Tim's latest, with some of my changes (also a TP suggestion) added:

instead of testing whether the list changed size after each
comparison, temporarily set the type of the list to an immutable list
type.  This should allow continued use of the list for legitimate
purposes but disallows all operations that can change it in any way.
(Changes to the internals of list items are not caught, of cause;
that's not possible to detect, and it's not necessary to protect the
sort code, either.)
This commit is contained in:
Guido van Rossum 1998-06-16 15:18:28 +00:00
parent 32490824b6
commit 4c4e7df755
1 changed files with 575 additions and 242 deletions

View File

@ -582,18 +582,15 @@ listappend(self, args)
/* Comparison function. Takes care of calling a user-supplied
comparison function (any callable Python object). Calls the
standard comparison function, cmpobject(), if the user-supplied
function is NULL. */
standard comparison function, PyObject_Compare(), if the user-
supplied function is NULL. */
static int
docompare(x, y, compare, list)
docompare(x, y, compare)
PyObject *x;
PyObject *y;
PyObject *compare;
PyListObject *list;
{
int size = list->ob_size; /* Number of elements to sort */
PyObject **array = list->ob_item; /* Start of array to sort */
PyObject *args, *res;
int i;
@ -601,11 +598,6 @@ docompare(x, y, compare, list)
i = PyObject_Compare(x, y);
if (i && PyErr_Occurred())
i = CMPERROR;
else if (size != list->ob_size || array != list->ob_item) {
PyErr_SetString(PyExc_SystemError,
"list changed size during sort");
i = CMPERROR;
}
return i;
}
@ -622,11 +614,6 @@ docompare(x, y, compare, list)
"comparison function should return int");
return CMPERROR;
}
if (size != list->ob_size || array != list->ob_item) {
PyErr_SetString(PyExc_SystemError,
"list changed size during sort");
return CMPERROR;
}
i = PyInt_AsLong(res);
Py_DECREF(res);
if (i < 0)
@ -636,248 +623,537 @@ docompare(x, y, compare, list)
return 0;
}
/* MINSIZE is the smallest array we care to partition; smaller arrays
are sorted using binary insertion. It must be at least 4 for the
quicksort implementation to work. Binary insertion always requires
fewer compares than quicksort, but does O(N**2) data movement. The
more expensive compares, the larger MINSIZE should be. */
#define MINSIZE 49
/* MINSIZE is the smallest array that will get a full-blown samplesort
treatment; smaller arrays are sorted using binary insertion. It must
be at least 7 for the samplesort implementation to work. Binary
insertion does fewer compares, but can suffer O(N**2) data movement.
The more expensive compares, the larger MINSIZE should be. */
#define MINSIZE 100
/* MINPARTITIONSIZE is the smallest array slice samplesort will bother to
partition; smaller slices are passed to binarysort. It must be at
least 2, and no larger than MINSIZE. Setting it higher reduces the #
of compares slowly, but increases the amount of data movement quickly.
The value here was chosen assuming a compare costs ~25x more than
swapping a pair of memory-resident pointers -- but under that assumption,
changing the value by a few dozen more or less has aggregate effect
under 1%. So the value is crucial, but not touchy <wink>. */
#define MINPARTITIONSIZE 40
/* MAXMERGE is the largest number of elements we'll always merge into
a known-to-be sorted chunk via binary insertion, regardless of the
size of that chunk. Given a chunk of N sorted elements, and a group
of K unknowns, the largest K for which it's better to do insertion
(than a full-blown sort) is a complicated function of N and K mostly
involving the expected number of compares and data moves under each
approach, and the relative cost of those operations on a specific
architecure. The fixed value here is conservative, and should be a
clear win regardless of architecture or N. */
#define MAXMERGE 15
/* STACKSIZE is the size of our work stack. A rough estimate is that
this allows us to sort arrays of MINSIZE * 2**STACKSIZE, or large
enough. (Because of the way we push the biggest partition first,
the worst case occurs when all subarrays are always partitioned
exactly in two.) */
#define STACKSIZE 64
this allows us to sort arrays of size N where
N / ln(N) = MINPARTITIONSIZE * 2**STACKSIZE, so 60 is more than enough
for arrays of size 2**64. Because we push the biggest partition
first, the worst case occurs when all subarrays are always partitioned
exactly in two. */
#define STACKSIZE 60
/* quicksort algorithm. Return -1 if an exception occurred; in this
case we leave the array partly sorted but otherwise in good health
(i.e. no items have been removed or duplicated). */
#define SETK(X,Y) if ((k = docompare(X,Y,compare))==CMPERROR) goto fail
/* binarysort is the best method for sorting small arrays: it does
few compares, but can do data movement quadratic in the number of
elements.
[lo, hi) is a contiguous slice of the list, and is sorted via
binary insertion.
On entry, must have lo <= start <= hi, and that [lo, start) is already
sorted (pass start == lo if you don't know!).
If docompare complains (returns CMPERROR) return -1, else 0.
Even in case of error, the output slice will be some permutation of
the input (nothing is lost or duplicated).
*/
static int
quicksort(list, compare)
PyListObject *list; /* List to sort */
binarysort(lo, hi, start, list, compare)
PyObject **lo;
PyObject **hi;
PyObject **start;
PyListObject *list; /* Needed by docompare for paranoia checks */
PyObject *compare;/* Comparison function object, or NULL for default */
{
register PyObject *tmp, *pivot;
register PyObject **l, **r, **p;
PyObject **lo, **hi, **notp;
int top, k, n, lisp, risp;
PyObject **lostack[STACKSIZE];
PyObject **histack[STACKSIZE];
/* Start out with the whole array on the work stack */
lostack[0] = list->ob_item;
histack[0] = list->ob_item + list->ob_size;
top = 1;
#define SETK(X,Y) if ((k = docompare(X,Y,compare,list))==CMPERROR) goto fail
/* Repeat until the work stack is empty */
while (--top >= 0) {
lo = lostack[top];
hi = histack[top];
n = hi - lo;
/* If it's a small one, use binary insertion sort */
if (n < MINSIZE) {
for (notp = lo+1; notp < hi; ++notp) {
/* set l to where *notp belongs */
l = lo;
r = notp;
pivot = *r;
do {
p = l + ((r - l) >> 1);
SETK(pivot, *p);
if (k < 0)
r = p;
else
l = p + 1;
} while (l < r);
/* Pivot should go at l -- slide over to
make room. Caution: using memmove
is much slower under MSVC 5; we're
not usually moving many slots. */
for (p = notp; p > l; --p)
*p = *(p-1);
*l = pivot;
}
continue;
}
/* Choose median of first, middle and last as pivot;
this is a simple unrolled 3-element insertion sort */
l = lo; /* First */
p = lo + (n>>1); /* Middle */
r = hi - 1; /* Last */
pivot = *p;
SETK(pivot, *l);
if (k < 0) {
*p = *l;
*l = pivot;
}
/* assert lo <= start <= hi
assert [lo, start) is sorted */
register int k;
register PyObject **l, **p, **r;
register PyObject *pivot;
if (lo == start)
++start;
for (; start < hi; ++start) {
/* set l to where *start belongs */
l = lo;
r = start;
pivot = *r;
SETK(pivot, *p);
if (k < 0) {
*r = *p;
*p = pivot; /* for consistency */
SETK(pivot, *l);
if (k < 0) {
*p = *l;
*l = pivot;
}
}
pivot = *p;
l++;
r--;
/* Partition the array */
for (;;) {
lisp = risp = 1; /* presumed guilty */
/* Move left index to element >= pivot */
while (l < p) {
SETK(*l, pivot);
if (k < 0)
l++;
else {
lisp = 0;
break;
}
}
/* Move right index to element <= pivot */
while (r > p) {
SETK(pivot, *r);
if (k < 0)
r--;
else {
risp = 0;
break;
}
}
if (lisp == risp) {
/* assert l < p < r or l == p == r
* This is the most common case, so we
* strive to get back to the top of the
* loop ASAP.
*/
tmp = *l; *l = *r; *r = tmp;
l++; r--;
if (l < r)
continue;
break;
}
/* One (exactly) of the pointers is at p */
/* assert (p == l) ^ (p == r) */
notp = lisp ? r : l;
*p = *notp;
k = (r - l) >> 1;
if (k) {
if (lisp) {
p = r - k;
l++;
}
else {
p = l + k;
r--;
}
*notp = *p;
*p = pivot; /* for consistency */
continue;
}
/* assert l+1 == r */
*notp = pivot;
p = notp;
break;
} /* end of partitioning loop */
/* assert *p == pivot
All in [lo,p) are <= pivot
At p == pivot
All in [p+1,hi) are >= pivot
*/
r = p;
l = p + 1;
/* Partitions are [lo,r) and [l,hi).
* See whether *l == pivot; we know *l >= pivot, so
* they're equal iff *l <= pivot too, or not pivot < *l.
* This wastes a compare if it fails, but can win big
* when there are runs of duplicates.
*/
SETK(pivot, *l);
if (!(k < 0)) {
/* Now extend as far as possible (around p) so that:
All in [lo,r) are <= pivot
All in [r,l) are == pivot
All in [l,hi) are >= pivot
Mildly tricky: continue using only "<" -- we
deduce equality indirectly.
*/
while (r > lo) {
/* because r-1 < p, *(r-1) <= pivot is known */
SETK(*(r-1), pivot);
if (k < 0)
break;
/* <= and not < implies == */
r--;
}
l++;
while (l < hi) {
/* because l > p, pivot <= *l is known */
SETK(pivot, *l);
if (k < 0)
break;
/* <= and not < implies == */
l++;
}
} /* end of checking for duplicates */
/* Push biggest partition first */
if (r - lo >= hi - l) {
/* First one is bigger */
lostack[top] = lo;
histack[top++] = r;
lostack[top] = l;
histack[top++] = hi;
} else {
/* Second one is bigger */
lostack[top] = l;
histack[top++] = hi;
lostack[top] = lo;
histack[top++] = r;
}
/* Should assert top <= STACKSIZE */
do {
p = l + ((r - l) >> 1);
SETK(pivot, *p);
if (k < 0)
r = p;
else
l = p + 1;
} while (l < r);
/* Pivot should go at l -- slide over to make room.
Caution: using memmove is much slower under MSVC 5;
we're not usually moving many slots. */
for (p = start; p > l; --p)
*p = *(p-1);
*l = pivot;
}
/* Success */
return 0;
fail:
return -1;
}
/* samplesortslice is the sorting workhorse.
[lo, hi) is a contiguous slice of the list, to be sorted in place.
On entry, must have lo <= hi,
If docompare complains (returns CMPERROR) return -1, else 0.
Even in case of error, the output slice will be some permutation of
the input (nothing is lost or duplicated).
samplesort is basically quicksort on steroids: a power of 2 close
to n/ln(n) is computed, and that many elements (less 1) are picked at
random from the array and sorted. These 2**k - 1 elements are then
used as preselected pivots for an equal number of quicksort
partitioning steps, partitioning the slice into 2**k chunks each of
size about ln(n). These small final chunks are then usually handled
by binarysort. Note that when k=1, this is roughly the same as an
ordinary quicksort using a random pivot, and when k=2 this is roughly
a median-of-3 quicksort. From that view, using k ~= lg(n/ln(n)) makes
this a "median of n/ln(n)" quicksort. You can also view it as a kind
of bucket sort, where 2**k-1 bucket boundaries are picked dynamically.
The large number of samples makes a quadratic-time case almost
impossible, and asymptotically drives the average-case number of
compares from quicksort's 2 N ln N (or 12/7 N ln N for the median-of-
3 variant) down to N lg N.
We also play lots of low-level tricks to cut the number of compares.
Very obscure: To avoid using extra memory, the PPs are stored in the
array and shuffled around as partitioning proceeds. At the start of a
partitioning step, we'll have 2**m-1 (for some m) PPs in sorted order,
adjacent (either on the left or the right!) to a chunk of X elements
that are to be partitioned: P X or X P. In either case we need to
shuffle things *in place* so that the 2**(m-1) smaller PPs are on the
left, followed by the PP to be used for this step (that's the middle
of the PPs), followed by X, followed by the 2**(m-1) larger PPs:
P X or X P -> Psmall pivot X Plarge
and the order of the PPs must not be altered. It can take a while
to realize this isn't trivial! It can take even longer <wink> to
understand why the simple code below works, using only 2**(m-1) swaps.
The key is that the order of the X elements isn't necessarily
preserved: X can end up as some cyclic permutation of its original
order. That's OK, because X is unsorted anyway. If the order of X
had to be preserved too, the simplest method I know of using O(1)
scratch storage requires len(X) + 2**(m-1) swaps, spread over 2 passes.
Since len(X) is typically several times larger than 2**(m-1), that
would slow things down.
*/
struct SamplesortStackNode {
/* Represents a slice of the array, from (& including) lo up
to (but excluding) hi. "extra" additional & adjacent elements
are pre-selected pivots (PPs), spanning [lo-extra, lo) if
extra > 0, or [hi, hi-extra) if extra < 0. The PPs are
already sorted, but nothing is known about the other elements
in [lo, hi). |extra| is always one less than a power of 2.
When extra is 0, we're out of PPs, and the slice must be
sorted by some other means. */
PyObject **lo;
PyObject **hi;
int extra;
};
/* The number of PPs we want is 2**k - 1, where 2**k is as close to
N / ln(N) as possible. So k ~= lg(N / ln(N). Calling libm routines
is undesirable, so cutoff values are canned in the "cutoff" table
below: cutoff[i] is the smallest N such that k == CUTOFFBASE + i. */
#define CUTOFFBASE 4
static int cutoff[] = {
43, /* smallest N such that k == 4 */
106, /* etc */
250,
576,
1298,
2885,
6339,
13805,
29843,
64116,
137030,
291554,
617916,
1305130,
2748295,
5771662,
12091672,
25276798,
52734615,
109820537,
228324027,
473977813,
982548444, /* smallest N such that k == 26 */
2034159050 /* largest N that fits in signed 32-bit; k == 27 */
};
static int
samplesortslice(lo, hi, list, compare)
PyObject **lo;
PyObject **hi;
PyListObject *list; /* Needed by docompare for paranoia checks */
PyObject *compare;/* Comparison function object, or NULL for default */
{
register PyObject **l, **r;
register PyObject *tmp, *pivot;
register int k;
int n, extra, top, extraOnRight;
struct SamplesortStackNode stack[STACKSIZE];
/* assert lo <= hi */
n = hi - lo;
/* ----------------------------------------------------------
* Special cases
* --------------------------------------------------------*/
if (n < 2)
return 0;
/* Set r to the largest value such that [lo,r) is sorted.
This catches the already-sorted case, the all-the-same
case, and the appended-a-few-elements-to-a-sorted-list case.
If the array is unsorted, we're very likely to get out of
the loop fast, so the test is cheap if it doesn't pay off.
*/
/* assert lo < hi */
for (r = lo+1; r < hi; ++r) {
SETK(*r, *(r-1));
if (k < 0)
break;
}
/* [lo,r) is sorted, [r,hi) unknown. Get out cheap if there are
few unknowns, or few elements in total. */
if (hi - r <= MAXMERGE || n < MINSIZE)
return binarysort(lo, hi, r, list, compare);
/* Check for the array already being reverse-sorted. Typical
benchmark-driven silliness <wink>. */
/* assert lo < hi */
for (r = lo+1; r < hi; ++r) {
SETK(*(r-1), *r);
if (k < 0)
break;
}
if (hi - r <= MAXMERGE) {
/* Reverse the reversed prefix, then insert the tail */
PyObject **originalr = r;
l = lo;
do {
--r;
tmp = *l; *l = *r; *r = tmp;
++l;
} while (l < r);
return binarysort(lo, hi, originalr, list, compare);
}
/* ----------------------------------------------------------
* Normal case setup: a large array without obvious pattern.
* --------------------------------------------------------*/
/* extra := a power of 2 ~= n/ln(n), less 1.
First find the smallest extra s.t. n < cutoff[extra] */
for (extra = 0;
extra < sizeof(cutoff) / sizeof(cutoff[0]);
++extra) {
if (n < cutoff[extra])
break;
/* note that if we fall out of the loop, the value of
extra still makes *sense*, but may be smaller than
we would like (but the array has more than ~= 2**31
elements in this case!) */
}
/* Now k == extra - 1 + CUTOFFBASE. The smallest value k can
have is CUTOFFBASE-1, so
assert MINSIZE >= 2**(CUTOFFBASE-1) - 1 */
extra = (1 << (extra - 1 + CUTOFFBASE)) - 1;
/* assert extra > 0 and n >= extra */
/* Swap that many values to the start of the array. The
selection of elements is pseudo-random, but the same on
every run (this is intentional! timing algorithm changes is
a pain if timing varies across runs). */
{
unsigned int seed = n / extra; /* arbitrary */
unsigned int i;
for (i = 0; i < (unsigned)extra; ++i) {
/* j := random int in [i, n) */
unsigned int j;
seed = seed * 69069 + 7;
j = i + seed % (n - i);
tmp = lo[i]; lo[i] = lo[j]; lo[j] = tmp;
}
}
/* Recursively sort the preselected pivots. */
if (samplesortslice(lo, lo + extra, list, compare) < 0)
goto fail;
top = 0; /* index of available stack slot */
lo += extra; /* point to first unknown */
extraOnRight = 0; /* the PPs are at the left end */
/* ----------------------------------------------------------
* Partition [lo, hi), and repeat until out of work.
* --------------------------------------------------------*/
for (;;) {
/* assert lo <= hi, so n >= 0 */
n = hi - lo;
/* We may not want, or may not be able, to partition:
If n is small, it's quicker to insert.
If extra is 0, we're out of pivots, and *must* use
another method.
*/
if (n < MINPARTITIONSIZE || extra == 0) {
if (n >= MINSIZE) {
/* assert extra == 0
This is rare, since the average size
of a final block is only about
ln(original n). */
if (samplesortslice(lo, hi, list,
compare) < 0)
goto fail;
}
else {
/* Binary insertion should be quicker,
and we can take advantage of the PPs
already being sorted. */
if (extraOnRight && extra) {
/* swap the PPs to the left end */
k = extra;
do {
tmp = *lo;
*lo = *hi;
*hi = tmp;
++lo; ++hi;
} while (--k);
}
if (binarysort(lo - extra, hi, lo,
list, compare) < 0)
goto fail;
}
/* Find another slice to work on. */
if (--top < 0)
break; /* no more -- done! */
lo = stack[top].lo;
hi = stack[top].hi;
extra = stack[top].extra;
extraOnRight = 0;
if (extra < 0) {
extraOnRight = 1;
extra = -extra;
}
continue;
}
/* Pretend the PPs are indexed 0, 1, ..., extra-1.
Then our preselected pivot is at (extra-1)/2, and we
want to move the PPs before that to the left end of
the slice, and the PPs after that to the right end.
The following section changes extra, lo, hi, and the
slice such that:
[lo-extra, lo) contains the smaller PPs.
*lo == our PP.
(lo, hi) contains the unknown elements.
[hi, hi+extra) contains the larger PPs.
*/
k = extra >>= 1; /* num PPs to move */
if (extraOnRight) {
/* Swap the smaller PPs to the left end.
Note that this loop actually moves k+1 items:
the last is our PP */
do {
tmp = *lo; *lo = *hi; *hi = tmp;
++lo; ++hi;
} while (k--);
}
else {
/* Swap the larger PPs to the right end. */
while (k--) {
--lo; --hi;
tmp = *lo; *lo = *hi; *hi = tmp;
}
}
--lo; /* *lo is now our PP */
pivot = *lo;
/* Now an almost-ordinary quicksort partition step.
Note that most of the time is spent here!
Only odd thing is that we partition into < and >=,
instead of the usual <= and >=. This helps when
there are lots of duplicates of different values,
because it eventually tends to make subfiles
"pure" (all duplicates), and we special-case for
duplicates later. */
l = lo + 1;
r = hi - 1;
/* assert lo < l < r < hi (small n weeded out above) */
do {
/* slide l right, looking for key >= pivot */
do {
SETK(*l, pivot);
if (k < 0)
++l;
else
break;
} while (l < r);
/* slide r left, looking for key < pivot */
while (l < r) {
SETK(*r, pivot);
if (k < 0)
break;
else
--r;
}
/* swap and advance both pointers */
if (l < r) {
tmp = *l; *l = *r; *r = tmp;
++l;
--r;
}
} while (l < r);
/* assert lo < r <= l < hi
assert r == l or r+1 == l
everything to the left of l is < pivot, and
everything to the right of r is >= pivot */
if (l == r) {
SETK(*r, pivot);
if (k < 0)
++l;
else
--r;
}
/* assert lo <= r and r+1 == l and l <= hi
assert r == lo or a[r] < pivot
assert a[lo] is pivot
assert l == hi or a[l] >= pivot
Swap the pivot into "the middle", so we can henceforth
ignore it.
*/
*lo = *r;
*r = pivot;
/* The following is true now, & will be preserved:
All in [lo,r) are < pivot
All in [r,l) == pivot (& so can be ignored)
All in [l,hi) are >= pivot */
/* Check for duplicates of the pivot. One compare is
wasted if there are no duplicates, but can win big
when there are.
Tricky: we're sticking to "<" compares, so deduce
equality indirectly. We know pivot <= *l, so they're
equal iff not pivot < *l.
*/
while (l < hi) {
/* pivot <= *l known */
SETK(pivot, *l);
if (k < 0)
break;
else
/* <= and not < implies == */
++l;
}
/* assert lo <= r < l <= hi
Partitions are [lo, r) and [l, hi) */
/* push fattest first; remember we still have extra PPs
to the left of the left chunk and to the right of
the right chunk! */
/* assert top < STACKSIZE */
if (r - lo <= hi - l) {
/* second is bigger */
stack[top].lo = l;
stack[top].hi = hi;
stack[top].extra = -extra;
hi = r;
extraOnRight = 0;
}
else {
/* first is bigger */
stack[top].lo = lo;
stack[top].hi = r;
stack[top].extra = extra;
lo = l;
extraOnRight = 1;
}
++top;
} /* end of partitioning loop */
return 0;
fail:
return -1;
}
#undef SETK
}
staticforward PyTypeObject immutable_list_type;
static PyObject *
listsort(self, compare)
PyListObject *self;
PyObject *compare;
{
if (quicksort(self, compare) < 0)
int err;
self->ob_type = &immutable_list_type;
err = samplesortslice(self->ob_item,
self->ob_item + self->ob_size,
self, compare);
self->ob_type = &PyList_Type;
if (err < 0)
return NULL;
Py_INCREF(Py_None);
return Py_None;
}
int
PyList_Sort(v)
PyObject *v;
{
if (v == NULL || !PyList_Check(v)) {
PyErr_BadInternalCall();
return -1;
}
v = listsort((PyListObject *)v, (PyObject *)NULL);
if (v == NULL)
return -1;
Py_DECREF(v);
return 0;
}
static PyObject *
listreverse(self, args)
PyListObject *self;
@ -919,17 +1195,6 @@ PyList_Reverse(v)
return 0;
}
int
PyList_Sort(v)
PyObject *v;
{
if (v == NULL || !PyList_Check(v)) {
PyErr_BadInternalCall();
return -1;
}
return quicksort((PyListObject *)v, (PyObject *)NULL);
}
PyObject *
PyList_AsTuple(v)
PyObject *v;
@ -1069,3 +1334,71 @@ PyTypeObject PyList_Type = {
&list_as_sequence, /*tp_as_sequence*/
0, /*tp_as_mapping*/
};
/* During a sort, we really can't have anyone modifying the list; it could
cause core dumps. Thus, we substitute a dummy type that raises an
explanatory exception when a modifying operation is used. Caveat:
comparisons may behave differently; but I guess it's a bad idea anyway to
compare a list that's being sorted... */
static PyObject *
immutable_list_op(/*No args!*/)
{
PyErr_SetString(PyExc_TypeError,
"a list cannot be modified while it is being sorted");
return NULL;
}
static PyMethodDef immutable_list_methods[] = {
{"append", (PyCFunction)immutable_list_op},
{"insert", (PyCFunction)immutable_list_op},
{"remove", (PyCFunction)immutable_list_op},
{"index", (PyCFunction)listindex},
{"count", (PyCFunction)listcount},
{"reverse", (PyCFunction)immutable_list_op},
{"sort", (PyCFunction)immutable_list_op},
{NULL, NULL} /* sentinel */
};
static PyObject *
immutable_list_getattr(f, name)
PyListObject *f;
char *name;
{
return Py_FindMethod(immutable_list_methods, (PyObject *)f, name);
}
static int
immutable_list_ass(/*No args!*/)
{
immutable_list_op();
return -1;
}
static PySequenceMethods immutable_list_as_sequence = {
(inquiry)list_length, /*sq_length*/
(binaryfunc)list_concat, /*sq_concat*/
(intargfunc)list_repeat, /*sq_repeat*/
(intargfunc)list_item, /*sq_item*/
(intintargfunc)list_slice, /*sq_slice*/
(intobjargproc)immutable_list_ass, /*sq_ass_item*/
(intintobjargproc)immutable_list_ass, /*sq_ass_slice*/
};
static PyTypeObject immutable_list_type = {
PyObject_HEAD_INIT(&PyType_Type)
0,
"list (immutable, during sort)",
sizeof(PyListObject),
0,
0, /*tp_dealloc*/ /* Cannot happen */
(printfunc)list_print, /*tp_print*/
(getattrfunc)immutable_list_getattr, /*tp_getattr*/
0, /*tp_setattr*/
0, /*tp_compare*/ /* Won't be called */
(reprfunc)list_repr, /*tp_repr*/
0, /*tp_as_number*/
&immutable_list_as_sequence, /*tp_as_sequence*/
0, /*tp_as_mapping*/
};