Capture exit code of exit command
I have this in a bash script:
exit 3;
exit_code="$?"
if [[ "$exit_code" != "0" ]]; then
echo -e "${r2g_magenta}Your r2g process is exiting with code $exit_code.${r2g_no_color}";
exit "$exit_code";
fi
It looks like it will exit right after the exit command, which makes sense.
I was wondering is there some simple command that can provide an exit code without exiting right away?
I was going to guess:
exec exit 3
but it gives an error message: exec: exit: not found
.
What can I do? :)
bash shell-script exec exit exit-code
|
show 1 more comment
I have this in a bash script:
exit 3;
exit_code="$?"
if [[ "$exit_code" != "0" ]]; then
echo -e "${r2g_magenta}Your r2g process is exiting with code $exit_code.${r2g_no_color}";
exit "$exit_code";
fi
It looks like it will exit right after the exit command, which makes sense.
I was wondering is there some simple command that can provide an exit code without exiting right away?
I was going to guess:
exec exit 3
but it gives an error message: exec: exit: not found
.
What can I do? :)
bash shell-script exec exit exit-code
1
Yeahexec exit 3
is no bueno, I get"exec: exit: not found"
– MrCholo
Dec 9 '18 at 8:34
6
I don't understand the question. Why not setexit_code=3
and eliminate theexit 3
line altogether?
– wjandrea
Dec 9 '18 at 18:25
@wjandrea is more a conceptual question than practical
– MrCholo
Dec 10 '18 at 1:03
2
It still makes no sense to me. Why would there be an exit code if you don't actually exit?
– Barmar
Dec 10 '18 at 5:09
1
@Barmar every process has an exit code. Most of the people trying to answer the question are interpreting the question to mean "what can I replace the 'exit 3' with in the script so it sets the$?
variable but doesn't exit this script"?
– icarus
Dec 10 '18 at 18:32
|
show 1 more comment
I have this in a bash script:
exit 3;
exit_code="$?"
if [[ "$exit_code" != "0" ]]; then
echo -e "${r2g_magenta}Your r2g process is exiting with code $exit_code.${r2g_no_color}";
exit "$exit_code";
fi
It looks like it will exit right after the exit command, which makes sense.
I was wondering is there some simple command that can provide an exit code without exiting right away?
I was going to guess:
exec exit 3
but it gives an error message: exec: exit: not found
.
What can I do? :)
bash shell-script exec exit exit-code
I have this in a bash script:
exit 3;
exit_code="$?"
if [[ "$exit_code" != "0" ]]; then
echo -e "${r2g_magenta}Your r2g process is exiting with code $exit_code.${r2g_no_color}";
exit "$exit_code";
fi
It looks like it will exit right after the exit command, which makes sense.
I was wondering is there some simple command that can provide an exit code without exiting right away?
I was going to guess:
exec exit 3
but it gives an error message: exec: exit: not found
.
What can I do? :)
bash shell-script exec exit exit-code
bash shell-script exec exit exit-code
edited Dec 9 '18 at 19:44
G-Man
12.9k93364
12.9k93364
asked Dec 9 '18 at 8:33
MrCholo
1566
1566
1
Yeahexec exit 3
is no bueno, I get"exec: exit: not found"
– MrCholo
Dec 9 '18 at 8:34
6
I don't understand the question. Why not setexit_code=3
and eliminate theexit 3
line altogether?
– wjandrea
Dec 9 '18 at 18:25
@wjandrea is more a conceptual question than practical
– MrCholo
Dec 10 '18 at 1:03
2
It still makes no sense to me. Why would there be an exit code if you don't actually exit?
– Barmar
Dec 10 '18 at 5:09
1
@Barmar every process has an exit code. Most of the people trying to answer the question are interpreting the question to mean "what can I replace the 'exit 3' with in the script so it sets the$?
variable but doesn't exit this script"?
– icarus
Dec 10 '18 at 18:32
|
show 1 more comment
1
Yeahexec exit 3
is no bueno, I get"exec: exit: not found"
– MrCholo
Dec 9 '18 at 8:34
6
I don't understand the question. Why not setexit_code=3
and eliminate theexit 3
line altogether?
– wjandrea
Dec 9 '18 at 18:25
@wjandrea is more a conceptual question than practical
– MrCholo
Dec 10 '18 at 1:03
2
It still makes no sense to me. Why would there be an exit code if you don't actually exit?
– Barmar
Dec 10 '18 at 5:09
1
@Barmar every process has an exit code. Most of the people trying to answer the question are interpreting the question to mean "what can I replace the 'exit 3' with in the script so it sets the$?
variable but doesn't exit this script"?
– icarus
Dec 10 '18 at 18:32
1
1
Yeah
exec exit 3
is no bueno, I get "exec: exit: not found"
– MrCholo
Dec 9 '18 at 8:34
Yeah
exec exit 3
is no bueno, I get "exec: exit: not found"
– MrCholo
Dec 9 '18 at 8:34
6
6
I don't understand the question. Why not set
exit_code=3
and eliminate the exit 3
line altogether?– wjandrea
Dec 9 '18 at 18:25
I don't understand the question. Why not set
exit_code=3
and eliminate the exit 3
line altogether?– wjandrea
Dec 9 '18 at 18:25
@wjandrea is more a conceptual question than practical
– MrCholo
Dec 10 '18 at 1:03
@wjandrea is more a conceptual question than practical
– MrCholo
Dec 10 '18 at 1:03
2
2
It still makes no sense to me. Why would there be an exit code if you don't actually exit?
– Barmar
Dec 10 '18 at 5:09
It still makes no sense to me. Why would there be an exit code if you don't actually exit?
– Barmar
Dec 10 '18 at 5:09
1
1
@Barmar every process has an exit code. Most of the people trying to answer the question are interpreting the question to mean "what can I replace the 'exit 3' with in the script so it sets the
$?
variable but doesn't exit this script"?– icarus
Dec 10 '18 at 18:32
@Barmar every process has an exit code. Most of the people trying to answer the question are interpreting the question to mean "what can I replace the 'exit 3' with in the script so it sets the
$?
variable but doesn't exit this script"?– icarus
Dec 10 '18 at 18:32
|
show 1 more comment
5 Answers
5
active
oldest
votes
If you have a script that runs some program
and looks at the program's exit status (with $?
),
and you want to test that script by doing something
that causes $?
to be set to some known value (e.g., 3
), just do
(exit 3)
The parentheses create a sub-shell.
Then the exit
command causes that sub-shell
to exit with the specified exit status.
Also, for debugging purposes it'd be just as simple to setexit_code="3"
for testing
– Centimane
Dec 10 '18 at 17:34
1
Yes, wjandrea pointed that out yesterday.
– G-Man
Dec 10 '18 at 20:25
add a comment |
exit
is a bash built-in, so you can't exec
it. Per bash's manual:
Bash's exit status is the exit status of the last command executed in the script. If no commands are executed, the exit status is 0.
Putting all this together, I'd say your only option is to store your desired exit status in a variable and then exit $MY_EXIT_STATUS
when appropriate.
hmmm what do you think about @G-man's idea?
– MrCholo
Dec 9 '18 at 8:52
2
Maybe I misunderstood what you're trying to accomplish. If you're just trying to set$?
(though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value,false
is another option.
– solarshado
Dec 9 '18 at 9:04
add a comment |
You can write a function that returns the status given as argument, or 255
if none given. (I call it ret
as it "returns" its value.)
ret() { return "${1:-255}"; }
and use ret
in place of your call to exit
. This is avoids the inefficiency of creating the sub-shell in the currently accepted answer.
Some measurements.
time bash -c 'for i in {1..10000} ; do (exit 3) ; done ; echo $?'
on my machine takes about 3.5 seconds.
time bash -c 'ret(){ return $1 ; } ; for i in {1..10000} ; do ret 3 ; done ; echo $?'
on my machine takes about 0.051 seconds, 70 times faster. Putting in the default handling still leaves it 60 times faster. Obviously the loop has some overhead. If I change the body of the loop to just be :
or true
then the time is halved to 0.025, a completely empty loop is invalid syntax. Adding ;:
to the loop shows that this minimal command takes 0.007 seconds, so the loop overhead is about 0.018. Subtracting this overhead from the two tests shows that the ret
solution is over 100 times faster.
Obviously this is a synthetic measurement, but things add up. If you make everything 100 times slower than they need to be then you end up with slow systems.
0.0
2
@iBug The extra space is not needed.
– icarus
Dec 9 '18 at 13:15
Good point about the inefficiency of creating the sub-shell. I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.
– G-Man
Dec 9 '18 at 19:38
add a comment |
About exec exit 3
... it would try to run an external command called exit
, but there isn't one, so you get the error. It has to be an external command instead of one built in to the shell, since exec
replaces the shell completely. Which also means that even if you had an external command called exit
, exec exit 3
would not return to continue your shell script, since the shell wouldn't be there any more.
1
I guess you could doexec bash -c "exit 3"
, but at the moment I can't think of any reason to do that as opposed to justexit 3
.
– David Z
Dec 9 '18 at 15:45
1
@DavidZ, in any case,exec
'ing or justexit
'ing will stop the script, which didn't seem like what the question wanted.
– ilkkachu
Dec 9 '18 at 17:16
add a comment |
You can do this with Awk:
awk 'BEGIN{exit 9}'
Or Sed:
sed Q9 /proc/stat
...or with a shell:sh -c 'exit 9'
– ilkkachu
Dec 9 '18 at 17:27
1
@ilkkachu if youre going to do that you might as well do the(exit 9)
in the accepted answer
– Steven Penny
Dec 9 '18 at 18:44
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you have a script that runs some program
and looks at the program's exit status (with $?
),
and you want to test that script by doing something
that causes $?
to be set to some known value (e.g., 3
), just do
(exit 3)
The parentheses create a sub-shell.
Then the exit
command causes that sub-shell
to exit with the specified exit status.
Also, for debugging purposes it'd be just as simple to setexit_code="3"
for testing
– Centimane
Dec 10 '18 at 17:34
1
Yes, wjandrea pointed that out yesterday.
– G-Man
Dec 10 '18 at 20:25
add a comment |
If you have a script that runs some program
and looks at the program's exit status (with $?
),
and you want to test that script by doing something
that causes $?
to be set to some known value (e.g., 3
), just do
(exit 3)
The parentheses create a sub-shell.
Then the exit
command causes that sub-shell
to exit with the specified exit status.
Also, for debugging purposes it'd be just as simple to setexit_code="3"
for testing
– Centimane
Dec 10 '18 at 17:34
1
Yes, wjandrea pointed that out yesterday.
– G-Man
Dec 10 '18 at 20:25
add a comment |
If you have a script that runs some program
and looks at the program's exit status (with $?
),
and you want to test that script by doing something
that causes $?
to be set to some known value (e.g., 3
), just do
(exit 3)
The parentheses create a sub-shell.
Then the exit
command causes that sub-shell
to exit with the specified exit status.
If you have a script that runs some program
and looks at the program's exit status (with $?
),
and you want to test that script by doing something
that causes $?
to be set to some known value (e.g., 3
), just do
(exit 3)
The parentheses create a sub-shell.
Then the exit
command causes that sub-shell
to exit with the specified exit status.
answered Dec 9 '18 at 8:49
G-Man
12.9k93364
12.9k93364
Also, for debugging purposes it'd be just as simple to setexit_code="3"
for testing
– Centimane
Dec 10 '18 at 17:34
1
Yes, wjandrea pointed that out yesterday.
– G-Man
Dec 10 '18 at 20:25
add a comment |
Also, for debugging purposes it'd be just as simple to setexit_code="3"
for testing
– Centimane
Dec 10 '18 at 17:34
1
Yes, wjandrea pointed that out yesterday.
– G-Man
Dec 10 '18 at 20:25
Also, for debugging purposes it'd be just as simple to set
exit_code="3"
for testing– Centimane
Dec 10 '18 at 17:34
Also, for debugging purposes it'd be just as simple to set
exit_code="3"
for testing– Centimane
Dec 10 '18 at 17:34
1
1
Yes, wjandrea pointed that out yesterday.
– G-Man
Dec 10 '18 at 20:25
Yes, wjandrea pointed that out yesterday.
– G-Man
Dec 10 '18 at 20:25
add a comment |
exit
is a bash built-in, so you can't exec
it. Per bash's manual:
Bash's exit status is the exit status of the last command executed in the script. If no commands are executed, the exit status is 0.
Putting all this together, I'd say your only option is to store your desired exit status in a variable and then exit $MY_EXIT_STATUS
when appropriate.
hmmm what do you think about @G-man's idea?
– MrCholo
Dec 9 '18 at 8:52
2
Maybe I misunderstood what you're trying to accomplish. If you're just trying to set$?
(though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value,false
is another option.
– solarshado
Dec 9 '18 at 9:04
add a comment |
exit
is a bash built-in, so you can't exec
it. Per bash's manual:
Bash's exit status is the exit status of the last command executed in the script. If no commands are executed, the exit status is 0.
Putting all this together, I'd say your only option is to store your desired exit status in a variable and then exit $MY_EXIT_STATUS
when appropriate.
hmmm what do you think about @G-man's idea?
– MrCholo
Dec 9 '18 at 8:52
2
Maybe I misunderstood what you're trying to accomplish. If you're just trying to set$?
(though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value,false
is another option.
– solarshado
Dec 9 '18 at 9:04
add a comment |
exit
is a bash built-in, so you can't exec
it. Per bash's manual:
Bash's exit status is the exit status of the last command executed in the script. If no commands are executed, the exit status is 0.
Putting all this together, I'd say your only option is to store your desired exit status in a variable and then exit $MY_EXIT_STATUS
when appropriate.
exit
is a bash built-in, so you can't exec
it. Per bash's manual:
Bash's exit status is the exit status of the last command executed in the script. If no commands are executed, the exit status is 0.
Putting all this together, I'd say your only option is to store your desired exit status in a variable and then exit $MY_EXIT_STATUS
when appropriate.
answered Dec 9 '18 at 8:38
solarshado
26714
26714
hmmm what do you think about @G-man's idea?
– MrCholo
Dec 9 '18 at 8:52
2
Maybe I misunderstood what you're trying to accomplish. If you're just trying to set$?
(though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value,false
is another option.
– solarshado
Dec 9 '18 at 9:04
add a comment |
hmmm what do you think about @G-man's idea?
– MrCholo
Dec 9 '18 at 8:52
2
Maybe I misunderstood what you're trying to accomplish. If you're just trying to set$?
(though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value,false
is another option.
– solarshado
Dec 9 '18 at 9:04
hmmm what do you think about @G-man's idea?
– MrCholo
Dec 9 '18 at 8:52
hmmm what do you think about @G-man's idea?
– MrCholo
Dec 9 '18 at 8:52
2
2
Maybe I misunderstood what you're trying to accomplish. If you're just trying to set
$?
(though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value, false
is another option.– solarshado
Dec 9 '18 at 9:04
Maybe I misunderstood what you're trying to accomplish. If you're just trying to set
$?
(though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value, false
is another option.– solarshado
Dec 9 '18 at 9:04
add a comment |
You can write a function that returns the status given as argument, or 255
if none given. (I call it ret
as it "returns" its value.)
ret() { return "${1:-255}"; }
and use ret
in place of your call to exit
. This is avoids the inefficiency of creating the sub-shell in the currently accepted answer.
Some measurements.
time bash -c 'for i in {1..10000} ; do (exit 3) ; done ; echo $?'
on my machine takes about 3.5 seconds.
time bash -c 'ret(){ return $1 ; } ; for i in {1..10000} ; do ret 3 ; done ; echo $?'
on my machine takes about 0.051 seconds, 70 times faster. Putting in the default handling still leaves it 60 times faster. Obviously the loop has some overhead. If I change the body of the loop to just be :
or true
then the time is halved to 0.025, a completely empty loop is invalid syntax. Adding ;:
to the loop shows that this minimal command takes 0.007 seconds, so the loop overhead is about 0.018. Subtracting this overhead from the two tests shows that the ret
solution is over 100 times faster.
Obviously this is a synthetic measurement, but things add up. If you make everything 100 times slower than they need to be then you end up with slow systems.
0.0
2
@iBug The extra space is not needed.
– icarus
Dec 9 '18 at 13:15
Good point about the inefficiency of creating the sub-shell. I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.
– G-Man
Dec 9 '18 at 19:38
add a comment |
You can write a function that returns the status given as argument, or 255
if none given. (I call it ret
as it "returns" its value.)
ret() { return "${1:-255}"; }
and use ret
in place of your call to exit
. This is avoids the inefficiency of creating the sub-shell in the currently accepted answer.
Some measurements.
time bash -c 'for i in {1..10000} ; do (exit 3) ; done ; echo $?'
on my machine takes about 3.5 seconds.
time bash -c 'ret(){ return $1 ; } ; for i in {1..10000} ; do ret 3 ; done ; echo $?'
on my machine takes about 0.051 seconds, 70 times faster. Putting in the default handling still leaves it 60 times faster. Obviously the loop has some overhead. If I change the body of the loop to just be :
or true
then the time is halved to 0.025, a completely empty loop is invalid syntax. Adding ;:
to the loop shows that this minimal command takes 0.007 seconds, so the loop overhead is about 0.018. Subtracting this overhead from the two tests shows that the ret
solution is over 100 times faster.
Obviously this is a synthetic measurement, but things add up. If you make everything 100 times slower than they need to be then you end up with slow systems.
0.0
2
@iBug The extra space is not needed.
– icarus
Dec 9 '18 at 13:15
Good point about the inefficiency of creating the sub-shell. I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.
– G-Man
Dec 9 '18 at 19:38
add a comment |
You can write a function that returns the status given as argument, or 255
if none given. (I call it ret
as it "returns" its value.)
ret() { return "${1:-255}"; }
and use ret
in place of your call to exit
. This is avoids the inefficiency of creating the sub-shell in the currently accepted answer.
Some measurements.
time bash -c 'for i in {1..10000} ; do (exit 3) ; done ; echo $?'
on my machine takes about 3.5 seconds.
time bash -c 'ret(){ return $1 ; } ; for i in {1..10000} ; do ret 3 ; done ; echo $?'
on my machine takes about 0.051 seconds, 70 times faster. Putting in the default handling still leaves it 60 times faster. Obviously the loop has some overhead. If I change the body of the loop to just be :
or true
then the time is halved to 0.025, a completely empty loop is invalid syntax. Adding ;:
to the loop shows that this minimal command takes 0.007 seconds, so the loop overhead is about 0.018. Subtracting this overhead from the two tests shows that the ret
solution is over 100 times faster.
Obviously this is a synthetic measurement, but things add up. If you make everything 100 times slower than they need to be then you end up with slow systems.
0.0
You can write a function that returns the status given as argument, or 255
if none given. (I call it ret
as it "returns" its value.)
ret() { return "${1:-255}"; }
and use ret
in place of your call to exit
. This is avoids the inefficiency of creating the sub-shell in the currently accepted answer.
Some measurements.
time bash -c 'for i in {1..10000} ; do (exit 3) ; done ; echo $?'
on my machine takes about 3.5 seconds.
time bash -c 'ret(){ return $1 ; } ; for i in {1..10000} ; do ret 3 ; done ; echo $?'
on my machine takes about 0.051 seconds, 70 times faster. Putting in the default handling still leaves it 60 times faster. Obviously the loop has some overhead. If I change the body of the loop to just be :
or true
then the time is halved to 0.025, a completely empty loop is invalid syntax. Adding ;:
to the loop shows that this minimal command takes 0.007 seconds, so the loop overhead is about 0.018. Subtracting this overhead from the two tests shows that the ret
solution is over 100 times faster.
Obviously this is a synthetic measurement, but things add up. If you make everything 100 times slower than they need to be then you end up with slow systems.
0.0
edited Dec 10 '18 at 17:55
answered Dec 9 '18 at 12:54
icarus
5,6611929
5,6611929
2
@iBug The extra space is not needed.
– icarus
Dec 9 '18 at 13:15
Good point about the inefficiency of creating the sub-shell. I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.
– G-Man
Dec 9 '18 at 19:38
add a comment |
2
@iBug The extra space is not needed.
– icarus
Dec 9 '18 at 13:15
Good point about the inefficiency of creating the sub-shell. I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.
– G-Man
Dec 9 '18 at 19:38
2
2
@iBug The extra space is not needed.
– icarus
Dec 9 '18 at 13:15
@iBug The extra space is not needed.
– icarus
Dec 9 '18 at 13:15
Good point about the inefficiency of creating the sub-shell. I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.
– G-Man
Dec 9 '18 at 19:38
Good point about the inefficiency of creating the sub-shell. I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.
– G-Man
Dec 9 '18 at 19:38
add a comment |
About exec exit 3
... it would try to run an external command called exit
, but there isn't one, so you get the error. It has to be an external command instead of one built in to the shell, since exec
replaces the shell completely. Which also means that even if you had an external command called exit
, exec exit 3
would not return to continue your shell script, since the shell wouldn't be there any more.
1
I guess you could doexec bash -c "exit 3"
, but at the moment I can't think of any reason to do that as opposed to justexit 3
.
– David Z
Dec 9 '18 at 15:45
1
@DavidZ, in any case,exec
'ing or justexit
'ing will stop the script, which didn't seem like what the question wanted.
– ilkkachu
Dec 9 '18 at 17:16
add a comment |
About exec exit 3
... it would try to run an external command called exit
, but there isn't one, so you get the error. It has to be an external command instead of one built in to the shell, since exec
replaces the shell completely. Which also means that even if you had an external command called exit
, exec exit 3
would not return to continue your shell script, since the shell wouldn't be there any more.
1
I guess you could doexec bash -c "exit 3"
, but at the moment I can't think of any reason to do that as opposed to justexit 3
.
– David Z
Dec 9 '18 at 15:45
1
@DavidZ, in any case,exec
'ing or justexit
'ing will stop the script, which didn't seem like what the question wanted.
– ilkkachu
Dec 9 '18 at 17:16
add a comment |
About exec exit 3
... it would try to run an external command called exit
, but there isn't one, so you get the error. It has to be an external command instead of one built in to the shell, since exec
replaces the shell completely. Which also means that even if you had an external command called exit
, exec exit 3
would not return to continue your shell script, since the shell wouldn't be there any more.
About exec exit 3
... it would try to run an external command called exit
, but there isn't one, so you get the error. It has to be an external command instead of one built in to the shell, since exec
replaces the shell completely. Which also means that even if you had an external command called exit
, exec exit 3
would not return to continue your shell script, since the shell wouldn't be there any more.
answered Dec 9 '18 at 10:50
ilkkachu
55.9k784155
55.9k784155
1
I guess you could doexec bash -c "exit 3"
, but at the moment I can't think of any reason to do that as opposed to justexit 3
.
– David Z
Dec 9 '18 at 15:45
1
@DavidZ, in any case,exec
'ing or justexit
'ing will stop the script, which didn't seem like what the question wanted.
– ilkkachu
Dec 9 '18 at 17:16
add a comment |
1
I guess you could doexec bash -c "exit 3"
, but at the moment I can't think of any reason to do that as opposed to justexit 3
.
– David Z
Dec 9 '18 at 15:45
1
@DavidZ, in any case,exec
'ing or justexit
'ing will stop the script, which didn't seem like what the question wanted.
– ilkkachu
Dec 9 '18 at 17:16
1
1
I guess you could do
exec bash -c "exit 3"
, but at the moment I can't think of any reason to do that as opposed to just exit 3
.– David Z
Dec 9 '18 at 15:45
I guess you could do
exec bash -c "exit 3"
, but at the moment I can't think of any reason to do that as opposed to just exit 3
.– David Z
Dec 9 '18 at 15:45
1
1
@DavidZ, in any case,
exec
'ing or just exit
'ing will stop the script, which didn't seem like what the question wanted.– ilkkachu
Dec 9 '18 at 17:16
@DavidZ, in any case,
exec
'ing or just exit
'ing will stop the script, which didn't seem like what the question wanted.– ilkkachu
Dec 9 '18 at 17:16
add a comment |
You can do this with Awk:
awk 'BEGIN{exit 9}'
Or Sed:
sed Q9 /proc/stat
...or with a shell:sh -c 'exit 9'
– ilkkachu
Dec 9 '18 at 17:27
1
@ilkkachu if youre going to do that you might as well do the(exit 9)
in the accepted answer
– Steven Penny
Dec 9 '18 at 18:44
add a comment |
You can do this with Awk:
awk 'BEGIN{exit 9}'
Or Sed:
sed Q9 /proc/stat
...or with a shell:sh -c 'exit 9'
– ilkkachu
Dec 9 '18 at 17:27
1
@ilkkachu if youre going to do that you might as well do the(exit 9)
in the accepted answer
– Steven Penny
Dec 9 '18 at 18:44
add a comment |
You can do this with Awk:
awk 'BEGIN{exit 9}'
Or Sed:
sed Q9 /proc/stat
You can do this with Awk:
awk 'BEGIN{exit 9}'
Or Sed:
sed Q9 /proc/stat
answered Dec 9 '18 at 17:07
Steven Penny
1
1
...or with a shell:sh -c 'exit 9'
– ilkkachu
Dec 9 '18 at 17:27
1
@ilkkachu if youre going to do that you might as well do the(exit 9)
in the accepted answer
– Steven Penny
Dec 9 '18 at 18:44
add a comment |
...or with a shell:sh -c 'exit 9'
– ilkkachu
Dec 9 '18 at 17:27
1
@ilkkachu if youre going to do that you might as well do the(exit 9)
in the accepted answer
– Steven Penny
Dec 9 '18 at 18:44
...or with a shell:
sh -c 'exit 9'
– ilkkachu
Dec 9 '18 at 17:27
...or with a shell:
sh -c 'exit 9'
– ilkkachu
Dec 9 '18 at 17:27
1
1
@ilkkachu if youre going to do that you might as well do the
(exit 9)
in the accepted answer– Steven Penny
Dec 9 '18 at 18:44
@ilkkachu if youre going to do that you might as well do the
(exit 9)
in the accepted answer– Steven Penny
Dec 9 '18 at 18:44
add a comment |
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1
Yeah
exec exit 3
is no bueno, I get"exec: exit: not found"
– MrCholo
Dec 9 '18 at 8:34
6
I don't understand the question. Why not set
exit_code=3
and eliminate theexit 3
line altogether?– wjandrea
Dec 9 '18 at 18:25
@wjandrea is more a conceptual question than practical
– MrCholo
Dec 10 '18 at 1:03
2
It still makes no sense to me. Why would there be an exit code if you don't actually exit?
– Barmar
Dec 10 '18 at 5:09
1
@Barmar every process has an exit code. Most of the people trying to answer the question are interpreting the question to mean "what can I replace the 'exit 3' with in the script so it sets the
$?
variable but doesn't exit this script"?– icarus
Dec 10 '18 at 18:32