Bash return code error handling when using heredoc input

Question

Motivation

I'm in a situation where I have to run multiple bash commands with a single bash invocation without the possibility to write a full script file (use case: Passing multiple commands to a container in Kubernetes). A common solution is to combine commands with ; or &&, for instance:

bash -c " \
echo \"Hello World\" ; \
ls -la ; \
run_some_command "

In practice writing bash scripts like that turns out to be error prone, because I often forget the semicolon leading to subtle bugs.

Inspired by this question, I was experiment with writing scripts in a more standard style by using a heredoc:

bash <<EOF
echo "Hello World"
ls -la
run_some_command
EOF

Unfortunately, I noticed that there is a difference in exit code error handling when using a heredoc. For instance:

bash -c " \
run_non_existing_command ; \
echo $? "

outputs (note that $? properly captures the exit code):

bash: run_non_existing_command: command not found
127

whereas

bash <<EOF
run_non_existing_command
echo $?
EOF

outputs (note that $? fails to capture the exit code compared to standard script execution):

bash: line 1: run_non_existing_command: command not found
0

Why is the heredoc version behaving differently? Is it possible to write the script in the heredoc style and maintaining normal exit code handling?

for instance: - your instance doesn't combine commands (at least not in a way I define "combine"), it runs bash -c echo "Hello World" and then runs ls -la and run_some_command, all in parent shell. I think you are missing \ slashes — KamilCuk
– KamilCuk, Commented Oct 15, 2019 at 12:12
@KamilCuk Correct fixed, thanks! Copy paste mistake, when deploying to Kubernetes the `` are handled via YAML multi line strings... The observations should still be valid. — bluenote10
– bluenote10, Commented Oct 15, 2019 at 12:25
No, it's not the same thing. Probably the same thing would be bash -c "echo \"Hello World\" ; ls -la ; run_some_command ;" — KamilCuk
– KamilCuk, Commented Oct 15, 2019 at 12:27

KamilCuk · Accepted Answer · 2019-10-15 12:14:56Z

4

Why is the heredoc version behaving differently?

Because $? is expanded before running the command.

The following will output 1, that is the exit status of false command:

false
bash <<EOF
run_non_existing_command
echo $?
EOF

It's the same in principle as the following, which will print 5:

variable=5
bash <<EOF
variable="This is ignored"
echo $variable
EOF

Is it possible to write the script in the heredoc style and maintaining normal exit code handling?

If you want to have the $? expanded inside the subshell, then:

bash <<EOF
run_non_existing_command
echo \$?
EOF

or

bash <<'EOF'
run_non_existing_command
echo $?
EOF

Also note that:

bash -c \
run_non_existing_command ;
echo $? ;

is just equal to:

bash -c run_non_existing_command
echo $?

The echo $? is not executed inside bash -c.

edited Oct 15, 2019 at 12:14

answered Oct 15, 2019 at 12:09

KamilCuk

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2 Comments

bluenote10 Over a year ago

Interesting, is there a simple explanation why quoting the EOF gets rid of the necessity to escape the $??

KamilCuk Over a year ago

Yes, it was designed this way. See posix shell 2.7.4 Here-Document If any character in word is quoted, .... the here-document lines shall not be expanded.

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