* </ul>
   *
   * <p>The computed result is within 1 ulp of the exact result.
   *
   * <p>If the result of this method will be immediately rounded to an {@code int}, {@link
   * #log2(double, RoundingMode)} is faster.
   */
  public static double log2(double x) {
    return log(x) / LN_2; // surprisingly within 1 ulp according to tests
  }

  /**

     *
     * 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[Integer.numberOfLeadingZeros(x)];
    /*

	trace.Stop()
	t2.End()
	t0.End()
}

func do(ctx context.Context, k int) {
	trace.Log(ctx, "log", "before do")

	var t *trace.Task
	ctx, t = trace.NewTask(ctx, "do")
	defer t.End()

	trace.Log(ctx, "log2", "do")

	// Create a region and spawn more tasks and more workers.
	trace.WithRegion(ctx, "fanout", func() {
		for i := 0; i < k; i++ {
			go func(i int) {
				trace.WithRegion(ctx, fmt.Sprintf("region%d", i), func() {

     *
     * 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[Integer.numberOfLeadingZeros(x)];
    /*

inline float CalculateBinWidth(const float min_value, const float max_value,
                               const int32_t num_bins) {
  const float raw_bin_width = (max_value - min_value) / num_bins;
  return std::pow(2, std::ceil(std::log2(raw_bin_width)));
}

// Calculates the lower bound of the histogram. The lower bound is in form of
// `N * bin_width`.
inline float CalculateLowerBound(const float min_value, const float bin_width) {

  /* Relies on the correctness of BigIntegerMath.log2 for all modes except UNNECESSARY. */
  public void testLog2MatchesBigInteger() {
    for (long x : POSITIVE_LONG_CANDIDATES) {
      for (RoundingMode mode : ALL_SAFE_ROUNDING_MODES) {
        // The BigInteger implementation is tested separately, use it as the reference.
        assertEquals(BigIntegerMath.log2(valueOf(x), mode), LongMath.log2(x, mode));
      }
    }
  }

  /* Relies on the correctness of BigIntegerMath.log2 for all modes except UNNECESSARY. */
  public void testLog2MatchesBigInteger() {
    for (long x : POSITIVE_LONG_CANDIDATES) {
      for (RoundingMode mode : ALL_SAFE_ROUNDING_MODES) {
        // The BigInteger implementation is tested separately, use it as the reference.
        assertEquals(BigIntegerMath.log2(valueOf(x), mode), LongMath.log2(x, mode));
      }
    }
  }

     *
     * 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)];
    /*

    }
    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;
  }

	"testing"
)

func TestMaxBase(t *testing.T) {
	if MaxBase != len(digits) {
		t.Fatalf("%d != %d", MaxBase, len(digits))
	}
}

// log2 computes the integer binary logarithm of x.
// The result is the integer n for which 2^n <= x < 2^(n+1).
// If x == 0, the result is -1.
func log2(x Word) int {
	return bits.Len(uint(x)) - 1
}

func itoa(x nat, base int) []byte {
	// special cases
	switch {
	case base < 2: