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bpo-46406: Faster single digit int division. (#30626) 

* bpo-46406: Faster single digit int division.

This expresses the algorithm in a more basic manner resulting in better
instruction generation by todays compilers.

See https://mail.python.org/archives/list/python-dev@python.org/thread/ZICIMX5VFCX4IOFH5NUPVHCUJCQ4Q7QM/#NEUNFZU3TQU4CPTYZNF3WCN7DOJBBTK5
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Gregory P. Smith 2022-01-23 02:00:41 -08:00 committed by GitHub
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Misc/NEWS.d/next/Core and Builtins/2022-01-16-15-40-11.bpo-46406.g0mke-.rst Normal file
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@ -0,0 +1,3 @@
The integer division ``//`` implementation has been optimized to better let the
compiler understand its constraints. It can be 20% faster on the amd64 platform
when dividing an int by a value smaller than ``2**30``.

34
Objects/longobject.c
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@ -1617,25 +1617,41 @@ v_rshift(digit *z, digit *a, Py_ssize_t m, int d)
in pout, and returning the remainder. pin and pout point at the LSD.
It's OK for pin == pout on entry, which saves oodles of mallocs/frees in
_PyLong_Format, but that should be done with great care since ints are
immutable. */
immutable.
This version of the code can be 20% faster than the pre-2022 version
on todays compilers on architectures like amd64. It evolved from Mark
Dickinson observing that a 128:64 divide instruction was always being
generated by the compiler despite us working with 30-bit digit values.
See the thread for full context:
https://mail.python.org/archives/list/python-dev@python.org/thread/ZICIMX5VFCX4IOFH5NUPVHCUJCQ4Q7QM/#NEUNFZU3TQU4CPTYZNF3WCN7DOJBBTK5
If you ever want to change this code, pay attention to performance using
different compilers, optimization levels, and cpu architectures. Beware of
PGO/FDO builds doing value specialization such as a fast path for //10. :)
Verify that 17 isn't specialized and this works as a quick test:
python -m timeit -s 'x = 10**1000; r=x//10; assert r == 10**999, r' 'x//17'
*/
static digit
inplace_divrem1(digit *pout, digit *pin, Py_ssize_t size, digit n)
{
twodigits rem = 0;
digit remainder = 0;
assert(n > 0 && n <= PyLong_MASK);
pin += size;
pout += size;
while (--size >= 0) {
digit hi;
rem = (rem << PyLong_SHIFT) | *--pin;
*--pout = hi = (digit)(rem / n);
rem -= (twodigits)hi * n;
twodigits dividend;
dividend = ((twodigits)remainder << PyLong_SHIFT) | pin[size];
digit quotient;
quotient = (digit)(dividend / n);
remainder = dividend % n;
pout[size] = quotient;
}
return (digit)rem;
return remainder;
}
/* Divide an integer by a digit, returning both the quotient
(as function result) and the remainder (through *prem).
The sign of a is ignored; n should not be zero. */