[linux] What does `set -x` do?

I have a shell script with the following line in it:

[ "$DEBUG" == 'true' ] && set -x

set -x

Prints a trace of simple commands, for commands, case commands, select commands, and arithmetic for commands and their arguments or associated word lists after they are expanded and before they are executed. The value of the PS4 variable is expanded and the resultant value is printed before the command and its expanded arguments.

[source]

Example

set -x
echo `expr 10 + 20 `
+ expr 10 + 20
+ echo 30
30

set +x
echo `expr 10 + 20 `
30

Above example illustrates the usage of set -x. When it is used, above arithmetic expression has been expanded. We could see how a singe line has been evaluated step by step.

• First step expr has been evaluated.
• Second step echo has been evaluated.

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when it comes to your shell script,

[ "$DEBUG" == 'true' ] && set -x

Your script might have been printing some additional lines of information when the execution mode selected as DEBUG. Traditionally people used to enable debug mode when a script called with optional argument such as -d

-u: disabled by default. When activated, an error message is displayed when using an unconfigured variable.

-v: inactive by default. After activation, the original content of the information will be displayed (without variable resolution) before the information is output.

-x: inactive by default. If activated, the command content will be displayed before the command is run (after variable resolution, there is a ++ symbol).

Compare the following differences:

/ # set -v && echo $HOME
/root
/ # set +v && echo $HOME
set +v && echo $HOME
/root

/ # set -x && echo $HOME
+ echo /root
/root
/ # set +x && echo $HOME
+ set +x
/root

/ # set -u && echo $NOSET
/bin/sh: NOSET: parameter not set
/ # set +u && echo $NOSET