mirror of https://github.com/python/cpython
Tim's revision of the previous patch. He also added some sparts to
the median-of-three code to get a few percent back.
This commit is contained in:
Guido van Rossum
parent
cc20b76ad0
commit
a119c0dd5e
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@ -659,8 +659,6 @@ quicksort(list, compare)
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PyListObject *list; /* List to sort */
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PyObject *compare;/* Comparison function object, or NULL for default */
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{
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int size = list->ob_size; /* Number of elements to sort */
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PyObject **array = list->ob_item; /* Start of array to sort */
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register PyObject *tmp, *pivot;
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register PyObject **l, **r, **p;
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PyObject **lo, **hi, **notp;
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@ -669,10 +667,12 @@ quicksort(list, compare)
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PyObject **histack[STACKSIZE];
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/* Start out with the whole array on the work stack */
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lostack[0] = array;
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histack[0] = array+size;
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lostack[0] = list->ob_item;
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histack[0] = list->ob_item + list->ob_size;
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top = 1;
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#define SETK(X,Y) if ((k = docompare(X,Y,compare,list))==CMPERROR) goto fail
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/* Repeat until the work stack is empty */
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while (--top >= 0) {
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lo = lostack[top];
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@ -688,10 +688,7 @@ quicksort(list, compare)
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pivot = *r;
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do {
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p = l + ((r - l) >> 1);
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k = docompare(pivot, *p,
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compare, list);
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if (k == CMPERROR)
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return -1;
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SETK(pivot, *p);
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if (k < 0)
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r = p;
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else
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@ -708,28 +705,30 @@ quicksort(list, compare)
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continue;
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}
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/* Choose median of first, middle and last as pivot */
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/* Choose median of first, middle and last as pivot;
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this is a simple unrolled 3-element insertion sort */
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l = lo; /* First */
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p = lo + (n>>1); /* Middle */
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r = hi - 1; /* Last */
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k = docompare(*p, *l, compare, list);
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if (k == CMPERROR)
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return -1;
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if (k < 0)
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{ tmp = *p; *p = *l; *l = tmp; }
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pivot = *p;
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SETK(pivot, *l);
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if (k < 0) {
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*p = *l;
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*l = pivot;
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}
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k = docompare(*r, *p, compare, list);
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if (k == CMPERROR)
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return -1;
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if (k < 0)
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{ tmp = *r; *r = *p; *p = tmp; }
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k = docompare(*p, *l, compare, list);
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if (k == CMPERROR)
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return -1;
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if (k < 0)
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{ tmp = *p; *p = *l; *l = tmp; }
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pivot = *r;
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SETK(pivot, *p);
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if (k < 0) {
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*r = *p;
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*p = pivot; /* for consistency */
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SETK(pivot, *l);
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if (k < 0) {
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*p = *l;
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*l = pivot;
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}
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}
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pivot = *p;
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l++;
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@ -741,9 +740,7 @@ quicksort(list, compare)
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/* Move left index to element >= pivot */
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while (l < p) {
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k = docompare(*l, pivot, compare, list);
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if (k == CMPERROR)
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return -1;
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SETK(*l, pivot);
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if (k < 0)
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l++;
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else {
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@ -753,9 +750,7 @@ quicksort(list, compare)
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}
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/* Move right index to element <= pivot */
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while (r > p) {
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k = docompare(pivot, *r, compare, list);
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if (k == CMPERROR)
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return -1;
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SETK(pivot, *r);
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if (k < 0)
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r--;
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else {
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@ -780,9 +775,9 @@ quicksort(list, compare)
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/* One (exactly) of the pointers is at p */
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/* assert (p == l) ^ (p == r) */
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notp = lisp ? r : l;
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*p = *notp;
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k = (r - l) >> 1;
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if (k) {
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*p = *notp;
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if (lisp) {
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p = r - k;
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l++;
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@ -791,14 +786,12 @@ quicksort(list, compare)
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p = l + k;
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r--;
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}
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/* assert l < p < r */
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*notp = *p;
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*p = pivot; /* for consistency */
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continue;
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}
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/* assert l+1 == r */
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*p = *notp;
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*notp = pivot;
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p = notp;
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break;
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@ -818,9 +811,7 @@ quicksort(list, compare)
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* This wastes a compare if it fails, but can win big
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* when there are runs of duplicates.
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*/
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k = docompare(pivot, *l, compare, list);
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if (k == CMPERROR)
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return -1;
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SETK(pivot, *l);
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if (!(k < 0)) {
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/* Now extend as far as possible (around p) so that:
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All in [lo,r) are <= pivot
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@ -831,9 +822,7 @@ quicksort(list, compare)
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*/
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while (r > lo) {
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/* because r-1 < p, *(r-1) <= pivot is known */
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k = docompare(*(r-1), pivot, compare, list);
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if (k == CMPERROR)
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return -1;
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SETK(*(r-1), pivot);
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if (k < 0)
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break;
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/* <= and not < implies == */
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@ -843,9 +832,7 @@ quicksort(list, compare)
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l++;
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while (l < hi) {
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/* because l > p, pivot <= *l is known */
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k = docompare(pivot, *l, compare, list);
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if (k == CMPERROR)
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return -1;
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SETK(pivot, *l);
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if (k < 0)
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break;
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/* <= and not < implies == */
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@ -873,6 +860,11 @@ quicksort(list, compare)
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/* Success */
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return 0;
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fail:
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return -1;
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#undef SETK
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}
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static PyObject *
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