Using export -f with xargs not working [duplicate] - macos

Question

I am trying to use xargs to call a more complex function in parallel.
#!/bin/bash
echo_var(){
echo $1
return 0
}
seq -f "n%04g" 1 100 |xargs -n 1 -P 10 -i echo_var {}
exit 0
This returns the error
xargs: echo_var: No such file or directory
Any ideas on how I can use xargs to accomplish this, or any other solution(s) would be welcome.

Answer 1

Exporting the function should do it (untested):
export -f echo_var
seq -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.

Answer 2

Using GNU Parallel is looks like this:
#!/bin/bash
echo_var(){
echo $1
return 0
}
export -f echo_var
seq -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 {}

Answer 3

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 # "

Answer 4

Maybe this is bad practice, but you if you are defining functions in a .bashrc or other script, you can wrap the file or at least the function definitions with a setting of allexport:
set -o allexport
function funcy_town {
echo 'this is a function'
}
function func_rock {
echo 'this is a function, but different'
}
function cyber_func {
echo 'this function does important things'
}
function the_man_from_funcle {
echo 'not gonna lie'
}
function funcle_wiggly {
echo 'at this point I\'m doing it for the funny names'
}
function extreme_function {
echo 'goodbye'
}
set +o allexport

Answer 5

Seems I can't make comments :-(
I was wondering about the focus on
bash -c 'echo_var "$#"' _ {}
vs
bash -c 'echo_var "{}"'
The 1st substitutes the {} as an arg to bash while the 2nd as an arg to the function. The fact that example 1 doesn't expand the $(date) is simply a a side effect.
If you don't want the functions args expanded , use single single quotes rather than double. To avoid messy nesting , use double quote (expand args on the other one)
$ echo '$(date)' | xargs -0 -L1 -I {} bash -c 'printit "{}"'
Fri 11 Sep 17:02:24 BST 2020
$ echo '$(date)' | xargs -0 -L1 -I {} bash -c "printit '{}'"
$(date)