This question already has answers here:
Are there any standard exit status codes in Linux?
(11 answers)
Closed 9 years ago.
I understand that EXIT command causes the shell or program to terminate.
But what is the difference between the below:
exit 2
exit 3
exit 4
exit $?
how is exit 2 different from exit 3 and so on
This is only an exit code. 0 is for fine exit, otherwise it's the error code. $? is a shell variable storing the previous exit value (so the program which ran before your one).
The exit command takes a single value which is the value of the process (e.g. shell) return code. $? is the return code from the last command executed by the shell.
For instance, the script which exits with a return code corresponding to the first argument:
#!/bin/sh
exit $1
Would give you:
# ./script 1
# echo $?
1
# ./script 2
# echo $?
2
Note on most UNIX systems, the return code is limited to a numeric value between 0 and 255, with 0 indicates success and 1-255 providing error information (specific to each process).
From the Advanced Bash-Scripting Guide, Chapter 6: Exit and Exit Status:
The exit command terminates a script, just as in a C program. It can also return a value, which is available to the script's parent process.
So the exit command lets you assign your own exit value, which you could describe in its man page, for example.
The $? will return the exit code of the previous command. For example; You write a script that executes cat example.txt, which results exit code 1. If you then do exit $?, your script will exit with the same code as cat example.txt
More info on exit codes here: Advanced Bash-Scripting Guide Chapter 6: Exit and Exit Status
Some info on $? here: What is the $? variable in shell scripting?
Related
I want to raise an error in a Bash script with message "Test cases Failed !!!". How to do this in Bash?
For example:
if [ condition ]; then
raise error "Test cases failed !!!"
fi
This depends on where you want the error message be stored.
You can do the following:
echo "Error!" > logfile.log
exit 125
Or the following:
echo "Error!" 1>&2
exit 64
When you raise an exception you stop the program's execution.
You can also use something like exit xxx where xxx is the error code you may want to return to the operating system (from 0 to 255). Here 125 and 64 are just random codes you can exit with. When you need to indicate to the OS that the program stopped abnormally (eg. an error occurred), you need to pass a non-zero exit code to exit.
As #chepner pointed out, you can do exit 1, which will mean an unspecified error.
Basic error handling
If your test case runner returns a non-zero code for failed tests, you can simply write:
test_handler test_case_x; test_result=$?
if ((test_result != 0)); then
printf '%s\n' "Test case x failed" >&2 # write error message to stderr
exit 1 # or exit $test_result
fi
Or even shorter:
if ! test_handler test_case_x; then
printf '%s\n' "Test case x failed" >&2
exit 1
fi
Or the shortest:
test_handler test_case_x || { printf '%s\n' "Test case x failed" >&2; exit 1; }
To exit with test_handler's exit code:
test_handler test_case_x || { ec=$?; printf '%s\n' "Test case x failed" >&2; exit $ec; }
Advanced error handling
If you want to take a more comprehensive approach, you can have an error handler:
exit_if_error() {
local exit_code=$1
shift
[[ $exit_code ]] && # do nothing if no error code passed
((exit_code != 0)) && { # do nothing if error code is 0
printf 'ERROR: %s\n' "$#" >&2 # we can use better logging here
exit "$exit_code" # we could also check to make sure
# error code is numeric when passed
}
}
then invoke it after running your test case:
run_test_case test_case_x
exit_if_error $? "Test case x failed"
or
run_test_case test_case_x || exit_if_error $? "Test case x failed"
The advantages of having an error handler like exit_if_error are:
we can standardize all the error handling logic such as logging, printing a stack trace, notification, doing cleanup etc., in one place
by making the error handler get the error code as an argument, we can spare the caller from the clutter of if blocks that test exit codes for errors
if we have a signal handler (using trap), we can invoke the error handler from there
Error handling and logging library
Here is a complete implementation of error handling and logging:
https://github.com/codeforester/base/blob/master/lib/stdlib.sh
Related posts
Error handling in Bash
The 'caller' builtin command on Bash Hackers Wiki
Are there any standard exit status codes in Linux?
BashFAQ/105 - Why doesn't set -e (or set -o errexit, or trap ERR) do what I expected?
Equivalent of __FILE__, __LINE__ in Bash
Is there a TRY CATCH command in Bash
To add a stack trace to the error handler, you may want to look at this post: Trace of executed programs called by a Bash script
Ignoring specific errors in a shell script
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How do I manage log verbosity inside a shell script?
How to log function name and line number in Bash?
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There are a couple more ways with which you can approach this problem. Assuming one of your requirement is to run a shell script/function containing a few shell commands and check if the script ran successfully and throw errors in case of failures.
The shell commands in generally rely on exit-codes returned to let the shell know if it was successful or failed due to some unexpected events.
So what you want to do falls upon these two categories
exit on error
exit and clean-up on error
Depending on which one you want to do, there are shell options available to use. For the first case, the shell provides an option with set -e and for the second you could do a trap on EXIT
Should I use exit in my script/function?
Using exit generally enhances readability In certain routines, once you know the answer, you want to exit to the calling routine immediately. If the routine is defined in such a way that it doesn’t require any further cleanup once it detects an error, not exiting immediately means that you have to write more code.
So in cases if you need to do clean-up actions on script to make the termination of the script clean, it is preferred to not to use exit.
Should I use set -e for error on exit?
No!
set -e was an attempt to add "automatic error detection" to the shell. Its goal was to cause the shell to abort any time an error occurred, but it comes with a lot of potential pitfalls for example,
The commands that are part of an if test are immune. In the example, if you expect it to break on the test check on the non-existing directory, it wouldn't, it goes through to the else condition
set -e
f() { test -d nosuchdir && echo no dir; }
f
echo survived
Commands in a pipeline other than the last one, are immune. In the example below, because the most recently executed (rightmost) command's exit code is considered ( cat) and it was successful. This could be avoided by setting by the set -o pipefail option but its still a caveat.
set -e
somecommand that fails | cat -
echo survived
Recommended for use - trap on exit
The verdict is if you want to be able to handle an error instead of blindly exiting, instead of using set -e, use a trap on the ERR pseudo signal.
The ERR trap is not to run code when the shell itself exits with a non-zero error code, but when any command run by that shell that is not part of a condition (like in if cmd, or cmd ||) exits with a non-zero exit status.
The general practice is we define an trap handler to provide additional debug information on which line and what cause the exit. Remember the exit code of the last command that caused the ERR signal would still be available at this point.
cleanup() {
exitcode=$?
printf 'error condition hit\n' 1>&2
printf 'exit code returned: %s\n' "$exitcode"
printf 'the command executing at the time of the error was: %s\n' "$BASH_COMMAND"
printf 'command present on line: %d' "${BASH_LINENO[0]}"
# Some more clean up code can be added here before exiting
exit $exitcode
}
and we just use this handler as below on top of the script that is failing
trap cleanup ERR
Putting this together on a simple script that contained false on line 15, the information you would be getting as
error condition hit
exit code returned: 1
the command executing at the time of the error was: false
command present on line: 15
The trap also provides options irrespective of the error to just run the cleanup on shell completion (e.g. your shell script exits), on signal EXIT. You could also trap on multiple signals at the same time. The list of supported signals to trap on can be found on the trap.1p - Linux manual page
Another thing to notice would be to understand that none of the provided methods work if you are dealing with sub-shells are involved in which case, you might need to add your own error handling.
On a sub-shell with set -e wouldn't work. The false is restricted to the sub-shell and never gets propagated to the parent shell. To do the error handling here, add your own logic to do (false) || false
set -e
(false)
echo survived
The same happens with trap also. The logic below wouldn't work for the reasons mentioned above.
trap 'echo error' ERR
(false)
Here's a simple trap that prints the last argument of whatever failed to STDERR, reports the line it failed on, and exits the script with the line number as the exit code. Note these are not always great ideas, but this demonstrates some creative application you could build on.
trap 'echo >&2 "$_ at $LINENO"; exit $LINENO;' ERR
I put that in a script with a loop to test it. I just check for a hit on some random numbers; you might use actual tests. If I need to bail, I call false (which triggers the trap) with the message I want to throw.
For elaborated functionality, have the trap call a processing function. You can always use a case statement on your arg ($_) if you need to do more cleanup, etc. Assign to a var for a little syntactic sugar -
trap 'echo >&2 "$_ at $LINENO"; exit $LINENO;' ERR
throw=false
raise=false
while :
do x=$(( $RANDOM % 10 ))
case "$x" in
0) $throw "DIVISION BY ZERO" ;;
3) $raise "MAGIC NUMBER" ;;
*) echo got $x ;;
esac
done
Sample output:
# bash tst
got 2
got 8
DIVISION BY ZERO at 6
# echo $?
6
Obviously, you could
runTest1 "Test1 fails" # message not used if it succeeds
Lots of room for design improvement.
The draw backs include the fact that false isn't pretty (thus the sugar), and other things tripping the trap might look a little stupid. Still, I like this method.
You have 2 options: Redirect the output of the script to a file, Introduce a log file in the script and
Redirecting output to a file:
Here you assume that the script outputs all necessary info, including warning and error messages. You can then redirect the output to a file of your choice.
./runTests &> output.log
The above command redirects both the standard output and the error output to your log file.
Using this approach you don't have to introduce a log file in the script, and so the logic is a tiny bit easier.
Introduce a log file to the script:
In your script add a log file either by hard coding it:
logFile='./path/to/log/file.log'
or passing it by a parameter:
logFile="${1}" # This assumes the first parameter to the script is the log file
It's a good idea to add the timestamp at the time of execution to the log file at the top of the script:
date '+%Y%-m%d-%H%M%S' >> "${logFile}"
You can then redirect your error messages to the log file
if [ condition ]; then
echo "Test cases failed!!" >> "${logFile}";
fi
This will append the error to the log file and continue execution. If you want to stop execution when critical errors occur, you can exit the script:
if [ condition ]; then
echo "Test cases failed!!" >> "${logFile}";
# Clean up if needed
exit 1;
fi
Note that exit 1 indicates that the program stop execution due to an unspecified error. You can customize this if you like.
Using this approach you can customize your logs and have a different log file for each component of your script.
If you have a relatively small script or want to execute somebody else's script without modifying it to the first approach is more suitable.
If you always want the log file to be at the same location, this is the better option of the 2. Also if you have created a big script with multiple components then you may want to log each part differently and the second approach is your only option.
I often find it useful to write a function to handle error messages so the code is cleaner overall.
# Usage: die [exit_code] [error message]
die() {
local code=$? now=$(date +%T.%N)
if [ "$1" -ge 0 ] 2>/dev/null; then # assume $1 is an error code if numeric
code="$1"
shift
fi
echo "$0: ERROR at ${now%???}${1:+: $*}" >&2
exit $code
}
This takes the error code from the previous command and uses it as the default error code when exiting the whole script. It also notes the time, with microseconds where supported (GNU date's %N is nanoseconds, which we truncate to microseconds later).
If the first option is zero or a positive integer, it becomes the exit code and we remove it from the list of options. We then report the message to standard error, with the name of the script, the word "ERROR", and the time (we use parameter expansion to truncate nanoseconds to microseconds, or for non-GNU times, to truncate e.g. 12:34:56.%N to 12:34:56). A colon and space are added after the word ERROR, but only when there is a provided error message. Finally, we exit the script using the previously determined exit code, triggering any traps as normal.
Some examples (assume the code lives in script.sh):
if [ condition ]; then die 123 "condition not met"; fi
# exit code 123, message "script.sh: ERROR at 14:58:01.234564: condition not met"
$command |grep -q condition || die 1 "'$command' lacked 'condition'"
# exit code 1, "script.sh: ERROR at 14:58:55.825626: 'foo' lacked 'condition'"
$command || die
# exit code comes from command's, message "script.sh: ERROR at 14:59:15.575089"
I've seen a lot of strange quirks in CentOS 6.5's init.d scripts, but one pattern I've seen at the end of most of these scripts is
case "$1" in
# ... commands here
esac
exit $?
What is the purpose of "exit $?" here?
It makes the script return the return code of the last significant command to the calling init system. Whenever a command exits, its return code is stored on $? by the shell.
It's actually not really necessary to explicitly specify $? but scripters probably just include it to be clear about what it intends to do.
exit: exit [n]
Exit the shell.
Exits the shell with a status of N. If N is omitted, the exit status
is that of the last command executed.
I also hope you don't actually mean eend $? of OpenRC:
eend retval [string ]
If retval does not equal 0 then output the string using eerror and !! in square > brackets at the end of the line. Otherwise output ok in square brackets at the end of the line. The value of retval is returned.
See source.
I am confused about what error code the command will return when executing a variable assignment plainly and with command substitution:
a=$(false); echo $?
It outputs 1, which let me think that variable assignment doesn't sweep or produce new error code upon the last one. But when I tried this:
false; a=""; echo $?
It outputs 0, obviously this is what a="" returns and it override 1 returned by false.
I want to know why this happens, is there any particularity in variable assignment that differs from other normal commands? Or just be cause a=$(false) is considered to be a single command and only command substitution part make sense?
-- UPDATE --
Thanks everyone, from the answers and comments I got the point "When you assign a variable using command substitution, the exit status is the status of the command." (by #Barmar), this explanation is excellently clear and easy to understand, but speak doesn't precise enough for programmers, I want to see the reference of this point from authorities such as TLDP or GNU man page, please help me find it out, thanks again!
Upon executing a command as $(command) allows the output of the command to replace itself.
When you say:
a=$(false) # false fails; the output of false is stored in the variable a
the output produced by the command false is stored in the variable a. Moreover, the exit code is the same as produced by the command. help false would tell:
false: false
Return an unsuccessful result.
Exit Status:
Always fails.
On the other hand, saying:
$ false # Exit code: 1
$ a="" # Exit code: 0
$ echo $? # Prints 0
causes the exit code for the assignment to a to be returned which is 0.
EDIT:
Quoting from the manual:
If one of the expansions contained a command substitution, the exit
status of the command is the exit status of the last command
substitution performed.
Quoting from BASHFAQ/002:
How can I store the return value and/or output of a command in a
variable?
...
output=$(command)
status=$?
The assignment to output has no effect on command's exit status, which
is still in $?.
Note that this isn't the case when combined with local, as in local variable="$(command)". That form will exit successfully even if command failed.
Take this Bash script for example:
#!/bin/bash
function funWithLocalAndAssignmentTogether() {
local output="$(echo "Doing some stuff.";exit 1)"
local exitCode=$?
echo "output: $output"
echo "exitCode: $exitCode"
}
function funWithLocalAndAssignmentSeparate() {
local output
output="$(echo "Doing some stuff.";exit 1)"
local exitCode=$?
echo "output: $output"
echo "exitCode: $exitCode"
}
funWithLocalAndAssignmentTogether
funWithLocalAndAssignmentSeparate
Here is the output of this:
nick.parry#nparry-laptop1:~$ ./tmp.sh
output: Doing some stuff.
exitCode: 0
output: Doing some stuff.
exitCode: 1
This is because local is actually a builtin command, and a command like local variable="$(command)" calls local after substituting the output of command. So you get the exit status from local.
I came across the same problem yesterday (Aug 29 2018).
In addition to local mentioned in Nick P.'s answer and #sevko's comment in the accepted answer, declare in global scope also has the same behavior.
Here's my Bash code:
#!/bin/bash
func1()
{
ls file_not_existed
local local_ret1=$?
echo "local_ret1=$local_ret1"
local local_var2=$(ls file_not_existed)
local local_ret2=$?
echo "local_ret2=$local_ret2"
local local_var3
local_var3=$(ls file_not_existed)
local local_ret3=$?
echo "local_ret3=$local_ret3"
}
func1
ls file_not_existed
global_ret1=$?
echo "global_ret1=$global_ret1"
declare global_var2=$(ls file_not_existed)
global_ret2=$?
echo "global_ret2=$global_ret2"
declare global_var3
global_var3=$(ls file_not_existed)
global_ret3=$?
echo "global_ret3=$global_ret3"
The output:
$ ./declare_local_command_substitution.sh 2>/dev/null
local_ret1=2
local_ret2=0
local_ret3=2
global_ret1=2
global_ret2=0
global_ret3=2
Note the values of local_ret2 and global_ret2 in the output above. The exit codes are overwritten by local and declare.
My Bash version:
$ echo $BASH_VERSION
4.4.19(1)-release
(not an answer to original question but too long for comment)
Note that export A=$(false); echo $? outputs 0! Apparently the rules quoted in devnull's answer no longer apply. To add a bit of context to that quote (emphasis mine):
3.7.1 Simple Command Expansion
...
If there is a command name left after expansion, execution proceeds as described below. Otherwise, the command exits. If one of the expansions contained a command substitution, the exit status of the command is the exit status of the last command substitution performed. If there were no command substitutions, the command exits with a status of zero.
3.7.2 Command Search and Execution [ — this is the "below" case]
IIUC the manual describes var=foo as special case of var=foo command... syntax (pretty confusing!). The "exit status of the last command substitution" rule only applies to the no-command case.
While it's tempting to think of export var=foo as a "modified assignment syntax", it isn't — export is a builtin command (that just happens to take assignment-like args).
=> If you want to export a var AND capture command substitution status, do it in 2 stages:
A=$(false)
# ... check $?
export A
This way also works in set -e mode — exits immediately if the command substitution return non-0.
As others have said, the exit code of the command substitution is the exit code of the substituted command, so
FOO=$(false)
echo $?
---
1
However, unexpectedly, adding export to the beginning of that produces a different result:
export FOO=$(false)
echo $?
---
0
This is because, while the substituted command false fails, the export command succeeds, and that is the exit code returned by the statement.
I know the return code will be contained in $? after a command was executed, but what does $? mean after a script was executed? The return code of the last command in that script?
Can I tell if a script has been excuted from head to tail and not interrupted by some unexpected system halt or something?
If I have a script like below excuted,
Command A;
if [ $? -eq 0]
then
echo "OK" >> log
else
echo "failed" >> log
fi
and the system halted while A was running, what will I find in that log file? "OK", "failed" or nothing?
Yes, or the value passed after exit, e.g. exit 31.
Not without taking measures within the other script to make it explicit.
$? reads the exit status of the last command executed. After a function returns, $? gives the exit status of the last command executed in the function. This is Bash's way of giving functions a "return value.
Example
#!/bin/bash
echo hello
echo $? # Exit status 0 returned because command executed successfully.
lskdf # Unrecognized command.
echo $? # Non-zero exit status returned because command failed to execute.
echo
exit 113 # Will return 113 to shell.
# To verify this, type "echo $?" after script terminates.
# By convention, an 'exit 0' indicates success,
#+ while a non-zero exit value means an error or anomalous condition
the return code of the script is indeed the return code of the last command executed, some commands allow you to finish execution at any point and arbitrarily set the return code; those are exit for scripts and return for functions but in both cases if you omit the argument they'll just use the return code of the previous command.
This question already has answers here:
Aborting a shell script if any command returns a non-zero value
(10 answers)
Closed 1 year ago.
I am a noob in shell-scripting. I want to print a message and exit my script if a command fails. I've tried:
my_command && (echo 'my_command failed; exit)
but it does not work. It keeps executing the instructions following this line in the script. I'm using Ubuntu and bash.
Try:
my_command || { echo 'my_command failed' ; exit 1; }
Four changes:
Change && to ||
Use { } in place of ( )
Introduce ; after exit and
spaces after { and before }
Since you want to print the message and exit only when the command fails ( exits with non-zero value) you need a || not an &&.
cmd1 && cmd2
will run cmd2 when cmd1 succeeds(exit value 0). Where as
cmd1 || cmd2
will run cmd2 when cmd1 fails(exit value non-zero).
Using ( ) makes the command inside them run in a sub-shell and calling a exit from there causes you to exit the sub-shell and not your original shell, hence execution continues in your original shell.
To overcome this use { }
The last two changes are required by bash.
The other answers have covered the direct question well, but you may also be interested in using set -e. With that, any command that fails (outside of specific contexts like if tests) will cause the script to abort. For certain scripts, it's very useful.
If you want that behavior for all commands in your script, just add
set -e
set -o pipefail
at the beginning of the script. This pair of options tell the bash interpreter to exit whenever a command returns with a non-zero exit code. (For more details about why pipefail is needed, see http://petereisentraut.blogspot.com/2010/11/pipefail.html)
This does not allow you to print an exit message, though.
Note also, each command's exit status is stored in the shell variable $?, which you can check immediately after running the command. A non-zero status indicates failure:
my_command
if [ $? -eq 0 ]
then
echo "it worked"
else
echo "it failed"
fi
I've hacked up the following idiom:
echo "Generating from IDL..."
idlj -fclient -td java/src echo.idl
if [ $? -ne 0 ]; then { echo "Failed, aborting." ; exit 1; } fi
echo "Compiling classes..."
javac *java
if [ $? -ne 0 ]; then { echo "Failed, aborting." ; exit 1; } fi
echo "Done."
Precede each command with an informative echo, and follow each command with that same
if [ $? -ne 0 ];... line. (Of course, you can edit that error message if you want to.)
Provided my_command is canonically designed, ie returns 0 when succeeds, then && is exactly the opposite of what you want. You want ||.
Also note that ( does not seem right to me in bash, but I cannot try from where I am. Tell me.
my_command || {
echo 'my_command failed' ;
exit 1;
}
You can also use, if you want to preserve exit error status, and have a readable file with one command per line:
my_command1 || exit
my_command2 || exit
This, however will not print any additional error message. But in some cases, the error will be printed by the failed command anyway.
The trap shell builtin allows catching signals, and other useful conditions, including failed command execution (i.e., a non-zero return status). So if you don't want to explicitly test return status of every single command you can say trap "your shell code" ERR and the shell code will be executed any time a command returns a non-zero status. For example:
trap "echo script failed; exit 1" ERR
Note that as with other cases of catching failed commands, pipelines need special treatment; the above won't catch false | true.
Using exit directly may be tricky as the script may be sourced from other places (e.g. from terminal). I prefer instead using subshell with set -e (plus errors should go into cerr, not cout) :
set -e
ERRCODE=0
my_command || ERRCODE=$?
test $ERRCODE == 0 ||
(>&2 echo "My command failed ($ERRCODE)"; exit $ERRCODE)