Question

This function works:

source foo.bash && foo -n "a b c.txt"

The problem is, no matter what I’ve tried, I couldn’t get the last line echo "$CMD" (or echo $CMD) to generate exactly this output:

cat -n "a b c.txt"

How to achieve that?

# foo.bash
function foo() {
    local argv=("$@");
    local OUT=`cat "${argv[@]}"`
    local CMD=`echo cat "${argv[@]}"`
    echo "--------------------------"
    echo "$OUT"
    echo "--------------------------"
    echo "$CMD"
}

The output is instead:

cat -n a b c.txt

With this command: foo -n \"a b c.txt\" it does work for the display of the command, but it gives errors for the execution via the backtick.

The file "a b c.txt" is a valid, small, text file.

Answer 1

There you go, with the help of number of tokens in bash variable I’ve come up with the right solution.

I’ve almost forgot WHY we actually need quoting for one argument, it’s because it has multiple words!

function foo() {
    local argv=( "$@" );
    local OUT=`cat "${argv[@]}"`
    echo "--------------------------"
    echo "$OUT"
    echo "--------------------------"
    local CMD="cat"
    for word in "${argv[@]}"; do
        words="${word//[^\ ]} "
        if [[ ${#words} > 1 ]]; then
            local CMD="$CMD \"${word}\""
        else
            local CMD="$CMD $word"
        fi
    done
    echo "$CMD"
}

Hope it helps someone.