bdiff: extend matches across popular lines stable
authorMatt Mackall <mpm@selenic.com>
Thu, 02 Jun 2016 17:09:06 -0500
branchstable
changeset 29322 66dbdd3cc2b9
parent 29295 9b4f0ad02f51
child 29323 d29cb5e735e9
bdiff: extend matches across popular lines For very large diffs that have large numbers of identical lines (JSON dumps) that also have large blocks of identical text, bdiff could become confused about which block matches which because it can only match very limited regions. The result is very large diffs for small sets of edits. The earlier recursion rebalancing fix made this behavior more frequent because it's now more prone to match block 1 to block 2. One frequent user of large JSON files reported being unable to pass the resulting diffs through their code review system. Prior to this change, bdiff would calculate the length of a match at (i, j) as 1 + length found at (i-1, j-1). With large number of popular (ignored) lines, this often meant matches couldn't be extended backwards at all and thus all matching regions were very small. Disabling the popularity threshold is not an option because it brings back quadratic behavior. Instead, we extend a match backwards until we either found a previously discovered match or we find a mismatching line. This thus successfully bridges over any popular lines inside and before a matching region. The larger regions then significant reduce the probability of confusion.
mercurial/bdiff.c
--- a/mercurial/bdiff.c	Thu Jun 02 16:18:44 2016 -0700
+++ b/mercurial/bdiff.c	Thu Jun 02 17:09:06 2016 -0500
@@ -166,10 +166,17 @@
 		/* loop through all lines match a[i] in b */
 		for (; j >= b1; j = b[j].n) {
 			/* does this extend an earlier match? */
-			if (i > a1 && j > b1 && pos[j - 1].pos == i - 1)
-				k = pos[j - 1].len + 1;
-			else
-				k = 1;
+			for (k = 1; j - k >= b1 && i - k >= a1; k++) {
+				/* reached an earlier match? */
+				if (pos[j - k].pos == i - k) {
+					k += pos[j - k].len;
+					break;
+				}
+				/* previous line mismatch? */
+				if (a[i - k].e != b[j - k].e)
+					break;
+			}
+
 			pos[j].pos = i;
 			pos[j].len = k;