commit - 26c524e8b8f4e1182ecc904b22d03fc6b69609e8
commit + 23d855f8c87933919c482c1e90e8807af8556ab6
blob - cba927a4a5813ed3a1afa1247227a9637271e176
blob + 3cd504beb19cf6cc9b4c82cede0854839951f2c4
--- lib/diff_myers.c
+++ lib/diff_myers.c
*
* So in the forward path, we can only match up diagonals when the delta is odd.
*/
+ if ((delta & 1) == 0)
+ continue;
/* Forwards is done first, so the backwards one was still at d - 1. Can't do this for d == 0. */
int backwards_d = d - 1;
- if ((delta & 1) && (backwards_d >= 0)) {
- debug("backwards_d = %d\n", backwards_d);
+ if (backwards_d < 0)
+ continue;
- /* If both sides have the same length, forward and backward start on the same diagonal, meaning the
- * backwards state index c == k.
- * As soon as the lengths are not the same, the backwards traversal starts on a different diagonal, and
- * c = k shifted by the difference in length.
- */
- int c = k_to_c(k, delta);
+ debug("backwards_d = %d\n", backwards_d);
- /* When the file sizes are very different, the traversal trees start on far distant diagonals.
- * They don't necessarily meet straight on. See whether this forward value is on a diagonal that
- * is also valid in kd_backward[], and match them if so. */
- if (c >= -backwards_d && c <= backwards_d) {
- /* Current k is on a diagonal that exists in kd_backward[]. If the two x positions have
- * met or passed (forward walked onto or past backward), then we've found a midpoint / a
- * mid-box.
- *
- * But we need to avoid matching a situation like this:
- * 0 1
- * x y
- * 0 o-o-o
- * x |\| |
- * 1 o-o-o
- * y | |\|
- * 2 (B)o-o <--(B) backwards traversal reached here
- * a | | |
- * 3 o-o-o<-- prev_x, prev_y
- * b | | |
- * 4 o-o(F) <--(F) forwards traversal reached here
- * x |\| | Now both are on the same diagonal and look like they passed,
- * 5 o-o-o but actually they have sneaked past each other and have not met.
- * y | |\|
- * 6 o-o-o
- *
- * The solution is to notice that prev_x,prev_y were also already past (B).
- */
- int backward_x = kd_backward[c];
- int backward_y = xc_to_y(backward_x, c, delta);
- debug(" prev_x,y = (%d,%d) c%d:backward_x,y = (%d,%d) k%d:x,y = (%d,%d)\n",
- prev_x, prev_y, c, backward_x, backward_y, k, x, xk_to_y(x, k));
- if (prev_x <= backward_x && prev_y <= backward_y
- && x >= backward_x) {
- *meeting_snake = (struct diff_box){
- .left_start = backward_x,
- .left_end = x,
- .right_start = xc_to_y(backward_x, c, delta),
- .right_end = xk_to_y(x, k),
- };
- debug("HIT x=(%u,%u) - y=(%u,%u)\n",
- meeting_snake->left_start,
- meeting_snake->right_start,
- meeting_snake->left_end,
- meeting_snake->right_end);
- return;
- }
+ /* If both sides have the same length, forward and backward start on the same diagonal, meaning the
+ * backwards state index c == k.
+ * As soon as the lengths are not the same, the backwards traversal starts on a different diagonal, and
+ * c = k shifted by the difference in length.
+ */
+ int c = k_to_c(k, delta);
+
+ /* When the file sizes are very different, the traversal trees start on far distant diagonals.
+ * They don't necessarily meet straight on. See whether this forward value is on a diagonal that
+ * is also valid in kd_backward[], and match them if so. */
+ if (c >= -backwards_d && c <= backwards_d) {
+ /* Current k is on a diagonal that exists in kd_backward[]. If the two x positions have
+ * met or passed (forward walked onto or past backward), then we've found a midpoint / a
+ * mid-box.
+ *
+ * But we need to avoid matching a situation like this:
+ * 0 1
+ * x y
+ * 0 o-o-o
+ * x |\| |
+ * 1 o-o-o
+ * y | |\|
+ * 2 (B)o-o <--(B) backwards traversal reached here
+ * a | | |
+ * 3 o-o-o<-- prev_x, prev_y
+ * b | | |
+ * 4 o-o(F) <--(F) forwards traversal reached here
+ * x |\| | Now both are on the same diagonal and look like they passed,
+ * 5 o-o-o but actually they have sneaked past each other and have not met.
+ * y | |\|
+ * 6 o-o-o
+ *
+ * The solution is to notice that prev_x,prev_y were also already past (B).
+ */
+ int backward_x = kd_backward[c];
+ int backward_y = xc_to_y(backward_x, c, delta);
+ debug(" prev_x,y = (%d,%d) c%d:backward_x,y = (%d,%d) k%d:x,y = (%d,%d)\n",
+ prev_x, prev_y, c, backward_x, backward_y, k, x, xk_to_y(x, k));
+ if (prev_x <= backward_x && prev_y <= backward_y
+ && x >= backward_x) {
+ *meeting_snake = (struct diff_box){
+ .left_start = backward_x,
+ .left_end = x,
+ .right_start = xc_to_y(backward_x, c, delta),
+ .right_end = xk_to_y(x, k),
+ };
+ debug("HIT x=(%u,%u) - y=(%u,%u)\n",
+ meeting_snake->left_start,
+ meeting_snake->right_start,
+ meeting_snake->left_end,
+ meeting_snake->right_end);
+ return;
}
}
}
