assign variable inside if condition in bash 4? - bash

is it possible to assign variable inside if conditional in bash 4? ie. in the function below I want to assign output of executing cmd to output and check whether it is an empty string - both inside test conditional. The function should output
"command returned: bar"
myfunc() {
local cmd="echo bar"
local output=
while [[ -z output=`$cmd` ]];
do
#cmd is failing so far, wait and try again
sleep 5
done
# great success
echo "command returned: $output"
}
why the above?
i prefer to run scripts with 'set -e' - which will cause script to terminate on first non-0 return/exit code that's not in an if/loop conditional.
with that in mind, imagine cmd is an unstable command that may exit with > 1 from time to time, and I want to keep calling it until it succeeds and i get some output.

You can try something like this:
myfunc() {
local cmd="echo bar"
local output=
while ! output=$($cmd) || [[ -z output ]];
do
#cmd is failing so far, wait and try again
sleep 5
done
# great success
echo "command returned: $output"
}
Note that it is strongly recommended to avoid the use of set -e.

I don't think you would be able to do it in your conditional
As yi_H pointed out, the if is equivalent to
if [[ ! -z output=bar ]];
which in turn is basically
if [[ ! -z "output=bar" ]];
So, all you are checking is if the string "output=bar" is empty or not...
So, output=bar could actually be anything like !##!#%=== and it would still do the same thing (that is, the expression isn't evaluated). You might have to assign the variable in a subshell somehow, but I'm not sure that would work.

Since assignment won't work there, you need some workaroudn.
You could temporary do a set +e...

You could use this way ...
$cmd
exit_status=$?
while [[ $exit_status -gt 0 ]];
do
#cmd is failing so far, wait and try again
sleep 5
$cmd
exit_status=$?
done
EDIT: This won't work with 'set -e' or other way around, don't use 'set -e' to begin with.

Related

Command online argument in shell script

Hi I have written small shell script, I am not able to understand the behavior of that script. can any one help me to understand that script.
Script:
#!/bin/bash
if [ -z $1 ]
then
echo "fail"
else
echo "success"
fi
While executing the script .
./test.sh one
It exuting the else statement instead of main statement , even though its passing the argument.
can any one explain me this behavior to understand
The -z test in bash is checking if a string is an empty (zero length) value.
Since you're passing an argument to the script $1 is not empty and therefore -z $1 evaluates to false, executing the else portion of your script.
Side note: Since you're working with strings I recommend you to quote variables as follows:
if [ -z "$1" ]; then
echo "String is empty / No argument given"
else
echo "String is not empty / Argument given"
fi
Edit:
As pointed out by user1934428 it's probably better to use [[ instead of [. This, among others, eliminates the need for quoting. See more differences here.
if [[ -z $1 ]]; then
...
However, be aware that this is a bash extension and won't work in sh scripts.

Bash setting a global variable inside a loop and retaining its value -- Or process substituion for dummies

I'm a C/C++ programmer and quite stupid in general (or at least the way bash does things it makes me feel confused). I can't wrap my head around process substitution.
I need to define a global boolean, set it somewhere in a loop, and make use of it in global scope. Could someone please explain in the simplest way possible how to adapt the code below to allow me to achieve my use case, simple enough so that I don't have to contort my brain again tomorrow to try and grasp process substitution .
# DEFINE HERE
for i in `seq 0 ${DAEMON_COUNT}`;
do
if [ ! -d "data$i" ]; then
# SET HERE
echo "data$i does not exist. Creating...";
mkdir data$i
fi
done
# TEST AND USE HERE
to be honest, I don't think bash is up to the task.... the next block looks like this.
echo "-------------------------------------------------------------------------------"
echo "checking the state of potentially running daemons"
for i in `seq 0 ${DAEMON_COUNT}`;
do
if [ ! -e "data$i/mongod.lock" ] ; then
echo "[no lock file] mongod process $i does not exist"
else
echo "[lock file exists] process $i lock file exists "
I_PID=`cat data$i/mongod.lock`
if [ ! ${I_PID} ]; then
echo " [GOOD] lock pid empty"
elif [ "`ps -p ${I_PID} | grep ${I_PID}`" ]; then
echo " [GOOD] data1 pid: ${I_PID} running"
else
echo "[PROBABLY FATAL] data1 pid: ${I_PID} not running."
fi
fi
done
echo "-------------------------------------------------------------------------------"
What I now need is a global array of structs so that I can loop over them and take conditional action to initialize my daemons correctly :/.
Might just use libc and do this stuff in lua, the only reason I hold back is having to install rocks, I don't like ad-hoc code repositories vomiting whatever they want onto my machine :D
The important thing to understand is this: child process is born with its own environment and cannot affect the variables of its parent. If you set a variable in a child process, then the value of the variable in the parent is not affected. These are actually two different variables which just happen to have the same name.
The second thing to understand is when bash runs a command as a child process. There are two cases relevant to the question:
Each process connected by a pipe | is a child of the current shell.
Running a single builtin command with a redirection (e.g. <) will not spawn a child process.
Here is a simple session which demonstrates these ideas:
$ somevar=initial
$ echo test1 | read somevar
$ echo $somevar
initial
$ read somevar < <(echo test2)
$ echo $somevar
test2
The first read is a child process and therefore somevar in the main shell does not change. The second read is executed by the main shell itself and hence somevar is updated.
This means that your code will work as you expect unless you add a pipe in front of or after the for loop, i.e. this works as you want it to:
# DEFINE HERE
SOMEVAR=0
DAEMON_COUNT=10
for i in `seq 0 ${DAEMON_COUNT}`;
do
if [ ! -d "data$i" ]; then
# SET HERE
SOMEVAR=10
echo "data$i does not exist. Creating...";
mkdir data$i
fi
done
# TEST AND USE HERE
echo ${SOMEVAR} # This displays 10
I might have misundestood but...
bool=false;
for i in `seq 0 ${DAEMON_COUNT}`;
do
if [ ! -d "data$i" ]; then
bool=true;
echo "data$i does not exist. Creating...";
mkdir data$i
fi
done
if [ $bool = true ]; then
...
fi
Is this what you want?

“set -e” in a function [duplicate]

set -e (or a script starting with #!/bin/sh -e) is extremely useful to automatically bomb out if there is a problem. It saves me having to error check every single command that might fail.
How do I get the equivalent of this inside a function?
For example, I have the following script that exits immediately on error with an error exit status:
#!/bin/sh -e
echo "the following command could fail:"
false
echo "this is after the command that fails"
The output is as expected:
the following command could fail:
Now I'd like to wrap this into a function:
#!/bin/sh -e
my_function() {
echo "the following command could fail:"
false
echo "this is after the command that fails"
}
if ! my_function; then
echo "dealing with the problem"
fi
echo "run this all the time regardless of the success of my_function"
Expected output:
the following command could fail:
dealing with the problem
run this all the time regardless of the success of my_function
Actual output:
the following output could fail:
this is after the command that fails
run this all the time regardless of the success of my_function
(ie. the function is ignoring set -e)
This presumably is expected behaviour. My question is: how do I get the effect and usefulness of set -e inside a shell function? I'd like to be able to set something up such that I don't have to individually error check every call, but the script will stop on encountering an error. It should unwind the stack as far as is needed until I do check the result, or exit the script itself if I haven't checked it. This is what set -e does already, except it doesn't nest.
I've found the same question asked outside Stack Overflow but no suitable answer.
I eventually went with this, which apparently works. I tried the export method at first, but then found that I needed to export every global (constant) variable the script uses.
Disable set -e, then run the function call inside a subshell that has set -e enabled. Save the exit status of the subshell in a variable, re-enable set -e, then test the var.
f() { echo "a"; false; echo "Should NOT get HERE"; }
# Don't pipe the subshell into anything or we won't be able to see its exit status
set +e ; ( set -e; f ) ; err_status=$?
set -e
## cleaner syntax which POSIX sh doesn't support. Use bash/zsh/ksh/other fancy shells
if ((err_status)) ; then
echo "f returned false: $err_status"
fi
## POSIX-sh features only (e.g. dash, /bin/sh)
if test "$err_status" -ne 0 ; then
echo "f returned false: $err_status"
fi
echo "always print this"
You can't run f as part of a pipeline, or as part of a && of || command list (except as the last command in the pipe or list), or as the condition in an if or while, or other contexts that ignore set -e. This code also can't be in any of those contexts, so if you use this in a function, callers have to use the same subshell / save-exit-status trickery. This use of set -e for semantics similar to throwing/catching exceptions is not really suitable for general use, given the limitations and hard-to-read syntax.
trap err_handler_function ERR has the same limitations as set -e, in that it won't fire for errors in contexts where set -e won't exit on failed commands.
You might think the following would work, but it doesn't:
if ! ( set -e; f );then ##### doesn't work, f runs ignoring -e
echo "f returned false: $?"
fi
set -e doesn't take effect inside the subshell because it remembers that it's inside the condition of an if. I thought being a subshell would change that, but only being in a separate file and running a whole separate shell on it would work.
From documentation of set -e:
When this option is on, if a simple command fails for any of the
reasons listed in Consequences of
Shell Errors or returns an exit status
value > 0, and is not part of the
compound list following a while,
until, or if keyword, and is not a
part of an AND or OR list, and is not
a pipeline preceded by the ! reserved
word, then the shell shall immediately
exit.
In your case, false is a part of a pipeline preceded by ! and a part of if. So the solution is to rewrite your code so that it isn't.
In other words, there's nothing special about functions here. Try:
set -e
! { false; echo hi; }
You may directly use a subshell as your function definition and set it to exit immediately with set -e. This would limit the scope of set -e to the function subshell only and would later avoid switching between set +e and set -e.
In addition, you can use a variable assignment in the if test and then echo the result in an additional else statement.
# use subshell for function definition
f() (
set -exo pipefail
echo a
false
echo Should NOT get HERE
exit 0
)
# next line also works for non-subshell function given by agsamek above
#if ret="$( set -e && f )" ; then
if ret="$( f )" ; then
true
else
echo "$ret"
fi
# prints
# ++ echo a
# ++ false
# a
This is a bit of a kludge, but you can do:
export -f f
if sh -ec f; then
...
This will work if your shell supports export -f (bash does).
Note that this will not terminate the script. The echo
after the false in f will not execute, nor will the body
of the if, but statements after the if will be executed.
If you are using a shell that does not support export -f, you can
get the semantics you want by running sh in the function:
f() { sh -ec '
echo This will execute
false
echo This will not
'
}
Note/Edit: As a commenter pointed out, this answer uses bash, and not sh like the OP used in his question. I missed that detail when I originaly posted an answer. I will leave this answer up anyway since it might be interested to some passerby.
Y'aaaaaaaaaaaaaaaaaaallll ready for this?
Here's a way to do it with leveraging the DEBUG trap, which runs before each command, and sort of makes errors like the whole exception/try/catch idioms from other languages. Take a look. I've made your example one more 'call' deep.
#!/bin/bash
# Get rid of that disgusting set -e. We don't need it anymore!
# functrace allows RETURN and DEBUG traps to be inherited by each
# subshell and function. Plus, it doesn't suffer from that horrible
# erasure problem that -e and -E suffer from when the command
# is used in a conditional expression.
set -o functrace
# A trap to bubble up the error unless our magic command is encountered
# ('catch=$?' in this case) at which point it stops. Also don't try to
# bubble the error if were not in a function.
trap '{
code=$?
if [[ $code != 0 ]] && [[ $BASH_COMMAND != '\''catch=$?'\'' ]]; then
# If were in a function, return, else exit.
[[ $FUNCNAME ]] && return $code || exit $code
fi
}' DEBUG
my_function() {
my_function2
}
my_function2() {
echo "the following command could fail:"
false
echo "this is after the command that fails"
}
# the || isn't necessary, but the 'catch=$?' is.
my_function || catch=$?
echo "Dealing with the problem with errcode=$catch (⌐■_■)"
echo "run this all the time regardless of the success of my_function"
and the output:
the following command could fail:
Dealing with the problem with errcode=1 (⌐■_■)
run this all the time regardless of the success of my_function
I haven't tested this in the wild, but off the top of my head, there are a bunch of pros:
It's actually not that slow. I've ran the script in a tight loop with and without the functrace option, and times are very close to each other under 10 000 iterations.
You could expand on this DEBUG trap to print a stack trace without doing that whole looping over $FUNCNAME and $BASH_LINENO nonsense. You kinda get it for free (besides actually doing an echo line).
Don't have to worry about that shopt -s inherit_errexit gotcha.
Join all commands in your function with the && operator. It's not too much trouble and will give the result you want.
This is by design and POSIX specification. We can read in man bash:
If a compound command or shell function executes in a context where -e is being ignored, none of the commands executed within the compound command or function body will be affected by the -e setting, even if -e is set and a command returns a failure status. If a compound command or shell function sets -e while executing in a context where -e is ignored, that setting will not have any effect until the compound command or the command containing the function call completes.
therefore you should avoid relying on set -e within functions.
Given the following exampleAustin Group:
set -e
start() {
some_server
echo some_server started successfully
}
start || echo >&2 some_server failed
the set -e is ignored within the function, because the function is a command in an AND-OR list other than the last.
The above behaviour is specified and required by POSIX (see: Desired Action):
The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline beginning with the ! reserved word, or any command of an AND-OR list other than the last.
I know this isn't what you asked, but you may or may not be aware that the behavior you seek is built into "make". Any part of a "make" process that fails aborts the run. It's a wholly different way of "programming", though, than shell scripting.
You will need to call your function in a sub shell (inside brackets ()) to achieve this.
I think you want to write your script like this:
#!/bin/sh -e
my_function() {
echo "the following command could fail:"
false
echo "this is after the command that fails"
}
(my_function)
if [ $? -ne 0 ] ; then
echo "dealing with the problem"
fi
echo "run this all the time regardless of the success of my_function"
Then the output is (as desired):
the following command could fail:
dealing with the problem
run this all the time regardless of the success of my_function
If a subshell isn't an option (say you need to do something crazy like set a variable) then you can just check every single command that might fail and deal with it by appending || return $?. This causes the function to return the error code on failure.
It's ugly, but it works
#!/bin/sh
set -e
my_function() {
echo "the following command could fail:"
false || return $?
echo "this is after the command that fails"
}
if ! my_function; then
echo "dealing with the problem"
fi
echo "run this all the time regardless of the success of my_function"
gives
the following command could fail:
dealing with the problem
run this all the time regardless of the success of my_function

How do I get the effect and usefulness of "set -e" inside a shell function?

set -e (or a script starting with #!/bin/sh -e) is extremely useful to automatically bomb out if there is a problem. It saves me having to error check every single command that might fail.
How do I get the equivalent of this inside a function?
For example, I have the following script that exits immediately on error with an error exit status:
#!/bin/sh -e
echo "the following command could fail:"
false
echo "this is after the command that fails"
The output is as expected:
the following command could fail:
Now I'd like to wrap this into a function:
#!/bin/sh -e
my_function() {
echo "the following command could fail:"
false
echo "this is after the command that fails"
}
if ! my_function; then
echo "dealing with the problem"
fi
echo "run this all the time regardless of the success of my_function"
Expected output:
the following command could fail:
dealing with the problem
run this all the time regardless of the success of my_function
Actual output:
the following output could fail:
this is after the command that fails
run this all the time regardless of the success of my_function
(ie. the function is ignoring set -e)
This presumably is expected behaviour. My question is: how do I get the effect and usefulness of set -e inside a shell function? I'd like to be able to set something up such that I don't have to individually error check every call, but the script will stop on encountering an error. It should unwind the stack as far as is needed until I do check the result, or exit the script itself if I haven't checked it. This is what set -e does already, except it doesn't nest.
I've found the same question asked outside Stack Overflow but no suitable answer.
I eventually went with this, which apparently works. I tried the export method at first, but then found that I needed to export every global (constant) variable the script uses.
Disable set -e, then run the function call inside a subshell that has set -e enabled. Save the exit status of the subshell in a variable, re-enable set -e, then test the var.
f() { echo "a"; false; echo "Should NOT get HERE"; }
# Don't pipe the subshell into anything or we won't be able to see its exit status
set +e ; ( set -e; f ) ; err_status=$?
set -e
## cleaner syntax which POSIX sh doesn't support. Use bash/zsh/ksh/other fancy shells
if ((err_status)) ; then
echo "f returned false: $err_status"
fi
## POSIX-sh features only (e.g. dash, /bin/sh)
if test "$err_status" -ne 0 ; then
echo "f returned false: $err_status"
fi
echo "always print this"
You can't run f as part of a pipeline, or as part of a && of || command list (except as the last command in the pipe or list), or as the condition in an if or while, or other contexts that ignore set -e. This code also can't be in any of those contexts, so if you use this in a function, callers have to use the same subshell / save-exit-status trickery. This use of set -e for semantics similar to throwing/catching exceptions is not really suitable for general use, given the limitations and hard-to-read syntax.
trap err_handler_function ERR has the same limitations as set -e, in that it won't fire for errors in contexts where set -e won't exit on failed commands.
You might think the following would work, but it doesn't:
if ! ( set -e; f );then ##### doesn't work, f runs ignoring -e
echo "f returned false: $?"
fi
set -e doesn't take effect inside the subshell because it remembers that it's inside the condition of an if. I thought being a subshell would change that, but only being in a separate file and running a whole separate shell on it would work.
From documentation of set -e:
When this option is on, if a simple command fails for any of the
reasons listed in Consequences of
Shell Errors or returns an exit status
value > 0, and is not part of the
compound list following a while,
until, or if keyword, and is not a
part of an AND or OR list, and is not
a pipeline preceded by the ! reserved
word, then the shell shall immediately
exit.
In your case, false is a part of a pipeline preceded by ! and a part of if. So the solution is to rewrite your code so that it isn't.
In other words, there's nothing special about functions here. Try:
set -e
! { false; echo hi; }
You may directly use a subshell as your function definition and set it to exit immediately with set -e. This would limit the scope of set -e to the function subshell only and would later avoid switching between set +e and set -e.
In addition, you can use a variable assignment in the if test and then echo the result in an additional else statement.
# use subshell for function definition
f() (
set -exo pipefail
echo a
false
echo Should NOT get HERE
exit 0
)
# next line also works for non-subshell function given by agsamek above
#if ret="$( set -e && f )" ; then
if ret="$( f )" ; then
true
else
echo "$ret"
fi
# prints
# ++ echo a
# ++ false
# a
This is a bit of a kludge, but you can do:
export -f f
if sh -ec f; then
...
This will work if your shell supports export -f (bash does).
Note that this will not terminate the script. The echo
after the false in f will not execute, nor will the body
of the if, but statements after the if will be executed.
If you are using a shell that does not support export -f, you can
get the semantics you want by running sh in the function:
f() { sh -ec '
echo This will execute
false
echo This will not
'
}
Note/Edit: As a commenter pointed out, this answer uses bash, and not sh like the OP used in his question. I missed that detail when I originaly posted an answer. I will leave this answer up anyway since it might be interested to some passerby.
Y'aaaaaaaaaaaaaaaaaaallll ready for this?
Here's a way to do it with leveraging the DEBUG trap, which runs before each command, and sort of makes errors like the whole exception/try/catch idioms from other languages. Take a look. I've made your example one more 'call' deep.
#!/bin/bash
# Get rid of that disgusting set -e. We don't need it anymore!
# functrace allows RETURN and DEBUG traps to be inherited by each
# subshell and function. Plus, it doesn't suffer from that horrible
# erasure problem that -e and -E suffer from when the command
# is used in a conditional expression.
set -o functrace
# A trap to bubble up the error unless our magic command is encountered
# ('catch=$?' in this case) at which point it stops. Also don't try to
# bubble the error if were not in a function.
trap '{
code=$?
if [[ $code != 0 ]] && [[ $BASH_COMMAND != '\''catch=$?'\'' ]]; then
# If were in a function, return, else exit.
[[ $FUNCNAME ]] && return $code || exit $code
fi
}' DEBUG
my_function() {
my_function2
}
my_function2() {
echo "the following command could fail:"
false
echo "this is after the command that fails"
}
# the || isn't necessary, but the 'catch=$?' is.
my_function || catch=$?
echo "Dealing with the problem with errcode=$catch (⌐■_■)"
echo "run this all the time regardless of the success of my_function"
and the output:
the following command could fail:
Dealing with the problem with errcode=1 (⌐■_■)
run this all the time regardless of the success of my_function
I haven't tested this in the wild, but off the top of my head, there are a bunch of pros:
It's actually not that slow. I've ran the script in a tight loop with and without the functrace option, and times are very close to each other under 10 000 iterations.
You could expand on this DEBUG trap to print a stack trace without doing that whole looping over $FUNCNAME and $BASH_LINENO nonsense. You kinda get it for free (besides actually doing an echo line).
Don't have to worry about that shopt -s inherit_errexit gotcha.
Join all commands in your function with the && operator. It's not too much trouble and will give the result you want.
This is by design and POSIX specification. We can read in man bash:
If a compound command or shell function executes in a context where -e is being ignored, none of the commands executed within the compound command or function body will be affected by the -e setting, even if -e is set and a command returns a failure status. If a compound command or shell function sets -e while executing in a context where -e is ignored, that setting will not have any effect until the compound command or the command containing the function call completes.
therefore you should avoid relying on set -e within functions.
Given the following exampleAustin Group:
set -e
start() {
some_server
echo some_server started successfully
}
start || echo >&2 some_server failed
the set -e is ignored within the function, because the function is a command in an AND-OR list other than the last.
The above behaviour is specified and required by POSIX (see: Desired Action):
The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline beginning with the ! reserved word, or any command of an AND-OR list other than the last.
I know this isn't what you asked, but you may or may not be aware that the behavior you seek is built into "make". Any part of a "make" process that fails aborts the run. It's a wholly different way of "programming", though, than shell scripting.
You will need to call your function in a sub shell (inside brackets ()) to achieve this.
I think you want to write your script like this:
#!/bin/sh -e
my_function() {
echo "the following command could fail:"
false
echo "this is after the command that fails"
}
(my_function)
if [ $? -ne 0 ] ; then
echo "dealing with the problem"
fi
echo "run this all the time regardless of the success of my_function"
Then the output is (as desired):
the following command could fail:
dealing with the problem
run this all the time regardless of the success of my_function
If a subshell isn't an option (say you need to do something crazy like set a variable) then you can just check every single command that might fail and deal with it by appending || return $?. This causes the function to return the error code on failure.
It's ugly, but it works
#!/bin/sh
set -e
my_function() {
echo "the following command could fail:"
false || return $?
echo "this is after the command that fails"
}
if ! my_function; then
echo "dealing with the problem"
fi
echo "run this all the time regardless of the success of my_function"
gives
the following command could fail:
dealing with the problem
run this all the time regardless of the success of my_function

Exit entire program from a function call in shell?

I have a function which I use in my shell script and based on a certain condition I'd like to call "exit 0" in that function so that the entire script exits. But for some reason the program still runs after calling exit 0 in the function. Why?
check()
{
printf "Would you like to try again?\n"
read S
if [ "$S" = "y" ]
then
./myprog
else
exit 0
fi
}
And I call it like this:
if test $WRONG -eq 7
then
printf "Sorry\n"
check
fi
What you have works for me:
$ cat test-shell-exit
#!/bin/sh
check()
{
printf "Would you like to try again?\n"
read S
if [ "$S" = "y" ]
then
echo Try again
else
echo Done
exit 0
fi
}
echo Before
check
echo After
$ ./test-shell-exit
Before
Would you like to try again?
y
Try again
After
$ ./test-shell-exit
Before
Would you like to try again?
n
Done
Could you try this test case and update your answer with any differences from it? It appears the problem you're running into is caused by something you haven't mentioned.
Update: Example of using a loop instead of calling your program again:
$ cat test-shell-exit
#!/bin/sh
check()
{
printf "Would you like to try again?\n"
read S
if [ "$S" = "y" ]
then
echo Try again
else
echo Done
exit 0
fi
}
while true; do
echo Before
check
echo After
done
$ ./test-shell-exit
Before
Would you like to try again?
y
Try again
After
Before
Would you like to try again?
y
Try again
After
Before
Would you like to try again?
n
Done
Without any code to go by, I'd guess that your function is invoking a sub-shell, and it's that sub-shell which is exit()ing, and not the parent shell running the main script.
I was expecting to find your code exiting a subshell, but I don't see any reason for that, the code looks OK, and it seems to run as expected for me.
So my guess is that either you are not reading a y when you think you are not, or you are really doing the exit 0 and have just missed noticing it in your test results.
When you say "the program still runs" do you mean "the shell program still runs after calling the check procedure"?
I agree with digitalross that something outside of what you're showing us is causing this behavior. exit 0 WILL exit the current process. so somehow check() is running in a separate child process that is exiting. Do you have code outside of this call that is invoking?
BTW, which OS and shell are you running?
You can force a parent to exit via something like (syntax depends on your shell)
check() || exit $?
Bill

Resources