The logic described in d020e27 (diff: Fix rename pretty-print when
suffix and prefix overlap, 2013-02-23) is wrong: The proof in the
comment is valid only if both strings are the same length. *One* of
old/new can reach a-1 (b-1, resp.) if 'a' is a suffix of 'b' (or vice
versa).
Since the intent was to let the loop run down to the '/' at the end of
the common prefix, fix it by making that distinction explicit: if
there is no prefix, allow no underrun.
Signed-off-by: Thomas Rast <tr...@student.ethz.ch>
---
Hi Antoine,
Unfortunately there's this bug in your patch. Luckily it was found by
valgrind on t4016 and others.
Cheers
Thomas
diff.c | 17 ++++++++++-------
1 file changed, 10 insertions(+), 7 deletions(-)
diff --git a/diff.c b/diff.c
index 7d49cab..8396527 100644
--- a/diff.c
+++ b/diff.c
@@ -1264,6 +1264,7 @@ static char *pprint_rename(const char *a, const char *b)
const char *new = b;
struct strbuf name = STRBUF_INIT;
int pfx_length, sfx_length;
+ int pfx_adjust_for_slash;
int len_a = strlen(a);
int len_b = strlen(b);
int a_midlen, b_midlen;
@@ -1291,14 +1292,16 @@ static char *pprint_rename(const char *a, const char *b)
new = b + len_b;
sfx_length = 0;
/*
- * Note:
- * if pfx_length is 0, old/new will never reach a - 1 because it
- * would mean the whole string is common suffix. But then, the
- * whole string would also be a common prefix, and we would not
- * have pfx_length equals 0.
+ * If there is a common prefix, it must end in a slash. In
+ * that case we let this loop run 1 into the prefix to see the
+ * same slash.
+ *
+ * If there is no common prefix, we cannot do this as it would
+ * underrun the input strings.
*/
- while (a + pfx_length - 1 <= old &&
- b + pfx_length - 1 <= new &&
+ pfx_adjust_for_slash = (pfx_length ? 1 : 0);
+ while (a + pfx_length - pfx_adjust_for_slash <= old &&
+ b + pfx_length - pfx_adjust_for_slash <= new &&
*old == *new) {
if (*old == '/')
sfx_length = len_a - (old - a);
--
1.8.2.rc1.307.ge0d2dea
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