* Note that when inverting, SvCUR shouldn't change */
}
+PERL_STATIC_INLINE IV*
+S_get_invlist_previous_index_addr(pTHX_ SV* invlist)
+{
+ /* Return the address of the UV that is reserved to hold the cached index
+ * */
+
+ PERL_ARGS_ASSERT_GET_INVLIST_PREVIOUS_INDEX_ADDR;
+
+ return (IV *) (SvPVX(invlist) + (INVLIST_PREVIOUS_INDEX_OFFSET * sizeof (UV)));
+}
+
+PERL_STATIC_INLINE IV
+S_invlist_previous_index(pTHX_ SV* const invlist)
+{
+ /* Returns cached index of previous search */
+
+ PERL_ARGS_ASSERT_INVLIST_PREVIOUS_INDEX;
+
+ return *get_invlist_previous_index_addr(invlist);
+}
+
+PERL_STATIC_INLINE void
+S_invlist_set_previous_index(pTHX_ SV* const invlist, const IV index)
+{
+ /* Caches <index> for later retrieval */
+
+ PERL_ARGS_ASSERT_INVLIST_SET_PREVIOUS_INDEX;
+
+ assert(index == 0 || index < (int) _invlist_len(invlist));
+
+ *get_invlist_previous_index_addr(invlist) = index;
+}
+
PERL_STATIC_INLINE UV
S_invlist_max(pTHX_ SV* const invlist)
{
* properly */
*get_invlist_zero_addr(new_list) = UV_MAX;
+ *get_invlist_previous_index_addr(new_list) = 0;
*get_invlist_version_id_addr(new_list) = INVLIST_VERSION_ID;
-#if HEADER_LENGTH != 4
+#if HEADER_LENGTH != 5
# error Need to regenerate VERSION_ID by running perl -E 'say int(rand 2**31-1)', and then changing the #if to the new length
#endif
* contains <cp> */
IV low = 0;
+ IV mid;
IV high = _invlist_len(invlist);
const IV highest_element = high - 1;
const UV* array;
* can't combine this with the test above, because we can't get the array
* unless we know the list is non-empty) */
array = invlist_array(invlist);
- if (cp < array[0]) {
- return -1;
+
+ mid = invlist_previous_index(invlist);
+ assert(mid >=0 && mid <= highest_element);
+
+ /* <mid> contains the cache of the result of the previous call to this
+ * function (0 the first time). See if this call is for the same result,
+ * or if it is for mid-1. This is under the theory that calls to this
+ * function will often be for related code points that are near each other.
+ * And benchmarks show that caching gives better results. We also test
+ * here if the code point is within the bounds of the list. These tests
+ * replace others that would have had to be made anyway to make sure that
+ * the array bounds were not exceeded, and give us extra information at the
+ * same time */
+ if (cp >= array[mid]) {
+ if (cp >= array[highest_element]) {
+ return highest_element;
+ }
+
+ /* Here, array[mid] <= cp < array[highest_element]. This means that
+ * the final element is not the answer, so can exclude it; it also
+ * means that <mid> is not the final element, so can refer to 'mid + 1'
+ * safely */
+ if (cp < array[mid + 1]) {
+ return mid;
+ }
+ high--;
+ low = mid + 1;
+ }
+ else { /* cp < aray[mid] */
+ if (cp < array[0]) { /* Fail if outside the array */
+ return -1;
+ }
+ high = mid;
+ if (cp >= array[mid - 1]) {
+ goto found_entry;
+ }
}
/* Binary search. What we are looking for is <i> such that
* The loop below converges on the i+1. Note that there may not be an
* (i+1)th element in the array, and things work nonetheless */
while (low < high) {
- IV mid = (low + high) / 2;
+ mid = (low + high) / 2;
assert(mid <= highest_element);
if (array[mid] <= cp) { /* cp >= array[mid] */
low = mid + 1;
}
}
- return high - 1;
+ found_entry:
+ high--;
+ invlist_set_previous_index(invlist, high);
+ return high;
}
void