Pass string as arguments with spaces in BASH

Question

I have a CLI program that generates arguments as one string for another CLI program. I call both from BASH. Here is simplified version of what I have:

produce_args() {
    echo "arg1 \"ar g2\" arg3"
}

consume_args() {
    for arg in "$@"; do
        echo "$arg"
    done
}

TMP="$(produce_args)"
consume_args $TMP

It prints:

arg1
"ar
g2"
arg3

So, argument 2 was split to two arguments. How to see an output like this instead?

arg1
ar g2
arg3

I can change my generator CLI program and BASH script but not consumer program. Also I can use zsh but I don't really think that matters in my case.

user594138 · Accepted Answer · 2019-10-08 10:52:58Z

2

I'd abstain from using eval as much as possible, albeit I don't mind using it if necessary. In your case I'd probably use a new-line as arg-separator, modify IFS accordingly, and make TMP an array:

#!/bin/bash
produce_args() {
        echo -e "arg1\nar g2\narg3"
}
consume_args() {
        for arg in "${@}"; do
                echo "$arg"
        done
}
NL="
"
OLDIFS="${IFS}"
IFS="${NL}"
declare -a TMP
TMP=($(produce_args))
IFS="${OLDIFS}"
consume_args "${TMP[@]}"

answered Oct 8, 2019 at 10:52

user594138

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1 Comment

Eugene Kosov Over a year ago

Thank you for your solution!

dash-o · Accepted Answer · 2019-10-08 10:31:12Z

0

You can use 'eval' to on the call to consume_args. It will allow the shell to process in the quoted string from the produce args as a single argument:

#! /bin/bash

produce_args() {
    echo "arg1 \"ar g2\" arg3"
}

consume_args() {
    for arg in "$@"; do
        echo "$arg"
    done
}

TMP="$(produce_args)"
eval consume_args $TMP

answered Oct 8, 2019 at 10:31

dash-o

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2 Comments

Eugene Kosov Over a year ago

Oh, gosh, man! That simple! It's a terrible answer :( Thank you a lot!

chepner Over a year ago

This is just asking for problems when produce_args changes. Consider, for example, if the output of produce_args produces an argument with more than one consecutive whitespace.

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