Re: [PATCH 10/19] git-submodule.sh: convert test -a/-o to && and ||
Am 5/20/2014 15:50, schrieb Elia Pinto:
> # If we don't already have a -f flag and the submodule has never been checked out
> - if test -z "$subsha1" -a -z "$force"
> + if test -z "$subsha1" || test -z "$force"
Should not be ||, but &&!
Show 10 quoted lines
> while read mod_src mod_dst sha1_src sha1_dst status sm_path
> do
> # Always show modules deleted or type-changed (blob<->module)
> - test $status = D -o $status = T && echo "$sm_path" && continue
> + {
> + test "$status" = D ||
> + test "$status" = T
> + } &&
> + echo "$sm_path"
> + && continueAs Matthieu noted, this is incorrect. It's not just a style violation, it's a syntax error. Why did your test runs not hickup on that?
In this case you could even leave the original code structure without changing the meaning:
test $status = D || test $status = T && echo "$sm_path" && continue
But a better idiom is
case "$status" in
[DT])
printf '%s\n' "$sm_path" &&
continue
esac
Show 6 quoted lines
> @@ -1233,7 +1238,7 @@ cmd_status()
> say "U$sha1 $displaypath"
> continue
> fi
> - if test -z "$url" || ! test -d "$sm_path"/.git -o -f "$sm_path"/.git
> + if test -z "$url" || ! test -d "$sm_path"/.git || test -f "$sm_path"/.git
Wrong grouping. This could be more correct (I didn't test):
if test -z "$url" ||
{
! test -d "$sm_path"/.git &&
! test -f "$sm_path"/.git
}-- Hannes
Elia Pinto
