Calling shell functions with xargs

Exporting the function should do it (untested):

export -f echo_varseq -f "n%04g" 1 100 | xargs -n 1 -P 10 -I {} bash -c 'echo_var "$@"' _ {}

You can use the builtin printf instead of the external seq:

printf "n%04g\n" {1..100} | xargs -n 1 -P 10 -I {} bash -c 'echo_var "$@"' _ {}

Also, using return 0 and exit 0 like that masks any error value that might be produced by the command preceding it. Also, if there's no error, it's the default and thus somewhat redundant.

@phobic mentions that the Bash command could be simplified to

bash -c 'echo_var "{}"'

moving the {} directly inside it. But it's vulnerable to command injection as pointed out by @Sasha.

Here is an example why you should not use the embedded format:

$ echo '$(date)' | xargs -I {} bash -c 'echo_var "{}"'Sun Aug 18 11:56:45 CDT 2019

Another example of why not:

echo '\"; date\"' | xargs -I {} bash -c 'echo_var "{}"'

This is what is output using the safe format:

$ echo '$(date)' | xargs -I {} bash -c 'echo_var "$@"' _ {}$(date)

This is comparable to using parameterized SQL queries to avoid injection.

I'm using date in a command substitution or in escaped quotes here instead of the rm command used in Sasha's comment since it's non-destructive.

Using GNU Parallel is looks like this:

#!/bin/bashecho_var(){    echo $1    return 0}export -f echo_varseq -f "n%04g" 1 100 | parallel -P 10 echo_var {} exit 0

If you use version 20170822 you do not even have to export -f as long as you have run this:

. `which env_parallel.bash`seq -f "n%04g" 1 100 | env_parallel -P 10 echo_var {} 

Something like this should work also:

function testing() { sleep $1 ; }echo {1..10} | xargs -n 1 | xargs -I@ -P4 bash -c "$(declare -f testing) ; testing @ ; echo @ "