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// This file is available under and governed by the GNU General Public
// License version 2 only, as published by the Free Software Foundation.
// However, the following notice accompanied the original version of this
// file:
//
// Copyright 2010 the V8 project authors. All rights reserved.
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// 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.
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package jdk.nashorn.internal.runtime.doubleconv;
import java.util.Arrays;
class Bignum {
// 3584 = 128 * 28. We can represent 2^3584 > 10^1000 accurately.
// This bignum can encode much bigger numbers, since it contains an
// exponent.
static final int kMaxSignificantBits = 3584;
static final int kChunkSize = 32; // size of int
static final int kDoubleChunkSize = 64; // size of long
// With bigit size of 28 we loose some bits, but a double still fits easily
// into two ints, and more importantly we can use the Comba multiplication.
static final int kBigitSize = 28;
static final int kBigitMask = (1 << kBigitSize) - 1;
// Every instance allocates kbigitLength ints on the stack. Bignums cannot
// grow. There are no checks if the stack-allocated space is sufficient.
static final int kBigitCapacity = kMaxSignificantBits / kBigitSize;
private int used_digits_;
// The Bignum's value equals value(bigits_) * 2^(exponent_ * kBigitSize).
private int exponent_;
private final int[] bigits_ = new int[kBigitCapacity];
Bignum() {}
void times10() { multiplyByUInt32(10); }
static boolean equal(final Bignum a, final Bignum b) {
return compare(a, b) == 0;
}
static boolean lessEqual(final Bignum a, final Bignum b) {
return compare(a, b) <= 0;
}
static boolean less(final Bignum a, final Bignum b) {
return compare(a, b) < 0;
}
// Returns a + b == c
static boolean plusEqual(final Bignum a, final Bignum b, final Bignum c) {
return plusCompare(a, b, c) == 0;
}
// Returns a + b <= c
static boolean plusLessEqual(final Bignum a, final Bignum b, final Bignum c) {
return plusCompare(a, b, c) <= 0;
}
// Returns a + b < c
static boolean plusLess(final Bignum a, final Bignum b, final Bignum c) {
return plusCompare(a, b, c) < 0;
}
private void ensureCapacity(final int size) {
if (size > kBigitCapacity) {
throw new RuntimeException();
}
}
// BigitLength includes the "hidden" digits encoded in the exponent.
int bigitLength() { return used_digits_ + exponent_; }
// Guaranteed to lie in one Bigit.
void assignUInt16(final char value) {
assert (kBigitSize >= 16);
zero();
if (value == 0) {
return;
}
ensureCapacity(1);
bigits_[0] = value;
used_digits_ = 1;
}
void assignUInt64(long value) {
final int kUInt64Size = 64;
zero();
if (value == 0) {
return;
}
final int needed_bigits = kUInt64Size / kBigitSize + 1;
ensureCapacity(needed_bigits);
for (int i = 0; i < needed_bigits; ++i) {
bigits_[i] = (int) (value & kBigitMask);
value = value >>> kBigitSize;
}
used_digits_ = needed_bigits;
clamp();
}
void assignBignum(final Bignum other) {
exponent_ = other.exponent_;
for (int i = 0; i < other.used_digits_; ++i) {
bigits_[i] = other.bigits_[i];
}
// Clear the excess digits (if there were any).
for (int i = other.used_digits_; i < used_digits_; ++i) {
bigits_[i] = 0;
}
used_digits_ = other.used_digits_;
}
static long readUInt64(final String str,
final int from,
final int digits_to_read) {
long result = 0;
for (int i = from; i < from + digits_to_read; ++i) {
final int digit = str.charAt(i) - '0';
assert (0 <= digit && digit <= 9);
result = result * 10 + digit;
}
return result;
}
void assignDecimalString(final String str) {
// 2^64 = 18446744073709551616 > 10^19
final int kMaxUint64DecimalDigits = 19;
zero();
int length = str.length();
int pos = 0;
// Let's just say that each digit needs 4 bits.
while (length >= kMaxUint64DecimalDigits) {
final long digits = readUInt64(str, pos, kMaxUint64DecimalDigits);
pos += kMaxUint64DecimalDigits;
length -= kMaxUint64DecimalDigits;
multiplyByPowerOfTen(kMaxUint64DecimalDigits);
addUInt64(digits);
}
final long digits = readUInt64(str, pos, length);
multiplyByPowerOfTen(length);
addUInt64(digits);
clamp();
}
static int hexCharValue(final char c) {
if ('0' <= c && c <= '9') return c - '0';
if ('a' <= c && c <= 'f') return 10 + c - 'a';
assert ('A' <= c && c <= 'F');
return 10 + c - 'A';
}
void assignHexString(final String str) {
zero();
final int length = str.length();
final int needed_bigits = length * 4 / kBigitSize + 1;
ensureCapacity(needed_bigits);
int string_index = length - 1;
for (int i = 0; i < needed_bigits - 1; ++i) {
// These bigits are guaranteed to be "full".
int current_bigit = 0;
for (int j = 0; j < kBigitSize / 4; j++) {
current_bigit += hexCharValue(str.charAt(string_index--)) << (j * 4);
}
bigits_[i] = current_bigit;
}
used_digits_ = needed_bigits - 1;
int most_significant_bigit = 0; // Could be = 0;
for (int j = 0; j <= string_index; ++j) {
most_significant_bigit <<= 4;
most_significant_bigit += hexCharValue(str.charAt(j));
}
if (most_significant_bigit != 0) {
bigits_[used_digits_] = most_significant_bigit;
used_digits_++;
}
clamp();
}
void addUInt64(final long operand) {
if (operand == 0) return;
final Bignum other = new Bignum();
other.assignUInt64(operand);
addBignum(other);
}
void addBignum(final Bignum other) {
assert (isClamped());
assert (other.isClamped());
// If this has a greater exponent than other append zero-bigits to this.
// After this call exponent_ <= other.exponent_.
align(other);
// There are two possibilities:
// aaaaaaaaaaa 0000 (where the 0s represent a's exponent)
// bbbbb 00000000
// ----------------
// ccccccccccc 0000
// or
// aaaaaaaaaa 0000
// bbbbbbbbb 0000000
// -----------------
// cccccccccccc 0000
// In both cases we might need a carry bigit.
ensureCapacity(1 + Math.max(bigitLength(), other.bigitLength()) - exponent_);
int carry = 0;
int bigit_pos = other.exponent_ - exponent_;
assert (bigit_pos >= 0);
for (int i = 0; i < other.used_digits_; ++i) {
final int sum = bigits_[bigit_pos] + other.bigits_[i] + carry;
bigits_[bigit_pos] = sum & kBigitMask;
carry = sum >>> kBigitSize;
bigit_pos++;
}
while (carry != 0) {
final int sum = bigits_[bigit_pos] + carry;
bigits_[bigit_pos] = sum & kBigitMask;
carry = sum >>> kBigitSize;
bigit_pos++;
}
used_digits_ = Math.max(bigit_pos, used_digits_);
assert (isClamped());
}
void subtractBignum(final Bignum other) {
assert (isClamped());
assert (other.isClamped());
// We require this to be bigger than other.
assert (lessEqual(other, this));
align(other);
final int offset = other.exponent_ - exponent_;
int borrow = 0;
int i;
for (i = 0; i < other.used_digits_; ++i) {
assert ((borrow == 0) || (borrow == 1));
final int difference = bigits_[i + offset] - other.bigits_[i] - borrow;
bigits_[i + offset] = difference & kBigitMask;
borrow = difference >>> (kChunkSize - 1);
}
while (borrow != 0) {
final int difference = bigits_[i + offset] - borrow;
bigits_[i + offset] = difference & kBigitMask;
borrow = difference >>> (kChunkSize - 1);
++i;
}
clamp();
}
void shiftLeft(final int shift_amount) {
if (used_digits_ == 0) return;
exponent_ += shift_amount / kBigitSize;
final int local_shift = shift_amount % kBigitSize;
ensureCapacity(used_digits_ + 1);
bigitsShiftLeft(local_shift);
}
void multiplyByUInt32(final int factor) {
if (factor == 1) return;
if (factor == 0) {
zero();
return;
}
if (used_digits_ == 0) return;
// The product of a bigit with the factor is of size kBigitSize + 32.
// Assert that this number + 1 (for the carry) fits into double int.
assert (kDoubleChunkSize >= kBigitSize + 32 + 1);
long carry = 0;
for (int i = 0; i < used_digits_; ++i) {
final long product = (factor & 0xFFFFFFFFL) * bigits_[i] + carry;
bigits_[i] = (int) (product & kBigitMask);
carry = product >>> kBigitSize;
}
while (carry != 0) {
ensureCapacity(used_digits_ + 1);
bigits_[used_digits_] = (int) (carry & kBigitMask);
used_digits_++;
carry >>>= kBigitSize;
}
}
void multiplyByUInt64(final long factor) {
if (factor == 1) return;
if (factor == 0) {
zero();
return;
}
assert (kBigitSize < 32);
long carry = 0;
final long low = factor & 0xFFFFFFFFL;
final long high = factor >>> 32;
for (int i = 0; i < used_digits_; ++i) {
final long product_low = low * bigits_[i];
final long product_high = high * bigits_[i];
final long tmp = (carry & kBigitMask) + product_low;
bigits_[i] = (int) (tmp & kBigitMask);
carry = (carry >>> kBigitSize) + (tmp >>> kBigitSize) +
(product_high << (32 - kBigitSize));
}
while (carry != 0) {
ensureCapacity(used_digits_ + 1);
bigits_[used_digits_] = (int) (carry & kBigitMask);
used_digits_++;
carry >>>= kBigitSize;
}
}
void multiplyByPowerOfTen(final int exponent) {
final long kFive27 = 0x6765c793fa10079dL;
final int kFive1 = 5;
final int kFive2 = kFive1 * 5;
final int kFive3 = kFive2 * 5;
final int kFive4 = kFive3 * 5;
final int kFive5 = kFive4 * 5;
final int kFive6 = kFive5 * 5;
final int kFive7 = kFive6 * 5;
final int kFive8 = kFive7 * 5;
final int kFive9 = kFive8 * 5;
final int kFive10 = kFive9 * 5;
final int kFive11 = kFive10 * 5;
final int kFive12 = kFive11 * 5;
final int kFive13 = kFive12 * 5;
final int kFive1_to_12[] =
{ kFive1, kFive2, kFive3, kFive4, kFive5, kFive6,
kFive7, kFive8, kFive9, kFive10, kFive11, kFive12 };
assert (exponent >= 0);
if (exponent == 0) return;
if (used_digits_ == 0) return;
// We shift by exponent at the end just before returning.
int remaining_exponent = exponent;
while (remaining_exponent >= 27) {
multiplyByUInt64(kFive27);
remaining_exponent -= 27;
}
while (remaining_exponent >= 13) {
multiplyByUInt32(kFive13);
remaining_exponent -= 13;
}
if (remaining_exponent > 0) {
multiplyByUInt32(kFive1_to_12[remaining_exponent - 1]);
}
shiftLeft(exponent);
}
void square() {
assert (isClamped());
final int product_length = 2 * used_digits_;
ensureCapacity(product_length);
// Comba multiplication: compute each column separately.
// Example: r = a2a1a0 * b2b1b0.
// r = 1 * a0b0 +
// 10 * (a1b0 + a0b1) +
// 100 * (a2b0 + a1b1 + a0b2) +
// 1000 * (a2b1 + a1b2) +
// 10000 * a2b2
//
// In the worst case we have to accumulate nb-digits products of digit*digit.
//
// Assert that the additional number of bits in a DoubleChunk are enough to
// sum up used_digits of Bigit*Bigit.
if ((1L << (2 * (kChunkSize - kBigitSize))) <= used_digits_) {
throw new RuntimeException("unimplemented");
}
long accumulator = 0;
// First shift the digits so we don't overwrite them.
final int copy_offset = used_digits_;
for (int i = 0; i < used_digits_; ++i) {
bigits_[copy_offset + i] = bigits_[i];
}
// We have two loops to avoid some 'if's in the loop.
for (int i = 0; i < used_digits_; ++i) {
// Process temporary digit i with power i.
// The sum of the two indices must be equal to i.
int bigit_index1 = i;
int bigit_index2 = 0;
// Sum all of the sub-products.
while (bigit_index1 >= 0) {
final int int1 = bigits_[copy_offset + bigit_index1];
final int int2 = bigits_[copy_offset + bigit_index2];
accumulator += ((long) int1) * int2;
bigit_index1--;
bigit_index2++;
}
bigits_[i] = (int) (accumulator & kBigitMask);
accumulator >>>= kBigitSize;
}
for (int i = used_digits_; i < product_length; ++i) {
int bigit_index1 = used_digits_ - 1;
int bigit_index2 = i - bigit_index1;
// Invariant: sum of both indices is again equal to i.
// Inner loop runs 0 times on last iteration, emptying accumulator.
while (bigit_index2 < used_digits_) {
final int int1 = bigits_[copy_offset + bigit_index1];
final int int2 = bigits_[copy_offset + bigit_index2];
accumulator += ((long) int1) * int2;
bigit_index1--;
bigit_index2++;
}
// The overwritten bigits_[i] will never be read in further loop iterations,
// because bigit_index1 and bigit_index2 are always greater
// than i - used_digits_.
bigits_[i] = (int) (accumulator & kBigitMask);
accumulator >>>= kBigitSize;
}
// Since the result was guaranteed to lie inside the number the
// accumulator must be 0 now.
assert (accumulator == 0);
// Don't forget to update the used_digits and the exponent.
used_digits_ = product_length;
exponent_ *= 2;
clamp();
}
void assignPowerUInt16(int base, final int power_exponent) {
assert (base != 0);
assert (power_exponent >= 0);
if (power_exponent == 0) {
assignUInt16((char) 1);
return;
}
zero();
int shifts = 0;
// We expect base to be in range 2-32, and most often to be 10.
// It does not make much sense to implement different algorithms for counting
// the bits.
while ((base & 1) == 0) {
base >>>= 1;
shifts++;
}
int bit_size = 0;
int tmp_base = base;
while (tmp_base != 0) {
tmp_base >>>= 1;
bit_size++;
}
final int final_size = bit_size * power_exponent;
// 1 extra bigit for the shifting, and one for rounded final_size.
ensureCapacity(final_size / kBigitSize + 2);
// Left to Right exponentiation.
int mask = 1;
while (power_exponent >= mask) mask <<= 1;
// The mask is now pointing to the bit above the most significant 1-bit of
// power_exponent.
// Get rid of first 1-bit;
mask >>>= 2;
long this_value = base;
boolean delayed_multiplication = false;
final long max_32bits = 0xFFFFFFFFL;
while (mask != 0 && this_value <= max_32bits) {
this_value = this_value * this_value;
// Verify that there is enough space in this_value to perform the
// multiplication. The first bit_size bits must be 0.
if ((power_exponent & mask) != 0) {
assert bit_size > 0;
final long base_bits_mask =
~((1L << (64 - bit_size)) - 1);
final boolean high_bits_zero = (this_value & base_bits_mask) == 0;
if (high_bits_zero) {
this_value *= base;
} else {
delayed_multiplication = true;
}
}
mask >>>= 1;
}
assignUInt64(this_value);
if (delayed_multiplication) {
multiplyByUInt32(base);
}
// Now do the same thing as a bignum.
while (mask != 0) {
square();
if ((power_exponent & mask) != 0) {
multiplyByUInt32(base);
}
mask >>>= 1;
}
// And finally add the saved shifts.
shiftLeft(shifts * power_exponent);
}
// Precondition: this/other < 16bit.
char divideModuloIntBignum(final Bignum other) {
assert (isClamped());
assert (other.isClamped());
assert (other.used_digits_ > 0);
// Easy case: if we have less digits than the divisor than the result is 0.
// Note: this handles the case where this == 0, too.
if (bigitLength() < other.bigitLength()) {
return 0;
}
align(other);
char result = 0;
// Start by removing multiples of 'other' until both numbers have the same
// number of digits.
while (bigitLength() > other.bigitLength()) {
// This naive approach is extremely inefficient if `this` divided by other
// is big. This function is implemented for doubleToString where
// the result should be small (less than 10).
assert (other.bigits_[other.used_digits_ - 1] >= ((1 << kBigitSize) / 16));
assert (bigits_[used_digits_ - 1] < 0x10000);
// Remove the multiples of the first digit.
// Example this = 23 and other equals 9. -> Remove 2 multiples.
result += (bigits_[used_digits_ - 1]);
subtractTimes(other, bigits_[used_digits_ - 1]);
}
assert (bigitLength() == other.bigitLength());
// Both bignums are at the same length now.
// Since other has more than 0 digits we know that the access to
// bigits_[used_digits_ - 1] is safe.
final int this_bigit = bigits_[used_digits_ - 1];
final int other_bigit = other.bigits_[other.used_digits_ - 1];
if (other.used_digits_ == 1) {
// Shortcut for easy (and common) case.
final int quotient = Integer.divideUnsigned(this_bigit, other_bigit);
bigits_[used_digits_ - 1] = this_bigit - other_bigit * quotient;
assert (Integer.compareUnsigned(quotient, 0x10000) < 0);
result += quotient;
clamp();
return result;
}
final int division_estimate = Integer.divideUnsigned(this_bigit, (other_bigit + 1));
assert (Integer.compareUnsigned(division_estimate, 0x10000) < 0);
result += division_estimate;
subtractTimes(other, division_estimate);
if (other_bigit * (division_estimate + 1) > this_bigit) {
// No need to even try to subtract. Even if other's remaining digits were 0
// another subtraction would be too much.
return result;
}
while (lessEqual(other, this)) {
subtractBignum(other);
result++;
}
return result;
}
static int sizeInHexChars(int number) {
assert (number > 0);
int result = 0;
while (number != 0) {
number >>>= 4;
result++;
}
return result;
}
static char hexCharOfValue(final int value) {
assert (0 <= value && value <= 16);
if (value < 10) return (char) (value + '0');
return (char) (value - 10 + 'A');
}
String toHexString() {
assert (isClamped());
// Each bigit must be printable as separate hex-character.
assert (kBigitSize % 4 == 0);
final int kHexCharsPerBigit = kBigitSize / 4;
if (used_digits_ == 0) {
return "0";
}
final int needed_chars = (bigitLength() - 1) * kHexCharsPerBigit +
sizeInHexChars(bigits_[used_digits_ - 1]);
final StringBuilder buffer = new StringBuilder(needed_chars);
buffer.setLength(needed_chars);
int string_index = needed_chars - 1;
for (int i = 0; i < exponent_; ++i) {
for (int j = 0; j < kHexCharsPerBigit; ++j) {
buffer.setCharAt(string_index--, '0');
}
}
for (int i = 0; i < used_digits_ - 1; ++i) {
int current_bigit = bigits_[i];
for (int j = 0; j < kHexCharsPerBigit; ++j) {
buffer.setCharAt(string_index--, hexCharOfValue(current_bigit & 0xF));
current_bigit >>>= 4;
}
}
// And finally the last bigit.
int most_significant_bigit = bigits_[used_digits_ - 1];
while (most_significant_bigit != 0) {
buffer.setCharAt(string_index--, hexCharOfValue(most_significant_bigit & 0xF));
most_significant_bigit >>>= 4;
}
return buffer.toString();
}
int bigitOrZero(final int index) {
if (index >= bigitLength()) return 0;
if (index < exponent_) return 0;
return bigits_[index - exponent_];
}
static int compare(final Bignum a, final Bignum b) {
assert (a.isClamped());
assert (b.isClamped());
final int bigit_length_a = a.bigitLength();
final int bigit_length_b = b.bigitLength();
if (bigit_length_a < bigit_length_b) return -1;
if (bigit_length_a > bigit_length_b) return +1;
for (int i = bigit_length_a - 1; i >= Math.min(a.exponent_, b.exponent_); --i) {
final int bigit_a = a.bigitOrZero(i);
final int bigit_b = b.bigitOrZero(i);
if (bigit_a < bigit_b) return -1;
if (bigit_a > bigit_b) return +1;
// Otherwise they are equal up to this digit. Try the next digit.
}
return 0;
}
static int plusCompare(final Bignum a, final Bignum b, final Bignum c) {
assert (a.isClamped());
assert (b.isClamped());
assert (c.isClamped());
if (a.bigitLength() < b.bigitLength()) {
return plusCompare(b, a, c);
}
if (a.bigitLength() + 1 < c.bigitLength()) return -1;
if (a.bigitLength() > c.bigitLength()) return +1;
// The exponent encodes 0-bigits. So if there are more 0-digits in 'a' than
// 'b' has digits, then the bigit-length of 'a'+'b' must be equal to the one
// of 'a'.
if (a.exponent_ >= b.bigitLength() && a.bigitLength() < c.bigitLength()) {
return -1;
}
int borrow = 0;
// Starting at min_exponent all digits are == 0. So no need to compare them.
final int min_exponent = Math.min(Math.min(a.exponent_, b.exponent_), c.exponent_);
for (int i = c.bigitLength() - 1; i >= min_exponent; --i) {
final int int_a = a.bigitOrZero(i);
final int int_b = b.bigitOrZero(i);
final int int_c = c.bigitOrZero(i);
final int sum = int_a + int_b;
if (sum > int_c + borrow) {
return +1;
} else {
borrow = int_c + borrow - sum;
if (borrow > 1) return -1;
borrow <<= kBigitSize;
}
}
if (borrow == 0) return 0;
return -1;
}
void clamp() {
while (used_digits_ > 0 && bigits_[used_digits_ - 1] == 0) {
used_digits_--;
}
if (used_digits_ == 0) {
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