On Mon, May 01, 2017 at 03:07:22PM -0700, Jonathan Tan wrote:
> > @@ -2298,8 +2296,12 @@ static uintmax_t do_change_note_fanout(
> > static uintmax_t change_note_fanout(struct tree_entry *root,
> > unsigned char fanout)
> > {
> > - char hex_sha1[40], path[60];
> > - return do_change_note_fanout(root, root, hex_sha1, 0, path, 0, fanout);
> > + /*
> > + * The size of path is due to one slash between every two hex digits,
> > + * plus the terminating NUL.
> > + */
> > + char hex_oid[GIT_MAX_HEXSZ], path[GIT_MAX_HEXSZ * 3 / 2];
>
> If your comment is correct, shouldn't the size of path be 61 (that is, add
> "+ 1")? I took a look at do_change_note_fanout() and your comment seems
> correct.
If you have 40 hex digits, then you have 20 hex pairs. But delimiting
them all takes only 19 slashes, since they only go in between pairs[1].
So the fully expanded formula is:
GIT_MAX_HEXSZ + (1) actual hex bytes
(GIT_MAX_HEXSZ / 2) - 1 + (2) internal slashes between pairs
1 (3) NUL terminator
which simplifies to 3/2 GIT_MAX_HEXSZ. It may be better to write it out
(the compiler can simplify) or explain that in the comment, though. It
took me a minute to figure out that it was correct, too.
-Peff
[1] This is sort of a reverse-fencepost error:
https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error