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Launch and kill a program in script and continue [duplicate]

Bash ignoring error for a particular commandGet the source directory of a Bash script from within the script itselfHow do I prompt for Yes/No/Cancel input in a Linux shell script?How to check if a program exists from a Bash script?What killed my process and why?YYYY-MM-DD format date in shell scriptFind (and kill) process locking port 3000 on MacHow do I run a node.js app as a background service?Check existence of input argument in a Bash shell scriptKilling a script launched in a Process via os.system()Xcode process launch failed: Security

1

This question already has an answer here:

• 
  Bash ignoring error for a particular command
  
  6 answers
  
  

I have a script that needs to launch a program in order for the program to create and populate some folders. The program only needs to run briefly and then close, before the script continues.

However when the program is killed, the script does not continue. How can I do this?

printf "before"

(timeout -sHUP 3s firefox)

printf "after"

I thought I was launching the program in a subshell but i am not sure if this is the case.

bash process

share|improve this question

asked Mar 8 at 11:07

myolmyol

1,94374077

marked as duplicate by John Kugelman bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.

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Mar 9 at 16:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  timeout returns an non-zero exit status when it kills the program that it runs. If set -e (or, equivalently, set -o errexit) is in effect when that happens, the script will stop immediately and silently. The parentheses ((...)) ensure that the program is run in a subshell. That doesn't change the effect of the errexit setting, and it's not clear why a subshell is being used here.
  
  – pjh
  Mar 8 at 19:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had set -o errexit set indeed. Any way to ignore the timeout error specifically but keep errexit?
  
  – myol
  Mar 9 at 15:13
  

  
  
  
  

add a comment | 

1

This question already has an answer here:

• 
  Bash ignoring error for a particular command
  
  6 answers
  
  

I have a script that needs to launch a program in order for the program to create and populate some folders. The program only needs to run briefly and then close, before the script continues.

However when the program is killed, the script does not continue. How can I do this?

printf "before"

(timeout -sHUP 3s firefox)

printf "after"

I thought I was launching the program in a subshell but i am not sure if this is the case.

bash process

share|improve this question

asked Mar 8 at 11:07

myolmyol

1,94374077

marked as duplicate by John Kugelman bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
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StackExchange.helpers.removeMessages();

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Mar 9 at 16:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  timeout returns an non-zero exit status when it kills the program that it runs. If set -e (or, equivalently, set -o errexit) is in effect when that happens, the script will stop immediately and silently. The parentheses ((...)) ensure that the program is run in a subshell. That doesn't change the effect of the errexit setting, and it's not clear why a subshell is being used here.
  
  – pjh
  Mar 8 at 19:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had set -o errexit set indeed. Any way to ignore the timeout error specifically but keep errexit?
  
  – myol
  Mar 9 at 15:13
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  Bash ignoring error for a particular command
  
  6 answers
  
  

I have a script that needs to launch a program in order for the program to create and populate some folders. The program only needs to run briefly and then close, before the script continues.

However when the program is killed, the script does not continue. How can I do this?

printf "before"

(timeout -sHUP 3s firefox)

printf "after"

I thought I was launching the program in a subshell but i am not sure if this is the case.

bash process

share|improve this question

asked Mar 8 at 11:07

myolmyol

1,94374077

printf "before"

(timeout -sHUP 3s firefox)

printf "after"

share|improve this question

asked Mar 8 at 11:07

myolmyol

1,94374077

share|improve this question

asked Mar 8 at 11:07

myolmyol

1,94374077

asked Mar 8 at 11:07

myolmyol

1,94374077

asked Mar 8 at 11:07

myolmyol

1,94374077

2 Answers
2

active

oldest

votes

1

I solved this with a hacky work around.

set -o errexit

printf "before"

timeout -sHUP 3s firefox || true

printf "after"

The alternative cleaner method would be

set -o errexit

printf "before"

set +e
timeout -sHUP 3s firefox
set -e

printf "after"

share|improve this answer

answered Mar 9 at 15:39

myolmyol

1,94374077

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not a hack at all. Appending || : is a common idiom to suppress the exit code of a failing command when you don't care.
  
  – John Kugelman
  Mar 9 at 15:59
  

  
  
  
  

add a comment | 

0

printf takes 2 arguments - a FORMAT string, then an ARGUMENT (i.e the text to be outputted).

When you call printf "before" you are actually passing a format string with no argument, which is not displayed properly.

The easiest fix is to instead use echo:

echo "before"
timeout -sHUP 3s firefox
echo "after"

Or if you need formatting, to instead do e.g.:

printf "%sn" "before"
timeout -sHUP 3s firefox
printf "%sn" "after"

share|improve this answer

answered Mar 8 at 11:25

matchmatch

3,1131925

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Addendum - printf will work fine with one argument - he just didn't include a newline, which unlike echo it doesn't add by default. printf "beforen"
  
  – Paul Hodges
  Mar 8 at 14:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stdout was just to make the example concise for stack overflow, the program has various other method calls and bash commands that get ignored after timeout. The issue seems to be in the use of set -o errexit
  
  – myol
  Mar 9 at 15:15
  

  
  
  
  

add a comment | 

1

I solved this with a hacky work around.

set -o errexit

printf "before"

timeout -sHUP 3s firefox || true

printf "after"

The alternative cleaner method would be

set -o errexit

printf "before"

set +e
timeout -sHUP 3s firefox
set -e

printf "after"

share|improve this answer

answered Mar 9 at 15:39

myolmyol

1,94374077

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not a hack at all. Appending || : is a common idiom to suppress the exit code of a failing command when you don't care.
  
  – John Kugelman
  Mar 9 at 15:59
  

  
  
  
  

add a comment | 

1

I solved this with a hacky work around.

set -o errexit

printf "before"

timeout -sHUP 3s firefox || true

printf "after"

The alternative cleaner method would be

set -o errexit

printf "before"

set +e
timeout -sHUP 3s firefox
set -e

printf "after"

share|improve this answer

answered Mar 9 at 15:39

myolmyol

1,94374077

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not a hack at all. Appending || : is a common idiom to suppress the exit code of a failing command when you don't care.
  
  – John Kugelman
  Mar 9 at 15:59
  

  
  
  
  

add a comment | 

I solved this with a hacky work around.

set -o errexit

printf "before"

timeout -sHUP 3s firefox || true

printf "after"

The alternative cleaner method would be

set -o errexit

printf "before"

set +e
timeout -sHUP 3s firefox
set -e

printf "after"

share|improve this answer

answered Mar 9 at 15:39

myolmyol

1,94374077

set -o errexit

printf "before"

timeout -sHUP 3s firefox || true

printf "after"

set -o errexit

printf "before"

set +e
timeout -sHUP 3s firefox
set -e

printf "after"

share|improve this answer

answered Mar 9 at 15:39

myolmyol

1,94374077

answered Mar 9 at 15:39

myolmyol

1,94374077

answered Mar 9 at 15:39

myolmyol

1,94374077

0

printf takes 2 arguments - a FORMAT string, then an ARGUMENT (i.e the text to be outputted).

When you call printf "before" you are actually passing a format string with no argument, which is not displayed properly.

The easiest fix is to instead use echo:

echo "before"
timeout -sHUP 3s firefox
echo "after"

Or if you need formatting, to instead do e.g.:

printf "%sn" "before"
timeout -sHUP 3s firefox
printf "%sn" "after"

share|improve this answer

answered Mar 8 at 11:25

matchmatch

3,1131925

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Addendum - printf will work fine with one argument - he just didn't include a newline, which unlike echo it doesn't add by default. printf "beforen"
  
  – Paul Hodges
  Mar 8 at 14:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stdout was just to make the example concise for stack overflow, the program has various other method calls and bash commands that get ignored after timeout. The issue seems to be in the use of set -o errexit
  
  – myol
  Mar 9 at 15:15
  

  
  
  
  

add a comment | 

0

printf takes 2 arguments - a FORMAT string, then an ARGUMENT (i.e the text to be outputted).

When you call printf "before" you are actually passing a format string with no argument, which is not displayed properly.

The easiest fix is to instead use echo:

echo "before"
timeout -sHUP 3s firefox
echo "after"

Or if you need formatting, to instead do e.g.:

printf "%sn" "before"
timeout -sHUP 3s firefox
printf "%sn" "after"

share|improve this answer

answered Mar 8 at 11:25

matchmatch

3,1131925

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Addendum - printf will work fine with one argument - he just didn't include a newline, which unlike echo it doesn't add by default. printf "beforen"
  
  – Paul Hodges
  Mar 8 at 14:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stdout was just to make the example concise for stack overflow, the program has various other method calls and bash commands that get ignored after timeout. The issue seems to be in the use of set -o errexit
  
  – myol
  Mar 9 at 15:15
  

  
  
  
  

add a comment | 

printf takes 2 arguments - a FORMAT string, then an ARGUMENT (i.e the text to be outputted).

When you call printf "before" you are actually passing a format string with no argument, which is not displayed properly.

The easiest fix is to instead use echo:

echo "before"
timeout -sHUP 3s firefox
echo "after"

Or if you need formatting, to instead do e.g.:

printf "%sn" "before"
timeout -sHUP 3s firefox
printf "%sn" "after"

share|improve this answer

answered Mar 8 at 11:25

matchmatch

3,1131925

echo "before"
timeout -sHUP 3s firefox
echo "after"

printf "%sn" "before"
timeout -sHUP 3s firefox
printf "%sn" "after"

share|improve this answer

answered Mar 8 at 11:25

matchmatch

3,1131925

answered Mar 8 at 11:25

matchmatch

3,1131925

answered Mar 8 at 11:25

matchmatch

3,1131925