Bash: test whether original script was run with `-x`

Question

Is it possible to test whether a bash script was executed with -x?

I'm writing a script which also runs some remote scripts, and I'd like to be able to do something like:

USE_X="$( run_with_x && "-x" || "" )"
ssh $host "bash $USE_X some_script.sh"

Answer 1

Just look at the $- variable.

USE_X=`case "$-" in *x*) echo "-x" ;; esac`

(Yup, I did manage to do that without invoking grep. Save the pids!)

Answer 2

After reading man pages and grepping for things, here's the solution:

$ cat show_x.sh
#!/bin/bash
if [ "$(set | grep xtrace)" ]; then
    echo "xtrace is on. :D"
else
    echo "xtrace is off. :("
fi

Result:

$ bash show_x.sh 
xtrace is off. :(

$ bash -x show_x.sh 
++ set
++ grep xtrace
+ '[' SHELLOPTS=braceexpand:hashall:interactive-comments:xtrace ']'
+ echo 'xtrace is on. :D'
xtrace is on. :D

Answer 3

The bash man page documents set -o xtrace as equivalent to set -x, which in turn is equivalent to giving -x as an argument to the shell. (So far, this is compatible with SUSv3 sh, with the caveat that -o option is required "if the system supports the User Portability Utilities option".)

Also, set -o without an option prints the current status of all of the -o options. (SUSv3 specifies this as well, but not usefully; the format is unspecified, so you can only depend on it if you know you're running a shell that uses a format similar to bash's.)

So, the pipeline set -o | grep xtrace | grep -q on should be what you're looking for.

Answer 4

Probably the simplest way of parsing $-, similar to dagbrown's method (also pure bash), but without using case and backticks (subshell):

if  [[ ${-/x} != $- ]] ; then USE_X="-x"; fi

or even shorter:

[[ ${-/x} != $- ]] && USE_X="-x"

Note that [[ is not an invocation of test builtin program (as is the case with [), but a bash syntax construct (according to man bash and this dev recommendation). So I claim it to be "purer bash" ;)