On Fri, Apr 30, 2021 at 11:40 PM Matheus Tavares <matheus.bernardino@xxxxxx> wrote: > > Note: previously, `checkout_all()` would not return on errors, but s/Note: previously/Previously/ > instead call `exit()` with a non-zero code. However, it only did that > after calling `checkout_entry()` for all index entries, thus not > stopping on the first error, but attempting to write the maximum number > of entries possible. In order to maintain this behavior we now propagate > `checkout_all()`s error status to `cmd_checkout_index()`, so that it can > call `run_parallel_checkout()` and attempt to write the queued entries > before exiting with the error code. > > Signed-off-by: Matheus Tavares <matheus.bernardino@xxxxxx> > @@ -142,11 +143,7 @@ static void checkout_all(const char *prefix, int prefix_length) > } > if (last_ce && to_tempfile) > write_tempfile_record(last_ce->name, prefix); > - if (errs) > - /* we have already done our error reporting. > - * exit with the same code as die(). > - */ > - exit(128); So when there were errors in checkout_all(), we used to exit() with error code 128 (same as die())... > @@ -275,12 +277,16 @@ int cmd_checkout_index(int argc, const char **argv, const char *prefix) > strbuf_release(&buf); > } > > + if (all) > + err |= checkout_all(prefix, prefix_length); > + > + if (pc_workers > 1) > + err |= run_parallel_checkout(&state, pc_workers, pc_threshold, > + NULL, NULL); > + > if (err) > return 1; ...but now it looks like we will exit with error code 1. I see that you already answered this comment in the previous round of review, but you didn't add the explanations to the commit message.
