#!/bin/sh
# make sure chgrp handles symlinks properly

if test "$VERBOSE" = yes; then
  set -x
  chgrp --version
fi

pwd=`pwd`
tmp=slink.$$
trap 'status=$?; cd $pwd; rm -rf $tmp && exit $status' 0
trap '(exit $?); exit' 1 2 13 15

framework_failure=0
mkdir $tmp || framework_failure=1
cd $tmp || framework_failure=1

if test $framework_failure = 1; then
  echo 'failure in testing framework' 1>&2
  (exit 1); exit
fi

groups=`id -nG 2>/dev/null`
case "$groups" in
  *' '*) ;;
  *) cat <<EOF 1>&2
$0: this test requires that you be a member of more than one group,
but running \`id -nG' either failed or found just one.
EOF
     (exit 77); exit
     ;;
esac

set _ $groups; shift
g1=$1
g2=$2

touch f
ln -s f symlink

chgrp $g2 symlink 2> /dev/null
set _ `ls -l symlink`
g=$5
test "$g" = $g2 || {
  cat <<EOF 1>&2
$0: skipping this test; your system doesn't support changing
the owner or group of a symbolic link.
EOF
  (exit 77); exit
}

fail=0

chgrp $g1 f
set _ `ls -l f`; g=$5; test "$g" = $g1 || fail=1

chgrp $g2 symlink || fail=1
set _ `ls -l f`; g=$5; test "$g" = $g1 || fail=1
set _ `ls -l symlink`; g=$5; test "$g" = $g2 || fail=1

# This should not change the group of f.
chgrp $g2 symlink || fail=1
set _ `ls -l f`; g=$5; test "$g" = $g1 || fail=1
set _ `ls -l symlink`; g=$5; test "$g" = $g2 || fail=1

chgrp $g2 f
set _ `ls -l f`; g=$5; test "$g" = $g2 || fail=1

# This *should* change the group of f.
# Though note that the diagnostic is misleading in that
# it says the `group of `symlink'' has been changed.
chgrp --dereference $g1 symlink
set _ `ls -l f`; g=$5; test "$g" = $g1 || fail=1
set _ `ls -l symlink`; g=$5; test "$g" = $g2 || fail=1

# See about traversing a symlink to a directory.
mkdir d e
touch e/f
ln -s ../e d/s
chgrp -R $g1 e/f
# Neither of the following should not change the group of e/f
chgrp -R $g2 d
set _ `ls -l e/f`; g=$5; test "$g" = $g1 || fail=1
chgrp --deref -R $g2 d
set _ `ls -l e/f`; g=$5; test "$g" = $g1 || fail=1

(exit $fail); exit