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|
from pypy.rlib.rarithmetic import LONG_BIT, intmask, r_uint, r_ulonglong
from pypy.rlib.rarithmetic import ovfcheck, r_longlong, widen
from pypy.rlib.debug import make_sure_not_resized
import math, sys
# note about digit sizes:
# In division, the native integer type must be able to hold
# a sign bit plus two digits plus 1 overflow bit.
#SHIFT = (LONG_BIT // 2) - 1
SHIFT = 31
MASK = int((1 << SHIFT) - 1)
FLOAT_MULTIPLIER = float(1 << SHIFT)
# Debugging digit array access.
#
# False == no checking at all
# True == check 0 <= value <= MASK
# For long multiplication, use the O(N**2) school algorithm unless
# both operands contain more than KARATSUBA_CUTOFF digits (this
# being an internal Python long digit, in base BASE).
USE_KARATSUBA = True # set to False for comparison
KARATSUBA_CUTOFF = 70
KARATSUBA_SQUARE_CUTOFF = 2 * KARATSUBA_CUTOFF
# For exponentiation, use the binary left-to-right algorithm
# unless the exponent contains more than FIVEARY_CUTOFF digits.
# In that case, do 5 bits at a time. The potential drawback is that
# a table of 2**5 intermediate results is computed.
FIVEARY_CUTOFF = 8
def mask_digit(x):
return intmask(x & MASK)
mask_digit._annspecialcase_ = 'specialize:argtype(0)'
def widen_digit(x):
if SHIFT <= 15:
return int(x)
return r_longlong(x)
class rbigint(object):
"""This is a reimplementation of longs using a list of digits."""
# XXX relace the list of ints with a list of rffi.INTs, maybe
def __init__(self, digits=None, sign=0):
if digits is None or len(digits) == 0:
digits = [0]
make_sure_not_resized(digits)
self.digits = digits
self.sign = sign
def _digit(self, x):
return widen_digit(self.digits[x])
def _setdigit(self, x, val):
val = mask_digit(val)
assert val >= 0
self.digits[x] = int(val)
_setdigit._annspecialcase_ = 'specialize:argtype(2)'
def _numdigits(self):
return len(self.digits)
def fromint(intval):
if intval < 0:
sign = -1
ival = r_uint(-intval)
elif intval > 0:
sign = 1
ival = r_uint(intval)
else:
return rbigint()
# Count the number of Python digits.
# We used to pick 5 ("big enough for anything"), but that's a
# waste of time and space given that 5*15 = 75 bits are rarely
# needed.
t = ival
ndigits = 0
while t:
ndigits += 1
t >>= SHIFT
v = rbigint([0] * ndigits, sign)
t = ival
p = 0
while t:
v._setdigit(p, t)
t >>= SHIFT
p += 1
return v
fromint = staticmethod(fromint)
def frombool(b):
if b:
return rbigint([1], 1)
return rbigint()
frombool = staticmethod(frombool)
def fromlong(l):
return rbigint(*args_from_long(l))
fromlong = staticmethod(fromlong)
def fromfloat(dval):
""" Create a new bigint object from a float """
neg = 0
if isinf(dval):
raise OverflowError
if dval < 0.0:
neg = 1
dval = -dval
frac, expo = math.frexp(dval) # dval = frac*2**expo; 0.0 <= frac < 1.0
if expo <= 0:
return rbigint()
ndig = (expo-1) // SHIFT + 1 # Number of 'digits' in result
v = rbigint([0] * ndig, 1)
frac = math.ldexp(frac, (expo-1) % SHIFT + 1)
for i in range(ndig-1, -1, -1):
bits = mask_digit(int(frac))
v._setdigit(i, bits)
frac -= float(bits)
frac = math.ldexp(frac, SHIFT)
if neg:
v.sign = -1
return v
fromfloat = staticmethod(fromfloat)
def fromrarith_int(i):
return rbigint(*args_from_rarith_int(i))
fromrarith_int._annspecialcase_ = "specialize:argtype(0)"
fromrarith_int = staticmethod(fromrarith_int)
def fromdecimalstr(s):
return _decimalstr_to_bigint(s)
fromdecimalstr = staticmethod(fromdecimalstr)
def toint(self):
"""
Get an integer from a bigint object.
Raises OverflowError if overflow occurs.
"""
x = self._touint_helper()
# Haven't lost any bits, but if the sign bit is set we're in
# trouble *unless* this is the min negative number. So,
# trouble iff sign bit set && (positive || some bit set other
# than the sign bit).
sign = self.sign
if intmask(x) < 0 and (sign > 0 or (x << 1) != 0):
raise OverflowError
return intmask(x * sign)
def tolonglong(self):
return _AsLongLong(self)
def tobool(self):
return self.sign != 0
def touint(self):
if self.sign == -1:
raise ValueError("cannot convert negative integer to unsigned int")
return self._touint_helper()
def _touint_helper(self):
x = r_uint(0)
i = self._numdigits() - 1
while i >= 0:
prev = x
x = r_uint((x << SHIFT) + self.digits[i])
if (x >> SHIFT) != prev:
raise OverflowError(
"long int too large to convert to unsigned int")
i -= 1
return x
def toulonglong(self):
if self.sign == -1:
raise ValueError("cannot convert negative integer to unsigned int")
return _AsULonglong_ignore_sign(self)
def uintmask(self):
return _AsUInt_mask(self)
def ulonglongmask(self):
"""Return r_ulonglong(self), truncating."""
return _AsULonglong_mask(self)
def tofloat(self):
return _AsDouble(self)
def _count_bits(self):
# return the number of bits in the digits
if self.sign == 0:
return 0
p = self._numdigits() - 1
bits = SHIFT * p
digit = self.digits[p]
while digit:
digit >>= 1
bits += 1
return bits
def is_odd(self):
# Note: this is a tiny optimization.
# Instead of implementing a general "get_bit" operation,
# which would be expensive for negative numbers,
# get_odd has the nice feature that it is always correct,
# no matter what the sign is (two's complement)
return self.digits[0] & 1
def format(self, digits, prefix='', suffix=''):
# 'digits' is a string whose length is the base to use,
# and where each character is the corresponding digit.
return _format(self, digits, prefix, suffix)
def repr(self):
return _format(self, BASE10, '', 'L')
def str(self):
return _format(self, BASE10)
def eq(self, other):
if (self.sign != other.sign or
self._numdigits() != other._numdigits()):
return False
i = 0
ld = self._numdigits()
while i < ld:
if self.digits[i] != other.digits[i]:
return False
i += 1
return True
def ne(self, other):
return not self.eq(other)
def lt(self, other):
if self.sign > other.sign:
return False
if self.sign < other.sign:
return True
ld1 = self._numdigits()
ld2 = other._numdigits()
if ld1 > ld2:
if other.sign > 0:
return False
else:
return True
elif ld1 < ld2:
if other.sign > 0:
return True
else:
return False
i = ld1 - 1
while i >= 0:
d1 = self.digits[i]
d2 = other.digits[i]
if d1 < d2:
if other.sign > 0:
return True
else:
return False
elif d1 > d2:
if other.sign > 0:
return False
else:
return True
i -= 1
return False
def le(self, other):
return not other.lt(self)
def gt(self, other):
return other.lt(self)
def ge(self, other):
return not self.lt(other)
def hash(self):
return _hash(self)
def add(self, other):
if self.sign == 0:
return other
if other.sign == 0:
return self
if self.sign == other.sign:
result = _x_add(self, other)
else:
result = _x_sub(other, self)
result.sign *= other.sign
return result
def sub(self, other):
if other.sign == 0:
return self
if self.sign == 0:
return rbigint(other.digits[:], -other.sign)
if self.sign == other.sign:
result = _x_sub(self, other)
else:
result = _x_add(self, other)
result.sign *= self.sign
result._normalize()
return result
def mul(self, other):
if USE_KARATSUBA:
result = _k_mul(self, other)
else:
result = _x_mul(self, other)
result.sign = self.sign * other.sign
return result
def truediv(self, other):
div = _bigint_true_divide(self, other)
return div
def floordiv(self, other):
div, mod = self.divmod(other)
return div
def div(self, other):
return self.floordiv(other)
def mod(self, other):
div, mod = self.divmod(other)
return mod
def divmod(v, w):
"""
The / and % operators are now defined in terms of divmod().
The expression a mod b has the value a - b*floor(a/b).
The _divrem function gives the remainder after division of
|a| by |b|, with the sign of a. This is also expressed
as a - b*trunc(a/b), if trunc truncates towards zero.
Some examples:
a b a rem b a mod b
13 10 3 3
-13 10 -3 7
13 -10 3 -7
-13 -10 -3 -3
So, to get from rem to mod, we have to add b if a and b
have different signs. We then subtract one from the 'div'
part of the outcome to keep the invariant intact.
"""
div, mod = _divrem(v, w)
if mod.sign * w.sign == -1:
mod = mod.add(w)
div = div.sub(rbigint([1], 1))
return div, mod
def pow(a, b, c=None):
negativeOutput = False # if x<0 return negative output
# 5-ary values. If the exponent is large enough, table is
# precomputed so that table[i] == a**i % c for i in range(32).
# python translation: the table is computed when needed.
if b.sign < 0: # if exponent is negative
if c is not None:
raise TypeError(
"pow() 2nd argument "
"cannot be negative when 3rd argument specified")
# XXX failed to implement
raise ValueError("bigint pow() too negative")
if c is not None:
if c.sign == 0:
raise ValueError("pow() 3rd argument cannot be 0")
# if modulus < 0:
# negativeOutput = True
# modulus = -modulus
if c.sign < 0:
negativeOutput = True
c = rbigint(c.digits, -c.sign)
# if modulus == 1:
# return 0
if c._numdigits() == 1 and c.digits[0] == 1:
return rbigint()
# if base < 0:
# base = base % modulus
# Having the base positive just makes things easier.
if a.sign < 0:
a, temp = a.divmod(c)
a = temp
# At this point a, b, and c are guaranteed non-negative UNLESS
# c is NULL, in which case a may be negative. */
z = rbigint([1], 1)
# python adaptation: moved macros REDUCE(X) and MULT(X, Y, result)
# into helper function result = _help_mult(x, y, c)
if b._numdigits() <= FIVEARY_CUTOFF:
# Left-to-right binary exponentiation (HAC Algorithm 14.79)
# http://www.cacr.math.uwaterloo.ca/hac/about/chap14.pdf
i = b._numdigits() - 1
while i >= 0:
bi = b.digits[i]
j = 1 << (SHIFT-1)
while j != 0:
z = _help_mult(z, z, c)
if bi & j:
z = _help_mult(z, a, c)
j >>= 1
i -= 1
else:
# Left-to-right 5-ary exponentiation (HAC Algorithm 14.82)
# z still holds 1L
table = [z] * 32
table[0] = z;
for i in range(1, 32):
table[i] = _help_mult(table[i-1], a, c)
i = b._numdigits() - 1
while i >= 0:
bi = b.digits[i]
j = SHIFT - 5
while j >= 0:
index = (bi >> j) & 0x1f
for k in range(5):
z = _help_mult(z, z, c)
if index:
z = _help_mult(z, table[index], c)
j -= 5
i -= 1
if negativeOutput and z.sign != 0:
z = z.sub(c)
return z
def neg(self):
return rbigint(self.digits, -self.sign)
def abs(self):
return rbigint(self.digits, abs(self.sign))
def invert(self): #Implement ~x as -(x + 1)
return self.add(rbigint([1], 1)).neg()
def lshift(self, int_other):
if int_other < 0:
raise ValueError("negative shift count")
elif int_other == 0:
return self
# wordshift, remshift = divmod(int_other, SHIFT)
wordshift = int_other // SHIFT
remshift = int_other - wordshift * SHIFT
oldsize = self._numdigits()
newsize = oldsize + wordshift
if remshift:
newsize += 1
z = rbigint([0] * newsize, self.sign)
accum = 0
i = wordshift
j = 0
while j < oldsize:
accum |= self._digit(j) << remshift
z._setdigit(i, accum)
accum >>= SHIFT
i += 1
j += 1
if remshift:
z._setdigit(newsize - 1, accum)
else:
assert not accum
z._normalize()
return z
def rshift(self, int_other):
if int_other < 0:
raise ValueError("negative shift count")
elif int_other == 0:
return self
if self.sign == -1:
a1 = self.invert()
a2 = a1.rshift(int_other)
return a2.invert()
wordshift = int_other // SHIFT
newsize = self._numdigits() - wordshift
if newsize <= 0:
return rbigint()
loshift = int_other % SHIFT
hishift = SHIFT - loshift
lomask = (1 << hishift) - 1
himask = MASK ^ lomask
z = rbigint([0] * newsize, self.sign)
i = 0
j = wordshift
while i < newsize:
z.digits[i] = (self.digits[j] >> loshift) & lomask
if i+1 < newsize:
z.digits[i] |= (self.digits[j+1] << hishift) & himask
i += 1
j += 1
z._normalize()
return z
def and_(self, other):
return _bitwise(self, '&', other)
def xor(self, other):
return _bitwise(self, '^', other)
def or_(self, other):
return _bitwise(self, '|', other)
def oct(self):
if self.sign == 0:
return '0L'
else:
return _format(self, BASE8, '0', 'L')
def hex(self):
return _format(self, BASE16, '0x', 'L')
def log(self, base):
# base is supposed to be positive or 0.0, which means we use e
if base == 10.0:
return _loghelper(math.log10, self)
ret = _loghelper(math.log, self)
if base != 0.0:
ret /= math.log(base)
return ret
def tolong(self): #YYYYYY
l = 0
digits = list(self.digits)
digits.reverse()
for d in digits:
l = l << SHIFT
l += long(d)
return l * self.sign
def _normalize(self):
if self._numdigits() == 0:
self.sign = 0
self.digits = [0]
return
i = self._numdigits()
while i > 1 and self.digits[i - 1] == 0:
i -= 1
assert i >= 1
if i != self._numdigits():
self.digits = self.digits[:i]
if self._numdigits() == 1 and self.digits[0] == 0:
self.sign = 0
def __repr__(self):
return "<rbigint digits=%s, sign=%s, %s>" % (self.digits, self.sign, self.str())
#_________________________________________________________________
# Helper Functions
def _help_mult(x, y, c):
"""
Multiply two values, then reduce the result:
result = X*Y % c. If c is None, skip the mod.
"""
res = x.mul(y)
# Perform a modular reduction, X = X % c, but leave X alone if c
# is NULL.
if c is not None:
res, temp = res.divmod(c)
res = temp
return res
def digits_from_nonneg_long(l):
digits = []
while True:
digits.append(mask_digit(l))
l = l >> SHIFT
if not l:
return digits[:] # to make it non-resizable
digits_from_nonneg_long._annspecialcase_ = "specialize:argtype(0)"
def digits_for_most_neg_long(l):
# This helper only works if 'l' is the most negative integer of its
# type, which in base 2 looks like: 1000000..0000
digits = []
while (mask_digit(l)) == 0:
digits.append(0)
l = l >> SHIFT
# now 'l' looks like: ...111100000
# turn it into: ...000100000
# to drop the extra unwanted 1's introduced by the signed right shift
l = -intmask(l)
assert l >= 0
digits.append(l)
return digits[:] # to make it non-resizable
digits_for_most_neg_long._annspecialcase_ = "specialize:argtype(0)"
def args_from_rarith_int1(x):
if x > 0:
return digits_from_nonneg_long(x), 1
elif x == 0:
return [0], 0
else:
try:
y = ovfcheck(-x)
except OverflowError:
y = -1
# be conservative and check again if the result is >= 0, even
# if no OverflowError was raised (e.g. broken CPython/GCC4.2)
if y >= 0:
# normal case
return digits_from_nonneg_long(y), -1
else:
# the most negative integer! hacks needed...
return digits_for_most_neg_long(x), -1
args_from_rarith_int1._annspecialcase_ = "specialize:argtype(0)"
def args_from_rarith_int(x):
return args_from_rarith_int1(widen(x))
args_from_rarith_int._annspecialcase_ = "specialize:argtype(0)"
# ^^^ specialized by the precise type of 'x', which is typically a r_xxx
# instance from rlib.rarithmetic
def args_from_long(x):
"NOT_RPYTHON"
if x >= 0:
if x == 0:
return [0], 0
else:
return digits_from_nonneg_long(x), 1
else:
return digits_from_nonneg_long(-long(x)), -1
def _x_add(a, b):
""" Add the absolute values of two bigint integers. """
size_a = a._numdigits()
size_b = b._numdigits()
# Ensure a is the larger of the two:
if size_a < size_b:
a, b = b, a
size_a, size_b = size_b, size_a
z = rbigint([0] * (a._numdigits() + 1), 1)
i = 0
carry = 0
while i < size_b:
carry += a._digit(i) + b._digit(i)
z._setdigit(i, carry)
carry >>= SHIFT
i += 1
while i < size_a:
carry += a._digit(i)
z._setdigit(i, carry)
carry >>= SHIFT
i += 1
z._setdigit(i, carry)
z._normalize()
return z
def _x_sub(a, b):
""" Subtract the absolute values of two integers. """
size_a = a._numdigits()
size_b = b._numdigits()
sign = 1
# Ensure a is the larger of the two:
if size_a < size_b:
sign = -1
a, b = b, a
size_a, size_b = size_b, size_a
elif size_a == size_b:
# Find highest digit where a and b differ:
i = size_a - 1
while i >= 0 and a.digits[i] == b.digits[i]:
i -= 1
if i < 0:
return rbigint([0], 0)
if a.digits[i] < b.digits[i]:
sign = -1
a, b = b, a
size_a = size_b = i+1
borrow = 0
z = rbigint([0] * size_a, 1)
i = 0
while i < size_b:
# The following assumes unsigned arithmetic
# works modulo 2**N for some N>SHIFT.
borrow = a._digit(i) - b._digit(i) - borrow
z._setdigit(i, borrow)
borrow >>= SHIFT
borrow &= 1 # Keep only one sign bit
i += 1
while i < size_a:
borrow = a._digit(i) - borrow
z._setdigit(i, borrow)
borrow >>= SHIFT
borrow &= 1 # Keep only one sign bit
i += 1
assert borrow == 0
if sign < 0:
z.sign = -1
z._normalize()
return z
def _x_mul(a, b):
"""
Grade school multiplication, ignoring the signs.
Returns the absolute value of the product, or None if error.
"""
size_a = a._numdigits()
size_b = b._numdigits()
z = rbigint([0] * (size_a + size_b), 1)
if a is b:
# Efficient squaring per HAC, Algorithm 14.16:
# http://www.cacr.math.uwaterloo.ca/hac/about/chap14.pdf
# Gives slightly less than a 2x speedup when a == b,
# via exploiting that each entry in the multiplication
# pyramid appears twice (except for the size_a squares).
i = 0
while i < size_a:
f = a._digit(i)
pz = i << 1
pa = i + 1
paend = size_a
carry = z._digit(pz) + f * f
z._setdigit(pz, carry)
pz += 1
carry >>= SHIFT
assert carry <= MASK
# Now f is added in twice in each column of the
# pyramid it appears. Same as adding f<<1 once.
f <<= 1
while pa < paend:
carry += z._digit(pz) + a._digit(pa) * f
pa += 1
z._setdigit(pz, carry)
pz += 1
carry >>= SHIFT
assert carry <= (widen_digit(MASK) << 1)
if carry:
carry += z._digit(pz)
z._setdigit(pz, carry)
pz += 1
carry >>= SHIFT
if carry:
z._setdigit(pz, z._digit(pz) + carry)
assert (carry >> SHIFT) == 0
i += 1
else:
# a is not the same as b -- gradeschool long mult
i = 0
while i < size_a:
carry = 0
f = a._digit(i)
pz = i
pb = 0
pbend = size_b
while pb < pbend:
carry += z._digit(pz) + b._digit(pb) * f
pb += 1
z._setdigit(pz, carry)
pz += 1
carry >>= SHIFT
assert carry <= MASK
if carry:
z._setdigit(pz, z._digit(pz) + carry)
assert (carry >> SHIFT) == 0
i += 1
z._normalize()
return z
def _kmul_split(n, size):
"""
A helper for Karatsuba multiplication (k_mul).
Takes a bigint "n" and an integer "size" representing the place to
split, and sets low and high such that abs(n) == (high << size) + low,
viewing the shift as being by digits. The sign bit is ignored, and
the return values are >= 0.
"""
size_n = n._numdigits()
size_lo = min(size_n, size)
lo = rbigint(n.digits[:size_lo], 1)
hi = rbigint(n.digits[size_lo:], 1)
lo._normalize()
hi._normalize()
return hi, lo
def _k_mul(a, b):
"""
Karatsuba multiplication. Ignores the input signs, and returns the
absolute value of the product (or raises if error).
See Knuth Vol. 2 Chapter 4.3.3 (Pp. 294-295).
"""
asize = a._numdigits()
bsize = b._numdigits()
# (ah*X+al)(bh*X+bl) = ah*bh*X*X + (ah*bl + al*bh)*X + al*bl
# Let k = (ah+al)*(bh+bl) = ah*bl + al*bh + ah*bh + al*bl
# Then the original product is
# ah*bh*X*X + (k - ah*bh - al*bl)*X + al*bl
# By picking X to be a power of 2, "*X" is just shifting, and it's
# been reduced to 3 multiplies on numbers half the size.
# We want to split based on the larger number; fiddle so that b
# is largest.
if asize > bsize:
a, b, asize, bsize = b, a, bsize, asize
# Use gradeschool math when either number is too small.
if a is b:
i = KARATSUBA_SQUARE_CUTOFF
else:
i = KARATSUBA_CUTOFF
if asize <= i:
if a.sign == 0:
return rbigint([0], 0)
else:
return _x_mul(a, b)
# If a is small compared to b, splitting on b gives a degenerate
# case with ah==0, and Karatsuba may be (even much) less efficient
# than "grade school" then. However, we can still win, by viewing
# b as a string of "big digits", each of width a->ob_size. That
# leads to a sequence of balanced calls to k_mul.
if 2 * asize <= bsize:
return _k_lopsided_mul(a, b)
# Split a & b into hi & lo pieces.
shift = bsize >> 1
ah, al = _kmul_split(a, shift)
assert ah.sign == 1 # the split isn't degenerate
if a == b:
bh = ah
bl = al
else:
bh, bl = _kmul_split(b, shift)
# The plan:
# 1. Allocate result space (asize + bsize digits: that's always
# enough).
# 2. Compute ah*bh, and copy into result at 2*shift.
# 3. Compute al*bl, and copy into result at 0. Note that this
# can't overlap with #2.
# 4. Subtract al*bl from the result, starting at shift. This may
# underflow (borrow out of the high digit), but we don't care:
# we're effectively doing unsigned arithmetic mod
# BASE**(sizea + sizeb), and so long as the *final* result fits,
# borrows and carries out of the high digit can be ignored.
# 5. Subtract ah*bh from the result, starting at shift.
# 6. Compute (ah+al)*(bh+bl), and add it into the result starting
# at shift.
# 1. Allocate result space.
ret = rbigint([0] * (asize + bsize), 1)
# 2. t1 <- ah*bh, and copy into high digits of result.
t1 = _k_mul(ah, bh)
assert t1.sign >= 0
assert 2*shift + t1._numdigits() <= ret._numdigits()
ret.digits[2*shift : 2*shift + t1._numdigits()] = t1.digits
# Zero-out the digits higher than the ah*bh copy. */
## ignored, assuming that we initialize to zero
##i = ret->ob_size - 2*shift - t1->ob_size;
##if (i)
## memset(ret->ob_digit + 2*shift + t1->ob_size, 0,
## i * sizeof(digit));
# 3. t2 <- al*bl, and copy into the low digits.
t2 = _k_mul(al, bl)
assert t2.sign >= 0
assert t2._numdigits() <= 2*shift # no overlap with high digits
ret.digits[:t2._numdigits()] = t2.digits
# Zero out remaining digits.
## ignored, assuming that we initialize to zero
##i = 2*shift - t2->ob_size; /* number of uninitialized digits */
##if (i)
## memset(ret->ob_digit + t2->ob_size, 0, i * sizeof(digit));
# 4 & 5. Subtract ah*bh (t1) and al*bl (t2). We do al*bl first
# because it's fresher in cache.
i = ret._numdigits() - shift # # digits after shift
_v_isub(ret, shift, i, t2, t2._numdigits())
_v_isub(ret, shift, i, t1, t1._numdigits())
del t1, t2
# 6. t3 <- (ah+al)(bh+bl), and add into result.
t1 = _x_add(ah, al)
del ah, al
if a == b:
t2 = t1
else:
t2 = _x_add(bh, bl)
del bh, bl
t3 = _k_mul(t1, t2)
del t1, t2
assert t3.sign >=0
# Add t3. It's not obvious why we can't run out of room here.
# See the (*) comment after this function.
_v_iadd(ret, shift, i, t3, t3._numdigits())
del t3
ret._normalize()
return ret
""" (*) Why adding t3 can't "run out of room" above.
Let f(x) mean the floor of x and c(x) mean the ceiling of x. Some facts
to start with:
1. For any integer i, i = c(i/2) + f(i/2). In particular,
bsize = c(bsize/2) + f(bsize/2).
2. shift = f(bsize/2)
3. asize <= bsize
4. Since we call k_lopsided_mul if asize*2 <= bsize, asize*2 > bsize in this
routine, so asize > bsize/2 >= f(bsize/2) in this routine.
We allocated asize + bsize result digits, and add t3 into them at an offset
of shift. This leaves asize+bsize-shift allocated digit positions for t3
to fit into, = (by #1 and #2) asize + f(bsize/2) + c(bsize/2) - f(bsize/2) =
asize + c(bsize/2) available digit positions.
bh has c(bsize/2) digits, and bl at most f(size/2) digits. So bh+hl has
at most c(bsize/2) digits + 1 bit.
If asize == bsize, ah has c(bsize/2) digits, else ah has at most f(bsize/2)
digits, and al has at most f(bsize/2) digits in any case. So ah+al has at
most (asize == bsize ? c(bsize/2) : f(bsize/2)) digits + 1 bit.
The product (ah+al)*(bh+bl) therefore has at most
c(bsize/2) + (asize == bsize ? c(bsize/2) : f(bsize/2)) digits + 2 bits
and we have asize + c(bsize/2) available digit positions. We need to show
this is always enough. An instance of c(bsize/2) cancels out in both, so
the question reduces to whether asize digits is enough to hold
(asize == bsize ? c(bsize/2) : f(bsize/2)) digits + 2 bits. If asize < bsize,
then we're asking whether asize digits >= f(bsize/2) digits + 2 bits. By #4,
asize is at least f(bsize/2)+1 digits, so this in turn reduces to whether 1
digit is enough to hold 2 bits. This is so since SHIFT=15 >= 2. If
asize == bsize, then we're asking whether bsize digits is enough to hold
c(bsize/2) digits + 2 bits, or equivalently (by #1) whether f(bsize/2) digits
is enough to hold 2 bits. This is so if bsize >= 2, which holds because
bsize >= KARATSUBA_CUTOFF >= 2.
Note that since there's always enough room for (ah+al)*(bh+bl), and that's
clearly >= each of ah*bh and al*bl, there's always enough room to subtract
ah*bh and al*bl too.
"""
def _k_lopsided_mul(a, b):
"""
b has at least twice the digits of a, and a is big enough that Karatsuba
would pay off *if* the inputs had balanced sizes. View b as a sequence
of slices, each with a->ob_size digits, and multiply the slices by a,
one at a time. This gives k_mul balanced inputs to work with, and is
also cache-friendly (we compute one double-width slice of the result
at a time, then move on, never bactracking except for the helpful
single-width slice overlap between successive partial sums).
"""
asize = a._numdigits()
bsize = b._numdigits()
# nbdone is # of b digits already multiplied
assert asize > KARATSUBA_CUTOFF
assert 2 * asize <= bsize
# Allocate result space, and zero it out.
ret = rbigint([0] * (asize + bsize), 1)
# Successive slices of b are copied into bslice.
#bslice = rbigint([0] * asize, 1)
# XXX we cannot pre-allocate, see comments below!
bslice = rbigint([0], 1)
nbdone = 0;
while bsize > 0:
nbtouse = min(bsize, asize)
# Multiply the next slice of b by a.
#bslice.digits[:nbtouse] = b.digits[nbdone : nbdone + nbtouse]
# XXX: this would be more efficient if we adopted CPython's
# way to store the size, instead of resizing the list!
# XXX change the implementation, encoding length via the sign.
bslice.digits = b.digits[nbdone : nbdone + nbtouse]
product = _k_mul(a, bslice)
# Add into result.
_v_iadd(ret, nbdone, ret._numdigits() - nbdone,
product, product._numdigits())
del product
bsize -= nbtouse
nbdone += nbtouse
ret._normalize()
return ret
def _inplace_divrem1(pout, pin, n, size=0):
"""
Divide bigint pin by non-zero digit n, storing quotient
in pout, and returning the remainder. It's OK for pin == pout on entry.
"""
rem = widen_digit(0)
assert n > 0 and n <= MASK
if not size:
size = pin._numdigits()
size -= 1
while size >= 0:
rem = (rem << SHIFT) + pin._digit(size)
hi = rem // n
pout._setdigit(size, hi)
rem -= hi * n
size -= 1
return mask_digit(rem)
def _divrem1(a, n):
"""
Divide a bigint integer by a digit, returning both the quotient
and the remainder as a tuple.
The sign of a is ignored; n should not be zero.
"""
assert n > 0 and n <= MASK
size = a._numdigits()
z = rbigint([0] * size, 1)
rem = _inplace_divrem1(z, a, n)
z._normalize()
return z, rem
def _v_iadd(x, xofs, m, y, n):
"""
x and y are rbigints, m >= n required. x.digits[0:n] is modified in place,
by adding y.digits[0:m] to it. Carries are propagated as far as
x[m-1], and the remaining carry (0 or 1) is returned.
Python adaptation: x is addressed relative to xofs!
"""
carry = 0;
assert m >= n
i = xofs
iend = xofs + n
while i < iend:
carry += x._digit(i) + y._digit(i-xofs)
x._setdigit(i, carry)
carry >>= SHIFT
assert (carry & 1) == carry
i += 1
iend = xofs + m
while carry and i < iend:
carry += x._digit(i)
x._setdigit(i, carry)
carry >>= SHIFT
assert (carry & 1) == carry
i += 1
return carry
def _v_isub(x, xofs, m, y, n):
"""
x and y are rbigints, m >= n required. x.digits[0:n] is modified in place,
by substracting y.digits[0:m] to it. Borrows are propagated as
far as x[m-1], and the remaining borrow (0 or 1) is returned.
Python adaptation: x is addressed relative to xofs!
"""
borrow = 0
assert m >= n
i = xofs
iend = xofs + n
while i < iend:
borrow = x._digit(i) - y._digit(i-xofs) - borrow
x._setdigit(i, borrow)
borrow >>= SHIFT
borrow &= 1 # keep only 1 sign bit
i += 1
iend = xofs + m
while borrow and i < iend:
borrow = x._digit(i) - borrow
x._setdigit(i, borrow)
borrow >>= SHIFT
borrow &= 1
i += 1
return borrow
def _muladd1(a, n, extra=widen_digit(0)):
"""Multiply by a single digit and add a single digit, ignoring the sign.
"""
size_a = a._numdigits()
z = rbigint([0] * (size_a+1), 1)
carry = extra
assert carry & MASK == carry
i = 0
while i < size_a:
carry += a._digit(i) * n
z._setdigit(i, carry)
carry >>= SHIFT
i += 1
z._setdigit(i, carry)
z._normalize()
return z
def _x_divrem(v1, w1):
""" Unsigned bigint division with remainder -- the algorithm """
size_w = w1._numdigits()
d = (widen_digit(MASK)+1) // (w1._digit(size_w-1) + 1)
v = _muladd1(v1, d)
w = _muladd1(w1, d)
size_v = v._numdigits()
size_w = w._numdigits()
assert size_v >= size_w and size_w > 1 # Assert checks by div()
size_a = size_v - size_w + 1
a = rbigint([0] * size_a, 1)
j = size_v
k = size_a - 1
while k >= 0:
if j >= size_v:
vj = 0
else:
vj = v._digit(j)
carry = 0
if vj == w._digit(size_w-1):
q = MASK
else:
q = ((vj << SHIFT) + v._digit(j-1)) // w._digit(size_w-1)
while (w._digit(size_w-2) * q >
((
(vj << SHIFT)
+ v._digit(j-1)
- q * w._digit(size_w-1)
) << SHIFT)
+ v._digit(j-2)):
q -= 1
i = 0
while i < size_w and i+k < size_v:
z = w._digit(i) * q
zz = z >> SHIFT
carry += v._digit(i+k) - z + (zz << SHIFT)
v._setdigit(i+k, carry)
carry >>= SHIFT
carry -= zz
i += 1
if i+k < size_v:
carry += v._digit(i+k)
v._setdigit(i+k, 0)
if carry == 0:
a._setdigit(k, q)
assert not q >> SHIFT
else:
assert carry == -1
q -= 1
a._setdigit(k, q)
assert not q >> SHIFT
carry = 0
i = 0
while i < size_w and i+k < size_v:
carry += v._digit(i+k) + w._digit(i)
v._setdigit(i+k, carry)
carry >>= SHIFT
i += 1
j -= 1
k -= 1
a._normalize()
rem, _ = _divrem1(v, d)
return a, rem
def _divrem(a, b):
""" Long division with remainder, top-level routine """
size_a = a._numdigits()
size_b = b._numdigits()
if b.sign == 0:
raise ZeroDivisionError("long division or modulo by zero")
if (size_a < size_b or
(size_a == size_b and
a.digits[size_a-1] < b.digits[size_b-1])):
# |a| < |b|
z = rbigint([0], 0)
rem = a
return z, rem
if size_b == 1:
z, urem = _divrem1(a, b._digit(0))
rem = rbigint([urem], int(urem != 0))
else:
z, rem = _x_divrem(a, b)
# Set the signs.
# The quotient z has the sign of a*b;
# the remainder r has the sign of a,
# so a = b*z + r.
if a.sign != b.sign:
z.sign = - z.sign
if a.sign < 0 and rem.sign != 0:
rem.sign = - rem.sign
return z, rem
# ______________ conversions to double _______________
def _AsScaledDouble(v):
"""
NBITS_WANTED should be > the number of bits in a double's precision,
but small enough so that 2**NBITS_WANTED is within the normal double
range. nbitsneeded is set to 1 less than that because the most-significant
Python digit contains at least 1 significant bit, but we don't want to
bother counting them (catering to the worst case cheaply).
57 is one more than VAX-D double precision; I (Tim) don't know of a double
format with more precision than that; it's 1 larger so that we add in at
least one round bit to stand in for the ignored least-significant bits.
"""
NBITS_WANTED = 57
if v.sign == 0:
return 0.0, 0
i = v._numdigits() - 1
sign = v.sign
x = float(v.digits[i])
nbitsneeded = NBITS_WANTED - 1
# Invariant: i Python digits remain unaccounted for.
while i > 0 and nbitsneeded > 0:
i -= 1
x = x * FLOAT_MULTIPLIER + float(v.digits[i])
nbitsneeded -= SHIFT
# There are i digits we didn't shift in. Pretending they're all
# zeroes, the true value is x * 2**(i*SHIFT).
exponent = i
assert x > 0.0
return x * sign, exponent
def isinf(x):
return x != 0.0 and x / 2 == x
##def ldexp(x, exp):
## assert type(x) is float
## lb1 = LONG_BIT - 1
## multiplier = float(1 << lb1)
## while exp >= lb1:
## x *= multiplier
## exp -= lb1
## if exp:
## x *= float(1 << exp)
## return x
# note that math.ldexp checks for overflows,
# while the C ldexp is not guaranteed to do.
# XXX make sure that we don't ignore this!
# YYY no, we decided to do ignore this!
def _AsDouble(v):
""" Get a C double from a bigint object. """
x, e = _AsScaledDouble(v)
if e <= sys.maxint / SHIFT:
x = math.ldexp(x, e * SHIFT)
#if not isinf(x):
# this is checked by math.ldexp
return x
raise OverflowError # can't say "long int too large to convert to float"
def _loghelper(func, arg):
"""
A decent logarithm is easy to compute even for huge bigints, but libm can't
do that by itself -- loghelper can. func is log or log10.
Note that overflow isn't possible: a bigint can contain
no more than INT_MAX * SHIFT bits, so has value certainly less than
2**(2**64 * 2**16) == 2**2**80, and log2 of that is 2**80, which is
small enough to fit in an IEEE single. log and log10 are even smaller.
"""
x, e = _AsScaledDouble(arg)
if x <= 0.0:
raise ValueError
# Value is ~= x * 2**(e*SHIFT), so the log ~=
# log(x) + log(2) * e * SHIFT.
# CAUTION: e*SHIFT may overflow using int arithmetic,
# so force use of double. */
return func(x) + (e * float(SHIFT) * func(2.0))
_loghelper._annspecialcase_ = 'specialize:arg(0)'
def _bigint_true_divide(a, b):
ad, aexp = _AsScaledDouble(a)
bd, bexp = _AsScaledDouble(b)
if bd == 0.0:
raise ZeroDivisionError("long division or modulo by zero")
# True value is very close to ad/bd * 2**(SHIFT*(aexp-bexp))
ad /= bd # overflow/underflow impossible here
aexp -= bexp
if aexp > sys.maxint / SHIFT:
raise OverflowError
elif aexp < -(sys.maxint / SHIFT):
return 0.0 # underflow to 0
ad = math.ldexp(ad, aexp * SHIFT)
##if isinf(ad): # ignore underflow to 0.0
## raise OverflowError
# math.ldexp checks and raises
return ad
BASE8 = '01234567'
BASE10 = '0123456789'
BASE16 = '0123456789abcdef'
def _format(a, digits, prefix='', suffix=''):
"""
Convert a bigint object to a string, using a given conversion base.
Return a string object.
"""
size_a = a._numdigits()
base = len(digits)
assert base >= 2 and base <= 36
# Compute a rough upper bound for the length of the string
i = base
bits = 0
while i > 1:
bits += 1
i >>= 1
i = 5 + len(prefix) + len(suffix) + (size_a*SHIFT + bits-1) // bits
s = [chr(0)] * i
p = i
j = len(suffix)
while j > 0:
p -= 1
j -= 1
s[p] = suffix[j]
if a.sign == 0:
p -= 1
s[p] = '0'
elif (base & (base - 1)) == 0:
# JRH: special case for power-of-2 bases
accum = 0
accumbits = 0 # # of bits in accum
basebits = 1 # # of bits in base-1
i = base
while 1:
i >>= 1
if i <= 1:
break
basebits += 1
for i in range(size_a):
accum |= a._digit(i) << accumbits
accumbits += SHIFT
assert accumbits >= basebits
while 1:
cdigit = accum & (base - 1)
p -= 1
assert p >= 0
s[p] = digits[cdigit]
accumbits -= basebits
accum >>= basebits
if i < size_a - 1:
if accumbits < basebits:
break
else:
if accum <= 0:
break
else:
# Not 0, and base not a power of 2. Divide repeatedly by
# base, but for speed use the highest power of base that
# fits in a digit.
size = size_a
pin = a # just for similarity to C source which uses the array
# powbase <- largest power of base that fits in a digit.
powbase = widen_digit(base) # powbase == base ** power
power = 1
while 1:
newpow = powbase * base
if newpow >> SHIFT: # doesn't fit in a digit
break
powbase = newpow
power += 1
# Get a scratch area for repeated division.
scratch = rbigint([0] * size, 1)
# Repeatedly divide by powbase.
while 1:
ntostore = power
rem = _inplace_divrem1(scratch, pin, powbase, size)
pin = scratch # no need to use a again
if pin.digits[size - 1] == 0:
size -= 1
# Break rem into digits.
assert ntostore > 0
while 1:
nextrem = rem // base
c = rem - nextrem * base
p -= 1
assert p >= 0
s[p] = digits[c]
rem = nextrem
ntostore -= 1
# Termination is a bit delicate: must not
# store leading zeroes, so must get out if
# remaining quotient and rem are both 0.
if not (ntostore and (size or rem)):
break
if size == 0:
break
j = len(prefix)
while j > 0:
p -= 1
j -= 1
s[p] = prefix[j]
if a.sign < 0:
p -= 1
s[p] = '-'
assert p >= 0 # otherwise, buffer overflow (this is also a
# hint for the annotator for the slice below)
return ''.join(s[p:])
def _bitwise(a, op, b): # '&', '|', '^'
""" Bitwise and/or/xor operations """
if a.sign < 0:
a = a.invert()
maska = MASK
else:
maska = 0
if b.sign < 0:
b = b.invert()
maskb = MASK
else:
maskb = 0
negz = 0
if op == '^':
if maska != maskb:
maska ^= MASK
negz = -1
elif op == '&':
if maska and maskb:
op = '|'
maska ^= MASK
maskb ^= MASK
negz = -1
elif op == '|':
if maska or maskb:
op = '&'
maska ^= MASK
maskb ^= MASK
negz = -1
# JRH: The original logic here was to allocate the result value (z)
# as the longer of the two operands. However, there are some cases
# where the result is guaranteed to be shorter than that: AND of two
# positives, OR of two negatives: use the shorter number. AND with
# mixed signs: use the positive number. OR with mixed signs: use the
# negative number. After the transformations above, op will be '&'
# iff one of these cases applies, and mask will be non-0 for operands
# whose length should be ignored.
size_a = a._numdigits()
size_b = b._numdigits()
if op == '&':
if maska:
size_z = size_b
else:
if maskb:
size_z = size_a
else:
size_z = min(size_a, size_b)
else:
size_z = max(size_a, size_b)
z = rbigint([0] * size_z, 1)
for i in range(size_z):
if i < size_a:
diga = a.digits[i] ^ maska
else:
diga = maska
if i < size_b:
digb = b.digits[i] ^ maskb
else:
digb = maskb
if op == '&':
z.digits[i] = diga & digb
elif op == '|':
z.digits[i] = diga | digb
elif op == '^':
z.digits[i] = diga ^ digb
z._normalize()
if negz == 0:
return z
return z.invert()
_bitwise._annspecialcase_ = "specialize:arg(1)"
ULONGLONG_BOUND = r_ulonglong(1L << (r_longlong.BITS-1))
LONGLONG_MIN = r_longlong(-(1L << (r_longlong.BITS-1)))
def _AsLongLong(v):
"""
Get a r_longlong integer from a bigint object.
Raises OverflowError if overflow occurs.
"""
x = _AsULonglong_ignore_sign(v)
# grr grr grr
if x >= ULONGLONG_BOUND:
if x == ULONGLONG_BOUND and v.sign < 0:
x = LONGLONG_MIN
else:
raise OverflowError
else:
x = r_longlong(x)
if v.sign < 0:
x = -x
return x
def _AsULonglong_ignore_sign(v):
x = r_ulonglong(0)
i = v._numdigits() - 1
while i >= 0:
prev = x
x = (x << SHIFT) + v._digit(i)
if (x >> SHIFT) != prev:
raise OverflowError(
"long int too large to convert to unsigned long long int")
i -= 1
return x
def make_unsigned_mask_conversion(T):
def _As_unsigned_mask(v):
x = T(0)
i = v._numdigits() - 1
while i >= 0:
prev = x
x = (x << SHIFT) + T(v._digit(i))
i -= 1
if v.sign < 0:
x = -x
return x
return _As_unsigned_mask
_AsULonglong_mask = make_unsigned_mask_conversion(r_ulonglong)
_AsUInt_mask = make_unsigned_mask_conversion(r_uint)
def _hash(v):
# This is designed so that Python ints and longs with the
# same value hash to the same value, otherwise comparisons
# of mapping keys will turn out weird
i = v._numdigits() - 1
sign = v.sign
x = 0
LONG_BIT_SHIFT = LONG_BIT - SHIFT
while i >= 0:
# Force a native long #-bits (32 or 64) circular shift
x = ((x << SHIFT) & ~MASK) | ((x >> LONG_BIT_SHIFT) & MASK)
x += v.digits[i]
i -= 1
x = intmask(x * sign)
return x
#_________________________________________________________________
# a few internal helpers
def digits_max_for_base(base):
dec_per_digit = 1
while base ** dec_per_digit < MASK:
dec_per_digit += 1
dec_per_digit -= 1
return base ** dec_per_digit
BASE_MAX = [0, 0] + [digits_max_for_base(_base) for _base in range(2, 37)]
DEC_MAX = digits_max_for_base(10)
assert DEC_MAX == BASE_MAX[10]
def _decimalstr_to_bigint(s):
# a string that has been already parsed to be decimal and valid,
# is turned into a bigint
p = 0
lim = len(s)
sign = False
if s[p] == '-':
sign = True
p += 1
elif s[p] == '+':
p += 1
a = rbigint.fromint(0)
tens = 1
dig = 0
ord0 = ord('0')
while p < lim:
dig = dig * 10 + ord(s[p]) - ord0
p += 1
tens *= 10
if tens == DEC_MAX or p == lim:
a = _muladd1(a, tens, dig)
tens = 1
dig = 0
if sign and a.sign == 1:
a.sign = -1
return a
def parse_digit_string(parser):
# helper for objspace.std.strutil
a = rbigint.fromint(0)
base = parser.base
digitmax = BASE_MAX[base]
tens, dig = 1, 0
while True:
digit = parser.next_digit()
if tens == digitmax or digit < 0:
a = _muladd1(a, tens, dig)
if digit < 0:
break
dig = digit
tens = base
else:
dig = dig * base + digit
tens *= base
a.sign *= parser.sign
return a
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