What's the meaning of a ! before a command in the shell?

TL;DR: This is just by-passing the set -e flag in the specific line where you are using it.

Adding add to hek2mgl's correct and useful answer.

You have:

set -e! command

Bash Reference Manual → Pipelines describes:

Each command in a pipeline is executed in its own subshell. The exit status of a pipeline is the exit status of the last command in the pipeline (...). If the reserved word ‘!’ precedes the pipeline, the exit status is the logical negation of the exit status as described above. The shell waits for all commands in the pipeline to terminate before returning a value.

This means that ! preceding a command is negating the exit status of it:

$ echo 2323$ echo $?0                # But$ ! echo 2323$ echo $?1

Or:

$ echo 23 && echo "true" || echo "fail"23true$ ! echo 23 && echo "true" || echo "fail"23fail

The exit status is useful in many ways. In your script, used together with set -e makes the script exit whenever a command returns a non-zero status.

Thus, when you have:

set -ecommand1command2

If command1 returns a non-zero status, the script will finish and won't proceed to command2.

However, there is also an interesting point to mention, described in 4.3.1 The Set Builtin:

-e

Exit immediately if a pipeline (see Pipelines), which may consist of a single simple command (see Simple Commands), a list (see Lists), or a compound command (see Compound Commands) returns a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in an if statement, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command’s return status is being inverted with !. If a compound command other than a subshell returns a non-zero status because a command failed while -e was being ignored, the shell does not exit. A trap on ERR, if set, is executed before the shell exits.

Taking all of these into consideration, when you have:

set -e! command1command2

What you are doing is to by-pass the set -e flag in the command1. Why?

• if command1 runs properly, it will return a zero status. ! will negate it, but set -e won't trigger an exit by the because it comes from a return status inverted with !, as described above.
• if command1 fails, it will return a non-zero status. ! will negate it, so the line will end up returning a zero status and the script will continue normally.

If you don't want the script to fail in both cases, error or success of the command, you can also use this alternative:

set -edocker stop foo || true

The boolean or true makes the pipeline always have 0 as the return value.