/* 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)];
    /*

	"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:

     *
     * 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)];
    /*

	// Go up to 60 to be large enough for 32bit and 64bit platforms.
	/*
		seq 60 | sed 's/^/5^/' | bc |
		awk 'BEGIN{ print "\t{ 0, \"\" }," }
		{
			log2 = log(2)/log(10)
			printf("\t{ %d, \"%s\" },\t// * %d\n",
				int(log2*NR+1), $0, 2**NR)
		}'
	*/
	{0, ""},
	{1, "5"},                                           // * 2
	{1, "25"},                                          // * 4

		var r nat
		index := len(table) - 1
		for len(q) > leafSize {
			// find divisor close to sqrt(q) if possible, but in any case < q
			maxLength := q.bitLen()     // ~= log2 q, or at of least largest possible q of this bit length
			minLength := maxLength >> 1 // ~= log2 sqrt(q)
			for index > 0 && table[index-1].nbits > minLength {
				index-- // desired
			}
			if table[index].nbits >= maxLength && table[index].bbb.cmp(q) >= 0 {

	"image/draw"
	"internal/byteorder"
	"io"
)

// Graphic control extension fields.
const (
	gcLabel     = 0xF9
	gcBlockSize = 0x04
)

var log2Lookup = [8]int{2, 4, 8, 16, 32, 64, 128, 256}

func log2(x int) int {
	for i, v := range log2Lookup {
		if x <= v {
			return i
		}
	}
	return -1
}

// writer is a buffered writer.
type writer interface {
	Flush() error
	io.Writer
	io.ByteWriter
}

// Rearranging,
//   2^e > x * delta * c
// x can be at most 2^n-1 and delta can be at most 1.
// So it is sufficient to have 2^e >= 2^n*c.
// So we'll choose e = n + s, with s = ⎡log2(c)⎤.
//
// An additional complication arises because m has n+1 bits in it.
// Hardware restricts us to n bit by n bit multiplies.
// We divide into 3 cases:
//
// Case 1: m is even.
//   ⎣x / c⎦ = ⎣x * m / 2^(n+s)⎦

				// which makes regenerating this test very annoying, since it will
				// likely break this test. Resolve this by making the order not matter.
				{Task: 12, Category: "log2", Message: "do"},
				{Task: 12, Category: "log", Message: "fanout region4"},
				{Task: 12, Category: "log", Message: "fanout region0"},
				{Task: 12, Category: "log", Message: "fanout region1"},

	// at most 2^(exp-mantbits).
	//
	// So the number is already shortest if 10^(dp-nd) > 2^(exp-mantbits),
	// or equivalently log2(10)*(dp-nd) > exp-mantbits.
	// It is true if 332/100*(dp-nd) >= exp-mantbits (log2(10) > 3.32).
	minexp := flt.bias + 1 // minimum possible exponent
	if exp > minexp && 332*(d.dp-d.nd) >= 100*(exp-int(flt.mantbits)) {
		// The number is already shortest.
		return