Bash set +x without it being printed

Question

Does anyone know if we can say set +x in bash without it being printed:

set -x
command
set +x

traces

+ command         


        

Answer 1

This is a combination of a few ideas that can enclose a block of code and preserves the exit status.

#!/bin/bash
shopt -s expand_aliases
alias trace_on='set -x'
alias trace_off='{ PREV_STATUS=$? ; set +x; } 2>/dev/null; (exit $PREV_STATUS)'

trace_on
echo hello
trace_off
echo "status: $?"

trace_on
(exit 56)
trace_off
echo "status: $?"

When executed:

$ ./test.sh 
+ echo hello
hello
status: 0
+ exit 56
status: 56

Answer 2

I hacked up a solution to this just recently when I became annoyed with it:

shopt -s expand_aliases
_xtrace() {
    case $1 in
        on) set -x ;;
        off) set +x ;;
    esac
}
alias xtrace='{ _xtrace $(cat); } 2>/dev/null <<<'

This allows you to enable and disable xtrace as in the following, where I'm logging how the arguments are assigned to variables:

xtrace on
ARG1=$1
ARG2=$2
xtrace off

And you get output that looks like:

$ ./script.sh one two
+ ARG1=one
+ ARG2=two

Answer 3

I had the same problem, and I was able to find a solution that doesn't use a subshell:

set -x
command
{ set +x; } 2>/dev/null

Answer 4

You can use a subshell. Upon exiting the subshell, the setting to x will be lost:

( set -x ; command )

Answer 5

How about a solution based on a simplified version of @user108471:

shopt -s expand_aliases
alias trace_on='set -x'
alias trace_off='{ set +x; } 2>/dev/null'

trace_on
...stuff...
trace_off