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/[email protected]/thread/ZICIMX5VFCX4IOFH5NUPVHCUJCQ4Q7QM/#NEUNFZU3TQU4CPTYZNF3WCN7DOJBBTK5

Author: gpshead

Misc/NEWS.d/next/Core and Builtins/2022-01-16-15-40-11.bpo-46406.g0mke-.rst

@@ -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``.

Objects/longobject.c

@@ -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/[email protected]/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;
-    }
-    return (digit)rem;
+        twodigits dividend;
+        dividend = ((twodigits)remainder << PyLong_SHIFT) | pin[size];
+        digit quotient;
+        quotient = (digit)(dividend / n);
+        remainder = dividend % n;
+        pout[size] = quotient;
+    }
+    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. */