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Home/ Questions/Q 7763777
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T14:44:39+00:00 2026-06-01T14:44:39+00:00

Inside a bash script function, I need to work with the command-line arguments of

  • 0

Inside a bash script function, I need to work with the command-line arguments of the script, and also with another list of arguments. So I’m trying to pass two argument lists to a function, the problem is that multi-word arguments get split.

function params()
{
    for PARAM in $1; do
        echo "$PARAM"
    done

    echo .

    for ITEM in $2; do
        echo "$ITEM"
    done
}

PARAMS="$@"
ITEMS="x y 'z t'"
params "$PARAMS" "$ITEMS"

calling the script gives me

myscript.sh a b 'c d'
a
b
c
d
.
x
y
'z
t'

Since there are two lists they must be passed as a whole to the function, the question is, how to iterate the elements while respecting multi-word items enclosed in single quotes ‘c d’ and ‘z t’?

The workaround that I have (see below) makes use of BASH_ARGV so I need to pass just a single list to the function. However I would like to get a better understanding of what’s going on and what’s needed to make the above work.

function params()
{
    for PARAM in "${BASH_ARGV[@]}"; do
        echo "$PARAM"
    done

    echo .

    for ITEM in "$@"; do
        echo "$ITEM"
    done
}

params x y 'z t'

calling the script gives me

myscript.sh a b 'c d'
c d
b
a
.
x
y
z t

… Which is how I need it (except that first list is reversed, but that would be tolerable I guess)

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T14:44:41+00:00Added an answer on June 1, 2026 at 2:44 pm
    function params()
    {
       arg=("$@")
    
       for ((i=1;i<=$1;i++)) ;do
           echo "${arg[i]}"
       done
    
       echo .
    
       for ((;i<=$#-$1+2;i++)) ;do
           echo "${arg[i]}"
       done
    }
    
    items=(w x 'z t')
    params $# "$@" "${items[@]}"
    

    Assuming you call your script with args a b 'c d', the output is:

    a
    b
    c d
    .
    x
    y
    z t
    
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