

    /*