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