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JDK 11 jdk.scripting.nashorn.jmod - Scripting Nashorn Module
JDK 11 jdk.scripting.nashorn.jmod is the JMOD file for JDK 11 Scripting Nashorn module.
JDK 11 Scripting Nashorn module compiled class files are stored in \fyicenter\jdk-11.0.1\jmods\jdk.scripting.nashorn.jmod.
JDK 11 Scripting Nashorn module compiled class files are also linked and stored in the \fyicenter\jdk-11.0.1\lib\modules JImage file.
JDK 11 Scripting Nashorn module source code files are stored in \fyicenter\jdk-11.0.1\lib\src.zip\jdk.scripting.nashorn.
You can click and view the content of each source code file in the list below.
✍: FYIcenter
⏎ jdk/nashorn/internal/runtime/doubleconv/BignumDtoa.java
/* * Copyright (c) 2015, Oracle and/or its affiliates. All rights reserved. * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms. * * * * * * * * * * * * * * * * * * * * */ // // // // // // Copyright 2010 the V8 project authors. All rights reserved. // Redistribution and use in source and binary forms, with or without // modification, are permitted provided that the following conditions are // met: // // * Redistributions of source code must retain the above copyright // notice, this list of conditions and the following disclaimer. // * Redistributions in binary form must reproduce the above // copyright notice, this list of conditions and the following // disclaimer in the documentation and/or other materials provided // with the distribution. // * Neither the name of Google Inc. nor the names of its // contributors may be used to endorse or promote products derived // from this software without specific prior written permission. // // THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS // "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT // LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR // A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT // OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, // SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT // LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, // DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY // THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT // (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE // OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. package jdk.nashorn.internal.runtime.doubleconv; // Dtoa implementation based on our own Bignum implementation, supporting // all conversion modes but slightly slower than the specialized implementations. class BignumDtoa { private static int normalizedExponent(long significand, int exponent) { assert (significand != 0); while ((significand & IeeeDouble.kHiddenBit) == 0) { significand = significand << 1; exponent = exponent - 1; } return exponent; } // Converts the given double 'v' to ascii. // The result should be interpreted as buffer * 10^(point-length). // The buffer will be null-terminated. // // The input v must be > 0 and different from NaN, and Infinity. // // The output depends on the given mode: // - SHORTEST: produce the least amount of digits for which the internal // identity requirement is still satisfied. If the digits are printed // (together with the correct exponent) then reading this number will give // 'v' again. The buffer will choose the representation that is closest to // 'v'. If there are two at the same distance, than the number is round up. // In this mode the 'requested_digits' parameter is ignored. // - FIXED: produces digits necessary to print a given number with // 'requested_digits' digits after the decimal point. The produced digits // might be too short in which case the caller has to fill the gaps with '0's. // Example: toFixed(0.001, 5) is allowed to return buffer="1", point=-2. // Halfway cases are rounded up. The call toFixed(0.15, 2) thus returns // buffer="2", point=0. // Note: the length of the returned buffer has no meaning wrt the significance // of its digits. That is, just because it contains '0's does not mean that // any other digit would not satisfy the internal identity requirement. // - PRECISION: produces 'requested_digits' where the first digit is not '0'. // Even though the length of produced digits usually equals // 'requested_digits', the function is allowed to return fewer digits, in // which case the caller has to fill the missing digits with '0's. // Halfway cases are again rounded up. // 'BignumDtoa' expects the given buffer to be big enough to hold all digits // and a terminating null-character. static void bignumDtoa(final double v, final DtoaMode mode, final int requested_digits, final DtoaBuffer buffer) { assert (v > 0); assert (!IeeeDouble.isSpecial(IeeeDouble.doubleToLong(v))); final long significand; final int exponent; final boolean lower_boundary_is_closer; final long l = IeeeDouble.doubleToLong(v); significand = IeeeDouble.significand(l); exponent = IeeeDouble.exponent(l); lower_boundary_is_closer = IeeeDouble.lowerBoundaryIsCloser(l); final boolean need_boundary_deltas = mode == DtoaMode.SHORTEST; final boolean is_even = (significand & 1) == 0; assert (significand != 0); final int normalizedExponent = normalizedExponent(significand, exponent); // estimated_power might be too low by 1. final int estimated_power = estimatePower(normalizedExponent); // Shortcut for Fixed. // The requested digits correspond to the digits after the point. If the // number is much too small, then there is no need in trying to get any // digits. if (mode == DtoaMode.FIXED && -estimated_power - 1 > requested_digits) { buffer.reset(); // Set decimal-point to -requested_digits. This is what Gay does. // Note that it should not have any effect anyways since the string is // empty. buffer.decimalPoint = -requested_digits; return; } final Bignum numerator = new Bignum(); final Bignum denominator = new Bignum(); final Bignum delta_minus = new Bignum(); final Bignum delta_plus = new Bignum(); // Make sure the bignum can grow large enough. The smallest double equals // 4e-324. In this case the denominator needs fewer than 324*4 binary digits. // The maximum double is 1.7976931348623157e308 which needs fewer than // 308*4 binary digits. assert (Bignum.kMaxSignificantBits >= 324*4); initialScaledStartValues(significand, exponent, lower_boundary_is_closer, estimated_power, need_boundary_deltas, numerator, denominator, delta_minus, delta_plus); // We now have v = (numerator / denominator) * 10^estimated_power. buffer.decimalPoint = fixupMultiply10(estimated_power, is_even, numerator, denominator, delta_minus, delta_plus); // We now have v = (numerator / denominator) * 10^(decimal_point-1), and // 1 <= (numerator + delta_plus) / denominator < 10 switch (mode) { case SHORTEST: generateShortestDigits(numerator, denominator, delta_minus, delta_plus, is_even, buffer); break; case FIXED: bignumToFixed(requested_digits, numerator, denominator, buffer); break; case PRECISION: generateCountedDigits(requested_digits, numerator, denominator, buffer); break; default: throw new RuntimeException(); } } // The procedure starts generating digits from the left to the right and stops // when the generated digits yield the shortest decimal representation of v. A // decimal representation of v is a number lying closer to v than to any other // double, so it converts to v when read. // // This is true if d, the decimal representation, is between m- and m+, the // upper and lower boundaries. d must be strictly between them if !is_even. // m- := (numerator - delta_minus) / denominator // m+ := (numerator + delta_plus) / denominator // // Precondition: 0 <= (numerator+delta_plus) / denominator < 10. // If 1 <= (numerator+delta_plus) / denominator < 10 then no leading 0 digit // will be produced. This should be the standard precondition. static void generateShortestDigits(final Bignum numerator, final Bignum denominator, final Bignum delta_minus, Bignum delta_plus, final boolean is_even, final DtoaBuffer buffer) { // Small optimization: if delta_minus and delta_plus are the same just reuse // one of the two bignums. if (Bignum.equal(delta_minus, delta_plus)) { delta_plus = delta_minus; } for (;;) { final char digit; digit = numerator.divideModuloIntBignum(denominator); assert (digit <= 9); // digit is a uint16_t and therefore always positive. // digit = numerator / denominator (integer division). // numerator = numerator % denominator. buffer.append((char) (digit + '0')); // Can we stop already? // If the remainder of the division is less than the distance to the lower // boundary we can stop. In this case we simply round down (discarding the // remainder). // Similarly we test if we can round up (using the upper boundary). final boolean in_delta_room_minus; final boolean in_delta_room_plus; if (is_even) { in_delta_room_minus = Bignum.lessEqual(numerator, delta_minus); } else { in_delta_room_minus = Bignum.less(numerator, delta_minus); } if (is_even) { in_delta_room_plus = Bignum.plusCompare(numerator, delta_plus, denominator) >= 0; } else { in_delta_room_plus = Bignum.plusCompare(numerator, delta_plus, denominator) > 0; } if (!in_delta_room_minus && !in_delta_room_plus) { // Prepare for next iteration. numerator.times10(); delta_minus.times10(); // We optimized delta_plus to be equal to delta_minus (if they share the // same value). So don't multiply delta_plus if they point to the same // object. if (delta_minus != delta_plus) { delta_plus.times10(); } } else if (in_delta_room_minus && in_delta_room_plus) { // Let's see if 2*numerator < denominator. // If yes, then the next digit would be < 5 and we can round down. final int compare = Bignum.plusCompare(numerator, numerator, denominator); if (compare < 0) { // Remaining digits are less than .5. -> Round down (== do nothing). } else if (compare > 0) { // Remaining digits are more than .5 of denominator. -> Round up. // Note that the last digit could not be a '9' as otherwise the whole // loop would have stopped earlier. // We still have an assert here in case the preconditions were not // satisfied. assert (buffer.chars[buffer.length - 1] != '9'); buffer.chars[buffer.length - 1]++; } else { // Halfway case. // TODO(floitsch): need a way to solve half-way cases. // For now let's round towards even (since this is what Gay seems to // do). if ((buffer.chars[buffer.length - 1] - '0') % 2 == 0) { // Round down => Do nothing. } else { assert (buffer.chars[buffer.length - 1] != '9'); buffer.chars[buffer.length - 1]++; } } return; } else if (in_delta_room_minus) { // Round down (== do nothing). return; } else { // in_delta_room_plus // Round up. // Note again that the last digit could not be '9' since this would have // stopped the loop earlier. // We still have an ASSERT here, in case the preconditions were not // satisfied. assert (buffer.chars[buffer.length -1] != '9'); buffer.chars[buffer.length - 1]++; return; } } } // Let v = numerator / denominator < 10. // Then we generate 'count' digits of d = x.xxxxx... (without the decimal point) // from left to right. Once 'count' digits have been produced we decide wether // to round up or down. Remainders of exactly .5 round upwards. Numbers such // as 9.999999 propagate a carry all the way, and change the // exponent (decimal_point), when rounding upwards. static void generateCountedDigits(final int count, final Bignum numerator, final Bignum denominator, final DtoaBuffer buffer) { assert (count >= 0); for (int i = 0; i < count - 1; ++i) { final char digit; digit = numerator.divideModuloIntBignum(denominator); assert (digit <= 9); // digit is a uint16_t and therefore always positive. // digit = numerator / denominator (integer division). // numerator = numerator % denominator. buffer.chars[i] = (char)(digit + '0'); // Prepare for next iteration. numerator.times10(); } // Generate the last digit. char digit; digit = numerator.divideModuloIntBignum(denominator); if (Bignum.plusCompare(numerator, numerator, denominator) >= 0) { digit++; } assert (digit <= 10); buffer.chars[count - 1] = (char) (digit + '0'); // Correct bad digits (in case we had a sequence of '9's). Propagate the // carry until we hat a non-'9' or til we reach the first digit. for (int i = count - 1; i > 0; --i) { if (buffer.chars[i] != '0' + 10) break; buffer.chars[i] = '0'; buffer.chars[i - 1]++; } if (buffer.chars[0] == '0' + 10) { // Propagate a carry past the top place. buffer.chars[0] = '1'; buffer.decimalPoint++; } buffer.length = count; } // Generates 'requested_digits' after the decimal point. It might omit // trailing '0's. If the input number is too small then no digits at all are // generated (ex.: 2 fixed digits for 0.00001). // // Input verifies: 1 <= (numerator + delta) / denominator < 10. static void bignumToFixed(final int requested_digits, final Bignum numerator, final Bignum denominator, final DtoaBuffer buffer) { // Note that we have to look at more than just the requested_digits, since // a number could be rounded up. Example: v=0.5 with requested_digits=0. // Even though the power of v equals 0 we can't just stop here. if (-buffer.decimalPoint > requested_digits) { // The number is definitively too small. // Ex: 0.001 with requested_digits == 1. // Set decimal-decimalPoint to -requested_digits. This is what Gay does. // Note that it should not have any effect anyways since the string is // empty. buffer.decimalPoint = -requested_digits; buffer.length = 0; // return; } else if (-buffer.decimalPoint == requested_digits) { // We only need to verify if the number rounds down or up. // Ex: 0.04 and 0.06 with requested_digits == 1. assert (buffer.decimalPoint == -requested_digits); // Initially the fraction lies in range (1, 10]. Multiply the denominator // by 10 so that we can compare more easily. denominator.times10(); if (Bignum.plusCompare(numerator, numerator, denominator) >= 0) { // If the fraction is >= 0.5 then we have to include the rounded // digit. buffer.chars[0] = '1'; buffer.length = 1; buffer.decimalPoint++; } else { // Note that we caught most of similar cases earlier. buffer.length = 0; } // return; } else { // The requested digits correspond to the digits after the point. // The variable 'needed_digits' includes the digits before the point. final int needed_digits = buffer.decimalPoint + requested_digits; generateCountedDigits(needed_digits, numerator, denominator, buffer); } } // Returns an estimation of k such that 10^(k-1) <= v < 10^k where // v = f * 2^exponent and 2^52 <= f < 2^53. // v is hence a normalized double with the given exponent. The output is an // approximation for the exponent of the decimal approimation .digits * 10^k. // // The result might undershoot by 1 in which case 10^k <= v < 10^k+1. // Note: this property holds for v's upper boundary m+ too. // 10^k <= m+ < 10^k+1. // (see explanation below). // // Examples: // EstimatePower(0) => 16 // EstimatePower(-52) => 0 // // Note: e >= 0 => EstimatedPower(e) > 0. No similar claim can be made for e<0. static int estimatePower(final int exponent) { // This function estimates log10 of v where v = f*2^e (with e == exponent). // Note that 10^floor(log10(v)) <= v, but v <= 10^ceil(log10(v)). // Note that f is bounded by its container size. Let p = 53 (the double's // significand size). Then 2^(p-1) <= f < 2^p. // // Given that log10(v) == log2(v)/log2(10) and e+(len(f)-1) is quite close // to log2(v) the function is simplified to (e+(len(f)-1)/log2(10)). // The computed number undershoots by less than 0.631 (when we compute log3 // and not log10). // // Optimization: since we only need an approximated result this computation // can be performed on 64 bit integers. On x86/x64 architecture the speedup is // not really measurable, though. // // Since we want to avoid overshooting we decrement by 1e10 so that // floating-point imprecisions don't affect us. // // Explanation for v's boundary m+: the computation takes advantage of // the fact that 2^(p-1) <= f < 2^p. Boundaries still satisfy this requirement // (even for denormals where the delta can be much more important). final double k1Log10 = 0.30102999566398114; // 1/lg(10) // For doubles len(f) == 53 (don't forget the hidden bit). final int kSignificandSize = IeeeDouble.kSignificandSize; final double estimate = Math.ceil((exponent + kSignificandSize - 1) * k1Log10 - 1e-10); return (int) estimate; } // See comments for InitialScaledStartValues. static void initialScaledStartValuesPositiveExponent( final long significand, final int exponent, final int estimated_power, final boolean need_boundary_deltas, final Bignum numerator, final Bignum denominator, final Bignum delta_minus, final Bignum delta_plus) { // A positive exponent implies a positive power. assert (estimated_power >= 0); // Since the estimated_power is positive we simply multiply the denominator // by 10^estimated_power. // numerator = v. numerator.assignUInt64(significand); numerator.shiftLeft(exponent); // denominator = 10^estimated_power. denominator.assignPowerUInt16(10, estimated_power); if (need_boundary_deltas) { // Introduce a common denominator so that the deltas to the boundaries are // integers. denominator.shiftLeft(1); numerator.shiftLeft(1); // Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common // denominator (of 2) delta_plus equals 2^e. delta_plus.assignUInt16((char) 1); delta_plus.shiftLeft(exponent); // Same for delta_minus. The adjustments if f == 2^p-1 are done later. delta_minus.assignUInt16((char) 1); delta_minus.shiftLeft(exponent); } } // See comments for InitialScaledStartValues static void initialScaledStartValuesNegativeExponentPositivePower( final long significand, final int exponent, final int estimated_power, final boolean need_boundary_deltas, final Bignum numerator, final Bignum denominator, final Bignum delta_minus, final Bignum delta_plus) { // v = f * 2^e with e < 0, and with estimated_power >= 0. // This means that e is close to 0 (have a look at how estimated_power is // computed). // numerator = significand // since v = significand * 2^exponent this is equivalent to // numerator = v * / 2^-exponent numerator.assignUInt64(significand); // denominator = 10^estimated_power * 2^-exponent (with exponent < 0) denominator.assignPowerUInt16(10, estimated_power); denominator.shiftLeft(-exponent); if (need_boundary_deltas) { // Introduce a common denominator so that the deltas to the boundaries are // integers. denominator.shiftLeft(1); numerator.shiftLeft(1); // Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common // denominator (of 2) delta_plus equals 2^e. // Given that the denominator already includes v's exponent the distance // to the boundaries is simply 1. delta_plus.assignUInt16((char) 1); // Same for delta_minus. The adjustments if f == 2^p-1 are done later. delta_minus.assignUInt16((char) 1); } } // See comments for InitialScaledStartValues static void initialScaledStartValuesNegativeExponentNegativePower( final long significand, final int exponent, final int estimated_power, final boolean need_boundary_deltas, final Bignum numerator, final Bignum denominator, final Bignum delta_minus, final Bignum delta_plus) { // Instead of multiplying the denominator with 10^estimated_power we // multiply all values (numerator and deltas) by 10^-estimated_power. // Use numerator as temporary container for power_ten. final Bignum power_ten = numerator; power_ten.assignPowerUInt16(10, -estimated_power); if (need_boundary_deltas) { // Since power_ten == numerator we must make a copy of 10^estimated_power // before we complete the computation of the numerator. // delta_plus = delta_minus = 10^estimated_power delta_plus.assignBignum(power_ten); delta_minus.assignBignum(power_ten); } // numerator = significand * 2 * 10^-estimated_power // since v = significand * 2^exponent this is equivalent to // numerator = v * 10^-estimated_power * 2 * 2^-exponent. // Remember: numerator has been abused as power_ten. So no need to assign it // to itself. assert (numerator == power_ten); numerator.multiplyByUInt64(significand); // denominator = 2 * 2^-exponent with exponent < 0. denominator.assignUInt16((char) 1); denominator.shiftLeft(-exponent); if (need_boundary_deltas) { // Introduce a common denominator so that the deltas to the boundaries are // integers. numerator.shiftLeft(1); denominator.shiftLeft(1); // With this shift the boundaries have their correct value, since // delta_plus = 10^-estimated_power, and // delta_minus = 10^-estimated_power. // These assignments have been done earlier. // The adjustments if f == 2^p-1 (lower boundary is closer) are done later. } } // Let v = significand * 2^exponent. // Computes v / 10^estimated_power exactly, as a ratio of two bignums, numerator // and denominator. The functions GenerateShortestDigits and // GenerateCountedDigits will then convert this ratio to its decimal // representation d, with the required accuracy. // Then d * 10^estimated_power is the representation of v. // (Note: the fraction and the estimated_power might get adjusted before // generating the decimal representation.) // // The initial start values consist of: // - a scaled numerator: s.t. numerator/denominator == v / 10^estimated_power. // - a scaled (common) denominator. // optionally (used by GenerateShortestDigits to decide if it has the shortest // decimal converting back to v): // - v - m-: the distance to the lower boundary. // - m+ - v: the distance to the upper boundary. // // v, m+, m-, and therefore v - m- and m+ - v all share the same denominator. // // Let ep == estimated_power, then the returned values will satisfy: // v / 10^ep = numerator / denominator. // v's boundarys m- and m+: // m- / 10^ep == v / 10^ep - delta_minus / denominator // m+ / 10^ep == v / 10^ep + delta_plus / denominator // Or in other words: // m- == v - delta_minus * 10^ep / denominator; // m+ == v + delta_plus * 10^ep / denominator; // // Since 10^(k-1) <= v < 10^k (with k == estimated_power) // or 10^k <= v < 10^(k+1) // we then have 0.1 <= numerator/denominator < 1 // or 1 <= numerator/denominator < 10 // // It is then easy to kickstart the digit-generation routine. // // The boundary-deltas are only filled if the mode equals BIGNUM_DTOA_SHORTEST // or BIGNUM_DTOA_SHORTEST_SINGLE. static void initialScaledStartValues(final long significand, final int exponent, final boolean lower_boundary_is_closer, final int estimated_power, final boolean need_boundary_deltas, final Bignum numerator, final Bignum denominator, final Bignum delta_minus, final Bignum delta_plus) { if (exponent >= 0) { initialScaledStartValuesPositiveExponent( significand, exponent, estimated_power, need_boundary_deltas, numerator, denominator, delta_minus, delta_plus); } else if (estimated_power >= 0) { initialScaledStartValuesNegativeExponentPositivePower( significand, exponent, estimated_power, need_boundary_deltas, numerator, denominator, delta_minus, delta_plus); } else { initialScaledStartValuesNegativeExponentNegativePower( significand, exponent, estimated_power, need_boundary_deltas, numerator, denominator, delta_minus, delta_plus); } if (need_boundary_deltas && lower_boundary_is_closer) { // The lower boundary is closer at half the distance of "normal" numbers. // Increase the common denominator and adapt all but the delta_minus. denominator.shiftLeft(1); // *2 numerator.shiftLeft(1); // *2 delta_plus.shiftLeft(1); // *2 } } // This routine multiplies numerator/denominator so that its values lies in the // range 1-10. That is after a call to this function we have: // 1 <= (numerator + delta_plus) /denominator < 10. // Let numerator the input before modification and numerator' the argument // after modification, then the output-parameter decimal_point is such that // numerator / denominator * 10^estimated_power == // numerator' / denominator' * 10^(decimal_point - 1) // In some cases estimated_power was too low, and this is already the case. We // then simply adjust the power so that 10^(k-1) <= v < 10^k (with k == // estimated_power) but do not touch the numerator or denominator. // Otherwise the routine multiplies the numerator and the deltas by 10. static int fixupMultiply10(final int estimated_power, final boolean is_even, final Bignum numerator, final Bignum denominator, final Bignum delta_minus, final Bignum delta_plus) { final boolean in_range; final int decimal_point; if (is_even) { // For IEEE doubles half-way cases (in decimal system numbers ending with 5) // are rounded to the closest floating-point number with even significand. in_range = Bignum.plusCompare(numerator, delta_plus, denominator) >= 0; } else { in_range = Bignum.plusCompare(numerator, delta_plus, denominator) > 0; } if (in_range) { // Since numerator + delta_plus >= denominator we already have // 1 <= numerator/denominator < 10. Simply update the estimated_power. decimal_point = estimated_power + 1; } else { decimal_point = estimated_power; numerator.times10(); if (Bignum.equal(delta_minus, delta_plus)) { delta_minus.times10(); delta_plus.assignBignum(delta_minus); } else { delta_minus.times10(); delta_plus.times10(); } } return decimal_point; } }
⏎ jdk/nashorn/internal/runtime/doubleconv/BignumDtoa.java
Or download all of them as a single archive file:
File name: jdk.scripting.nashorn-11.0.1-src.zip File size: 1390965 bytes Release date: 2018-11-04 Download
⇒ JDK 11 jdk.scripting.nashorn.shell.jmod - Scripting Nashorn Shell Module
2020-04-25, 108165👍, 0💬
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