Author: Armin Rigo <ar...@tunes.org>
Branch:
Changeset: r85556:676d2fb4524b
Date: 2016-07-05 13:10 +0200
http://bitbucket.org/pypy/pypy/changeset/676d2fb4524b/
Log: Found out how to reduce this division to a simple algorithm instead
of using rbigint. Avoids troubles if the full RPython program is
also using parts of rbigint.
diff --git a/rpython/jit/metainterp/optimizeopt/intdiv.py
b/rpython/jit/metainterp/optimizeopt/intdiv.py
--- a/rpython/jit/metainterp/optimizeopt/intdiv.py
+++ b/rpython/jit/metainterp/optimizeopt/intdiv.py
@@ -1,5 +1,5 @@
from rpython.rlib.rarithmetic import LONG_BIT, intmask, r_uint
-from rpython.rlib.rbigint import rbigint, ONERBIGINT
+
from rpython.jit.metainterp.history import ConstInt
from rpython.jit.metainterp.resoperation import ResOperation, rop
@@ -17,10 +17,19 @@
while (r_uint(1) << (i+1)) < r_uint(m):
i += 1
- # k = 2**(64+i) // m + 1, computed manually using rbigint
- # because that's the easiest
- k1 = ONERBIGINT.lshift(LONG_BIT + i).floordiv(rbigint.fromint(m))
- k = k1.touint() + r_uint(1)
+ # quotient = 2**(64+i) // m
+ high_word_dividend = r_uint(1) << i
+ quotient = r_uint(0)
+ for bit in range(LONG_BIT-1, -1, -1):
+ t = quotient + (r_uint(1) << bit)
+ # check: is 't * m' small enough to be < 2**(64+i), or not?
+ # note that we're really computing (2**(64+i)-1) // m, but the result
+ # is the same, because powers of two are not multiples of m.
+ if unsigned_mul_high(t, m) < high_word_dividend:
+ quotient = t # yes, small enough
+
+ # k = 2**(64+i) // m + 1
+ k = quotient + r_uint(1)
assert k != r_uint(0)
# Proof that k < 2**64 holds in all cases, even with the "+1":
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