xfs_filblks_t len2 = *indlen2;
xfs_filblks_t nres = len1 + len2; /* new total res. */
xfs_filblks_t stolen = 0;
+ xfs_filblks_t resfactor;
/*
* Steal as many blocks as we can to try and satisfy the worst case
* indlen for both new extents.
*/
- while (nres > ores && avail) {
- nres--;
- avail--;
- stolen++;
- }
+ if (ores < nres && avail)
+ stolen = XFS_FILBLKS_MIN(nres - ores, avail);
+ ores += stolen;
+
+ /* nothing else to do if we've satisfied the new reservation */
+ if (ores >= nres)
+ return stolen;
+
+ /*
+ * We can't meet the total required reservation for the two extents.
+ * Calculate the percent of the overall shortage between both extents
+ * and apply this percentage to each of the requested indlen values.
+ * This distributes the shortage fairly and reduces the chances that one
+ * of the two extents is left with nothing when extents are repeatedly
+ * split.
+ */
+ resfactor = (ores * 100);
+ do_div(resfactor, nres);
+ len1 *= resfactor;
+ do_div(len1, 100);
+ len2 *= resfactor;
+ do_div(len2, 100);
+ ASSERT(len1 + len2 <= ores);
+ ASSERT(len1 < *indlen1 && len2 < *indlen2);
/*
- * The only blocks available are those reserved for the original
- * extent and what we can steal from the extent being removed.
- * If this still isn't enough to satisfy the combined
- * requirements for the two new extents, skim blocks off of each
- * of the new reservations until they match what is available.
+ * Hand out the remainder to each extent. If one of the two reservations
+ * is zero, we want to make sure that one gets a block first. The loop
+ * below starts with len1, so hand len2 a block right off the bat if it
+ * is zero.
*/
- while (nres > ores) {
- if (len1) {
- len1--;
- nres--;
+ ores -= (len1 + len2);
+ ASSERT((*indlen1 - len1) + (*indlen2 - len2) >= ores);
+ if (ores && !len2 && *indlen2) {
+ len2++;
+ ores--;
+ }
+ while (ores) {
+ if (len1 < *indlen1) {
+ len1++;
+ ores--;
}
- if (nres == ores)
+ if (!ores)
break;
- if (len2) {
- len2--;
- nres--;
+ if (len2 < *indlen2) {
+ len2++;
+ ores--;
}
}