"lsof &> /dev/null" is apparently zsh (and maybe some other shells,
but not bash) for "redirect both stderr and stdout". Under bash,
I believe what this means is "execute the prior command in the
background, and then execute the meaningless "> /dev/null" command".
That wouldn't be a problem except that the return code becomes the
return code of the "> /dev/null", hiding the return code of the
'lsof' command.
Fix: use the more-standard "lsof >/dev/null 2>&1" syntax.
Signed-off-by: Dan Mick <dan.mick@xxxxxxxxxxx>
---
scripts/tgt-admin | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/scripts/tgt-admin b/scripts/tgt-admin
index d374172..183b0af 100755
--- a/scripts/tgt-admin
+++ b/scripts/tgt-admin
@@ -1319,7 +1319,7 @@ sub check_device {
# Check if userspace uses this device
my $lsof_check = check_exe("lsof");
if ($lsof_check ne 1) {
- system("lsof $backing_store &>/dev/null");
+ system("lsof $backing_store >/dev/null 2>&1");
my $exit_value = $? >> 8;
if ($exit_value eq 0) {
execute("# Device $backing_store is used (already tgtd target?).");
--
1.7.10.4
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