Kaydet (Commit) 212e614f authored tarafından Tim Peters's avatar Tim Peters

divrem1 & long_format: found a clean way to factor divrem1 so that

long_format can reuse a scratch area for its repeated divisions (instead
of malloc/free for every digit produced); speeds str(long)/repr(long).
üst c8a6b9b6
......@@ -15,7 +15,7 @@
static PyLongObject *long_normalize(PyLongObject *);
static PyLongObject *mul1(PyLongObject *, wdigit);
static PyLongObject *muladd1(PyLongObject *, wdigit, wdigit);
static PyLongObject *divrem1(PyLongObject *, wdigit, digit *);
static PyLongObject *divrem1(PyLongObject *, digit, digit *);
static PyObject *long_format(PyObject *aa, int base, int addL);
static int ticker; /* XXX Could be shared with ceval? */
......@@ -707,28 +707,44 @@ muladd1(PyLongObject *a, wdigit n, wdigit extra)
return long_normalize(z);
}
/* Divide long pin, w/ size digits, by non-zero digit n, storing quotient
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
long_format, but that should be done with great care since longs are
immutable. */
static digit
inplace_divrem1(digit *pout, digit *pin, int size, digit n)
{
twodigits rem = 0;
assert(n > 0 && n <= MASK);
pin += size;
pout += size;
while (--size >= 0) {
digit hi;
rem = (rem << SHIFT) + *--pin;
*--pout = hi = (digit)(rem / n);
rem -= hi * n;
}
return (digit)rem;
}
/* Divide a long 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. */
static PyLongObject *
divrem1(PyLongObject *a, wdigit n, digit *prem)
divrem1(PyLongObject *a, digit n, digit *prem)
{
int size = ABS(a->ob_size);
const int size = ABS(a->ob_size);
PyLongObject *z;
int i;
twodigits rem = 0;
assert(n > 0 && n <= MASK);
z = _PyLong_New(size);
if (z == NULL)
return NULL;
for (i = size; --i >= 0; ) {
rem = (rem << SHIFT) + a->ob_digit[i];
z->ob_digit[i] = (digit) (rem/n);
rem %= n;
}
*prem = (digit) rem;
*prem = inplace_divrem1(z->ob_digit, a->ob_digit, size, n);
return long_normalize(z);
}
......@@ -742,7 +758,7 @@ long_format(PyObject *aa, int base, int addL)
register PyLongObject *a = (PyLongObject *)aa;
PyStringObject *str;
int i;
int size_a = ABS(a->ob_size);
const int size_a = ABS(a->ob_size);
char *p;
int bits;
char sign = '\0';
......@@ -779,7 +795,7 @@ long_format(PyObject *aa, int base, int addL)
twodigits temp = a->ob_digit[0];
int bitsleft = SHIFT;
int rem;
int last = abs(a->ob_size);
int last = size_a;
int basebits = 1;
i = base;
while ((i >>= 1) > 1)
......@@ -814,6 +830,10 @@ long_format(PyObject *aa, int base, int addL)
/* 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. */
int size = size_a;
digit *pin = a->ob_digit;
PyLongObject *scratch;
/* powbasw <- largest power of base that fits in a digit. */
digit powbase = base; /* powbase == base ** power */
int power = 1;
for (;;) {
......@@ -823,23 +843,29 @@ long_format(PyObject *aa, int base, int addL)
powbase = (digit)newpow;
++power;
}
Py_INCREF(a);
/* Get a scratch area for repeated division. */
scratch = _PyLong_New(size);
if (scratch == NULL) {
Py_DECREF(str);
return NULL;
}
/* Repeatedly divide by powbase. */
do {
int ntostore = power;
digit rem;
PyLongObject *temp = divrem1(a, powbase, &rem);
Py_DECREF(a);
if (temp == NULL) {
Py_DECREF(str);
return NULL;
}
a = temp;
digit rem = inplace_divrem1(scratch->ob_digit,
pin, size, powbase);
pin = scratch->ob_digit; /* no need to use a again */
if (pin[size - 1] == 0)
--size;
SIGCHECK({
Py_DECREF(a);
Py_DECREF(scratch);
Py_DECREF(str);
return NULL;
})
/* Break rem into digits. */
assert(ntostore > 0);
do {
digit nextrem = (digit)(rem / base);
......@@ -851,10 +877,10 @@ long_format(PyObject *aa, int base, int addL)
--ntostore;
/* Termination is a bit delicate: must not
store leading zeroes, so must get out if
a and rem are both 0 now. */
} while (ntostore && (a->ob_size || rem));
} while (a->ob_size != 0);
Py_DECREF(a);
remaining quotient and rem are both 0. */
} while (ntostore && (size || rem));
} while (size != 0);
Py_DECREF(scratch);
}
if (base == 8) {
......
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