Rubén Justo <rjusto@xxxxxxxxx> writes:
> @@ -1448,10 +1448,15 @@ static int patch_update_file(struct add_p_state *s,
>
> strbuf_reset(&s->buf);
> if (file_diff->hunk_nr) {
> - render_hunk(s, hunk, 0, colored, &s->buf);
> - fputs(s->buf.buf, stdout);
> + if (rendered_hunk_index != hunk_index) {
So, the one previously rendered is compared with the current one,
which raises an obvious question, what happens to the first new hunk
resulting from splitting a hunk? The answer is below ...
> + render_hunk(s, hunk, 0, colored, &s->buf);
> + fputs(s->buf.buf, stdout);
> +
> + rendered_hunk_index = hunk_index;
> + }
>
> strbuf_reset(&s->buf);
> +
> if (undecided_previous >= 0) {
> permitted |= ALLOW_GOTO_PREVIOUS_UNDECIDED_HUNK;
> strbuf_addstr(&s->buf, ",k");
> @@ -1649,10 +1654,12 @@ static int patch_update_file(struct add_p_state *s,
> if (!(permitted & ALLOW_SPLIT))
> err(s, _("Sorry, cannot split this hunk"));
> else if (!split_hunk(s, file_diff,
> - hunk - file_diff->hunk))
> + hunk - file_diff->hunk)) {
> color_fprintf_ln(stdout, s->s.header_color,
> _("Split into %d hunks."),
> (int)splittable_into);
> + rendered_hunk_index = -1;
> + }
... we explicitly say "we always want to show the current one after
this operation", which makes sense.
> } else if (s->answer.buf[0] == 'e') {
> if (!(permitted & ALLOW_EDIT))
> err(s, _("Sorry, cannot edit this hunk"));
> @@ -1661,7 +1668,7 @@ static int patch_update_file(struct add_p_state *s,
> goto soft_increment;
> }
> } else if (s->answer.buf[0] == 'p') {
> - /* nothing special is needed */
> + rendered_hunk_index = -1;
And that matches what is done for 'p', which is the base case that
wants to say "no matter what, show the current one". Doubly makes
sense.
> } else {
> const char *p = _(help_patch_remainder), *eol = p;
Looking good. As we are not doing anything dynamic to the help
text, I think dropping "again" in [1/2] would make sense.
Will queue. Thanks.
