in rhino/src/main/java/org/mozilla/javascript/DToA.java [418:1043]
static int JS_dtoa(
double d, int mode, boolean biasUp, int ndigits, boolean[] sign, StringBuilder buf) {
/* Arguments ndigits, decpt, sign are similar to those
of ecvt and fcvt; trailing zeros are suppressed from
the returned string. If not null, *rve is set to point
to the end of the return value. If d is +-Infinity or NaN,
then *decpt is set to 9999.
mode:
0 ==> shortest string that yields d when read in
and rounded to nearest.
1 ==> like 0, but with Steele & White stopping rule;
e.g. with IEEE P754 arithmetic , mode 0 gives
1e23 whereas mode 1 gives 9.999999999999999e22.
2 ==> max(1,ndigits) significant digits. This gives a
return value similar to that of ecvt, except
that trailing zeros are suppressed.
3 ==> through ndigits past the decimal point. This
gives a return value similar to that from fcvt,
except that trailing zeros are suppressed, and
ndigits can be negative.
4-9 should give the same return values as 2-3, i.e.,
4 <= mode <= 9 ==> same return as mode
2 + (mode & 1). These modes are mainly for
debugging; often they run slower but sometimes
faster than modes 2-3.
4,5,8,9 ==> left-to-right digit generation.
6-9 ==> don't try fast floating-point estimate
(if applicable).
Values of mode other than 0-9 are treated as mode 0.
Sufficient space is allocated to the return value
to hold the suppressed trailing zeros.
*/
int b2, b5, i, ieps, ilim, ilim0, ilim1, j, j1, k, k0, m2, m5, s2, s5;
char dig;
long L;
long x;
BigInteger b, b1, delta, mlo, mhi, S;
int[] be = new int[1];
int[] bbits = new int[1];
double d2, ds, eps;
boolean spec_case, denorm, k_check, try_quick, leftright;
if ((word0(d) & Sign_bit) != 0) {
/* set sign for everything, including 0's and NaNs */
sign[0] = true;
// word0(d) &= ~Sign_bit; /* clear sign bit */
d = setWord0(d, word0(d) & ~Sign_bit);
} else sign[0] = false;
if ((word0(d) & Exp_mask) == Exp_mask) {
/* Infinity or NaN */
buf.append(((word1(d) == 0) && ((word0(d) & Frac_mask) == 0)) ? "Infinity" : "NaN");
return 9999;
}
if (d == 0) {
// no_digits:
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
}
b = d2b(d, be, bbits);
if ((i = (word0(d) >>> Exp_shift1 & (Exp_mask >> Exp_shift1))) != 0) {
d2 = setWord0(d, (word0(d) & Frac_mask1) | Exp_11);
/* log(x) ~=~ log(1.5) + (x-1.5)/1.5
* log10(x) = log(x) / log(10)
* ~=~ log(1.5)/log(10) + (x-1.5)/(1.5*log(10))
* log10(d) = (i-Bias)*log(2)/log(10) + log10(d2)
*
* This suggests computing an approximation k to log10(d) by
*
* k = (i - Bias)*0.301029995663981
* + ( (d2-1.5)*0.289529654602168 + 0.176091259055681 );
*
* We want k to be too large rather than too small.
* The error in the first-order Taylor series approximation
* is in our favor, so we just round up the constant enough
* to compensate for any error in the multiplication of
* (i - Bias) by 0.301029995663981; since |i - Bias| <= 1077,
* and 1077 * 0.30103 * 2^-52 ~=~ 7.2e-14,
* adding 1e-13 to the constant term more than suffices.
* Hence we adjust the constant term to 0.1760912590558.
* (We could get a more accurate k by invoking log10,
* but this is probably not worthwhile.)
*/
i -= Bias;
denorm = false;
} else {
/* d is denormalized */
i = bbits[0] + be[0] + (Bias + (P - 1) - 1);
x =
(i > 32)
? ((long) word0(d)) << (64 - i) | word1(d) >>> (i - 32)
: ((long) word1(d)) << (32 - i);
// d2 = x;
// word0(d2) -= 31*Exp_msk1; /* adjust exponent */
d2 = setWord0((double) x, word0((double) x) - 31 * Exp_msk1);
i -= (Bias + (P - 1) - 1) + 1;
denorm = true;
}
/* At this point d = f*2^i, where 1 <= f < 2. d2 is an approximation of f. */
ds = (d2 - 1.5) * 0.289529654602168 + 0.1760912590558 + i * 0.301029995663981;
k = (int) ds;
if (ds < 0.0 && ds != k) k--; /* want k = floor(ds) */
k_check = true;
if (k >= 0 && k <= Ten_pmax) {
if (d < tens[k]) k--;
k_check = false;
}
/* At this point floor(log10(d)) <= k <= floor(log10(d))+1.
If k_check is zero, we're guaranteed that k = floor(log10(d)). */
j = bbits[0] - i - 1;
/* At this point d = b/2^j, where b is an odd integer. */
if (j >= 0) {
b2 = 0;
s2 = j;
} else {
b2 = -j;
s2 = 0;
}
if (k >= 0) {
b5 = 0;
s5 = k;
s2 += k;
} else {
b2 -= k;
b5 = -k;
s5 = 0;
}
/* At this point d/10^k = (b * 2^b2 * 5^b5) / (2^s2 * 5^s5), where b is an odd integer,
b2 >= 0, b5 >= 0, s2 >= 0, and s5 >= 0. */
if (mode < 0 || mode > 9) mode = 0;
try_quick = true;
if (mode > 5) {
mode -= 4;
try_quick = false;
}
leftright = true;
ilim = ilim1 = 0;
switch (mode) {
case 0:
case 1:
ilim = ilim1 = -1;
i = 18;
ndigits = 0;
break;
case 2:
leftright = false;
/* fall through */
case 4:
if (ndigits <= 0) ndigits = 1;
ilim = ilim1 = i = ndigits;
break;
case 3:
leftright = false;
/* fall through */
case 5:
i = ndigits + k + 1;
ilim = i;
ilim1 = i - 1;
if (i <= 0) i = 1;
}
/* ilim is the maximum number of significant digits we want, based on k and ndigits. */
/* ilim1 is the maximum number of significant digits we want, based on k and ndigits,
when it turns out that k was computed too high by one. */
boolean fast_failed = false;
if (ilim >= 0 && ilim <= Quick_max && try_quick) {
/* Try to get by with floating-point arithmetic. */
i = 0;
d2 = d;
k0 = k;
ilim0 = ilim;
ieps = 2; /* conservative */
/* Divide d by 10^k, keeping track of the roundoff error and avoiding overflows. */
if (k > 0) {
ds = tens[k & 0xf];
j = k >> 4;
if ((j & Bletch) != 0) {
/* prevent overflows */
j &= Bletch - 1;
d /= bigtens[n_bigtens - 1];
ieps++;
}
for (; (j != 0); j >>= 1, i++)
if ((j & 1) != 0) {
ieps++;
ds *= bigtens[i];
}
d /= ds;
} else if ((j1 = -k) != 0) {
d *= tens[j1 & 0xf];
for (j = j1 >> 4; (j != 0); j >>= 1, i++)
if ((j & 1) != 0) {
ieps++;
d *= bigtens[i];
}
}
/* Check that k was computed correctly. */
if (k_check && d < 1.0 && ilim > 0) {
if (ilim1 <= 0) fast_failed = true;
else {
ilim = ilim1;
k--;
d *= 10.;
ieps++;
}
}
/* eps bounds the cumulative error. */
// eps = ieps*d + 7.0;
// word0(eps) -= (P-1)*Exp_msk1;
eps = ieps * d + 7.0;
eps = setWord0(eps, word0(eps) - (P - 1) * Exp_msk1);
if (ilim == 0) {
S = mhi = null;
d -= 5.0;
if (d > eps) {
buf.append('1');
k++;
return k + 1;
}
if (d < -eps) {
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
}
fast_failed = true;
}
if (!fast_failed) {
fast_failed = true;
if (leftright) {
/* Use Steele & White method of only
* generating digits needed.
*/
eps = 0.5 / tens[ilim - 1] - eps;
for (i = 0; ; ) {
L = (long) d;
d -= L;
buf.append((char) ('0' + L));
if (d < eps) {
return k + 1;
}
if (1.0 - d < eps) {
// goto bump_up;
char lastCh;
while (true) {
lastCh = buf.charAt(buf.length() - 1);
buf.setLength(buf.length() - 1);
if (lastCh != '9') break;
if (buf.length() == 0) {
k++;
lastCh = '0';
break;
}
}
buf.append((char) (lastCh + 1));
return k + 1;
}
if (++i >= ilim) break;
eps *= 10.0;
d *= 10.0;
}
} else {
/* Generate ilim digits, then fix them up. */
eps *= tens[ilim - 1];
for (i = 1; ; i++, d *= 10.0) {
L = (long) d;
d -= L;
buf.append((char) ('0' + L));
if (i == ilim) {
if (d > 0.5 + eps) {
// goto bump_up;
char lastCh;
while (true) {
lastCh = buf.charAt(buf.length() - 1);
buf.setLength(buf.length() - 1);
if (lastCh != '9') break;
if (buf.length() == 0) {
k++;
lastCh = '0';
break;
}
}
buf.append((char) (lastCh + 1));
return k + 1;
} else if (d < 0.5 - eps) {
stripTrailingZeroes(buf);
// while(*--s == '0') ;
// s++;
return k + 1;
}
break;
}
}
}
}
if (fast_failed) {
buf.setLength(0);
d = d2;
k = k0;
ilim = ilim0;
}
}
/* Do we have a "small" integer? */
if (be[0] >= 0 && k <= Int_max) {
/* Yes. */
ds = tens[k];
if (ndigits < 0 && ilim <= 0) {
S = mhi = null;
if (ilim < 0 || d < 5 * ds || (!biasUp && d == 5 * ds)) {
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
}
buf.append('1');
k++;
return k + 1;
}
for (i = 1; ; i++) {
L = (long) (d / ds);
d -= L * ds;
buf.append((char) ('0' + L));
if (i == ilim) {
d += d;
if ((d > ds) || (d == ds && (((L & 1) != 0) || biasUp))) {
// bump_up:
// while(*--s == '9')
// if (s == buf) {
// k++;
// *s = '0';
// break;
// }
// ++*s++;
char lastCh;
while (true) {
lastCh = buf.charAt(buf.length() - 1);
buf.setLength(buf.length() - 1);
if (lastCh != '9') break;
if (buf.length() == 0) {
k++;
lastCh = '0';
break;
}
}
buf.append((char) (lastCh + 1));
}
break;
}
d *= 10.0;
if (d == 0) break;
}
return k + 1;
}
m2 = b2;
m5 = b5;
mhi = mlo = null;
if (leftright) {
if (mode < 2) {
i = denorm ? be[0] + (Bias + (P - 1) - 1 + 1) : 1 + P - bbits[0];
/* i is 1 plus the number of trailing zero bits in d's significand. Thus,
(2^m2 * 5^m5) / (2^(s2+i) * 5^s5) = (1/2 lsb of d)/10^k. */
} else {
j = ilim - 1;
if (m5 >= j) m5 -= j;
else {
s5 += j -= m5;
b5 += j;
m5 = 0;
}
if ((i = ilim) < 0) {
m2 -= i;
i = 0;
}
/* (2^m2 * 5^m5) / (2^(s2+i) * 5^s5) = (1/2 * 10^(1-ilim))/10^k. */
}
b2 += i;
s2 += i;
mhi = BigInteger.valueOf(1);
/* (mhi * 2^m2 * 5^m5) / (2^s2 * 5^s5) = one-half of last printed (when mode >= 2) or
input (when mode < 2) significant digit, divided by 10^k. */
}
/* We still have d/10^k = (b * 2^b2 * 5^b5) / (2^s2 * 5^s5). Reduce common factors in
b2, m2, and s2 without changing the equalities. */
if (m2 > 0 && s2 > 0) {
i = (m2 < s2) ? m2 : s2;
b2 -= i;
m2 -= i;
s2 -= i;
}
/* Fold b5 into b and m5 into mhi. */
if (b5 > 0) {
if (leftright) {
if (m5 > 0) {
mhi = pow5mult(mhi, m5);
b1 = mhi.multiply(b);
b = b1;
}
if ((j = b5 - m5) != 0) b = pow5mult(b, j);
} else b = pow5mult(b, b5);
}
/* Now we have d/10^k = (b * 2^b2) / (2^s2 * 5^s5) and
(mhi * 2^m2) / (2^s2 * 5^s5) = one-half of last printed or input significant digit, divided by 10^k. */
S = BigInteger.valueOf(1);
if (s5 > 0) S = pow5mult(S, s5);
/* Now we have d/10^k = (b * 2^b2) / (S * 2^s2) and
(mhi * 2^m2) / (S * 2^s2) = one-half of last printed or input significant digit, divided by 10^k. */
/* Check for special case that d is a normalized power of 2. */
spec_case = false;
if (mode < 2) {
if ((word1(d) == 0)
&& ((word0(d) & Bndry_mask) == 0)
&& ((word0(d) & (Exp_mask & Exp_mask << 1)) != 0)) {
/* The special case. Here we want to be within a quarter of the last input
significant digit instead of one half of it when the decimal output string's value is less than d. */
b2 += Log2P;
s2 += Log2P;
spec_case = true;
}
}
/* Arrange for convenient computation of quotients:
* shift left if necessary so divisor has 4 leading 0 bits.
*
* Perhaps we should just compute leading 28 bits of S once
* and for all and pass them and a shift to quorem, so it
* can do shifts and ors to compute the numerator for q.
*/
byte[] S_bytes = S.toByteArray();
int S_hiWord = 0;
for (int idx = 0; idx < 4; idx++) {
S_hiWord = (S_hiWord << 8);
if (idx < S_bytes.length) S_hiWord |= (S_bytes[idx] & 0xFF);
}
if ((i = (((s5 != 0) ? 32 - hi0bits(S_hiWord) : 1) + s2) & 0x1f) != 0) i = 32 - i;
/* i is the number of leading zero bits in the most significant word of S*2^s2. */
if (i > 4) {
i -= 4;
b2 += i;
m2 += i;
s2 += i;
} else if (i < 4) {
i += 28;
b2 += i;
m2 += i;
s2 += i;
}
/* Now S*2^s2 has exactly four leading zero bits in its most significant word. */
if (b2 > 0) b = b.shiftLeft(b2);
if (s2 > 0) S = S.shiftLeft(s2);
/* Now we have d/10^k = b/S and
(mhi * 2^m2) / S = maximum acceptable error, divided by 10^k. */
if (k_check) {
if (b.compareTo(S) < 0) {
k--;
b = b.multiply(BigInteger.valueOf(10)); /* we botched the k estimate */
if (leftright) mhi = mhi.multiply(BigInteger.valueOf(10));
ilim = ilim1;
}
}
/* At this point 1 <= d/10^k = b/S < 10. */
if (ilim <= 0 && mode > 2) {
/* We're doing fixed-mode output and d is less than the minimum nonzero output in this mode.
Output either zero or the minimum nonzero output depending on which is closer to d. */
if ((ilim < 0)
|| ((i = b.compareTo(S = S.multiply(BigInteger.valueOf(5)))) < 0)
|| (i == 0 && !biasUp)) {
/* Always emit at least one digit. If the number appears to be zero
using the current mode, then emit one '0' digit and set decpt to 1. */
/*no_digits:
k = -1 - ndigits;
goto ret; */
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
// goto no_digits;
}
// one_digit:
buf.append('1');
k++;
return k + 1;
}
if (leftright) {
if (m2 > 0) mhi = mhi.shiftLeft(m2);
/* Compute mlo -- check for special case
* that d is a normalized power of 2.
*/
mlo = mhi;
if (spec_case) {
mhi = mlo;
mhi = mhi.shiftLeft(Log2P);
}
/* mlo/S = maximum acceptable error, divided by 10^k, if the output is less than d. */
/* mhi/S = maximum acceptable error, divided by 10^k, if the output is greater than d. */
for (i = 1; ; i++) {
BigInteger[] divResult = b.divideAndRemainder(S);
b = divResult[1];
dig = (char) (divResult[0].intValue() + '0');
/* Do we yet have the shortest decimal string
* that will round to d?
*/
j = b.compareTo(mlo);
/* j is b/S compared with mlo/S. */
delta = S.subtract(mhi);
j1 = (delta.signum() <= 0) ? 1 : b.compareTo(delta);
/* j1 is b/S compared with 1 - mhi/S. */
if ((j1 == 0) && (mode == 0) && ((word1(d) & 1) == 0)) {
if (dig == '9') {
buf.append('9');
if (roundOff(buf)) {
k++;
buf.append('1');
}
return k + 1;
// goto round_9_up;
}
if (j > 0) dig++;
buf.append(dig);
return k + 1;
}
if ((j < 0) || ((j == 0) && (mode == 0) && ((word1(d) & 1) == 0))) {
if (j1 > 0) {
/* Either dig or dig+1 would work here as the least significant decimal digit.
Use whichever would produce a decimal value closer to d. */
b = b.shiftLeft(1);
j1 = b.compareTo(S);
if (((j1 > 0) || (j1 == 0 && (((dig & 1) == 1) || biasUp)))
&& (dig++ == '9')) {
buf.append('9');
if (roundOff(buf)) {
k++;
buf.append('1');
}
return k + 1;
// goto round_9_up;
}
}
buf.append(dig);
return k + 1;
}
if (j1 > 0) {
if (dig == '9') {
/* possible if i == 1 */
// round_9_up:
// *s++ = '9';
// goto roundoff;
buf.append('9');
if (roundOff(buf)) {
k++;
buf.append('1');
}
return k + 1;
}
buf.append((char) (dig + 1));
return k + 1;
}
buf.append(dig);
if (i == ilim) break;
b = b.multiply(BigInteger.valueOf(10));
if (Objects.equals(mlo, mhi)) mlo = mhi = mhi.multiply(BigInteger.valueOf(10));
else {
mlo = mlo.multiply(BigInteger.valueOf(10));
mhi = mhi.multiply(BigInteger.valueOf(10));
}
}
} else
for (i = 1; ; i++) {
// (char)(dig = quorem(b,S) + '0');
BigInteger[] divResult = b.divideAndRemainder(S);
b = divResult[1];
dig = (char) (divResult[0].intValue() + '0');
buf.append(dig);
if (i >= ilim) break;
b = b.multiply(BigInteger.valueOf(10));
}
/* Round off last digit */
b = b.shiftLeft(1);
j = b.compareTo(S);
if ((j > 0) || (j == 0 && (((dig & 1) == 1) || biasUp))) {
// roundoff:
// while(*--s == '9')
// if (s == buf) {
// k++;
// *s++ = '1';
// goto ret;
// }
// ++*s++;
if (roundOff(buf)) {
k++;
buf.append('1');
return k + 1;
}
} else {
stripTrailingZeroes(buf);
// while(*--s == '0') ;
// s++;
}
// ret:
// Bfree(S);
// if (mhi) {
// if (mlo && mlo != mhi)
// Bfree(mlo);
// Bfree(mhi);
// }
// ret1:
// Bfree(b);
// JS_ASSERT(s < buf + bufsize);
return k + 1;
}