Prefix every argument with -o in BASH

Question

How do I prefix -p to every argument passed to my function?

Modifying the arguments themselves and creating a new array are both fine.

Answer 1

This should work nicely even for complicated arguments with whitespace and worse:

#!/bin/bash
new_args=()
for arg
do
    new_args+=( '-p' )
    new_args+=( "$arg" )
done

for arg in "${new_args[@]}"
do
    echo "$arg"
done

Test:

$ ~/test.sh foo $'bar\n\tbaz bay'
-p
foo
-p
bar
    baz bay

Answer 2

Put the arguments in an array and use bash pattern substitution on them with array substitution and prefix matching:

ARGS=("$@")
echo ${ARGS[@]/#/-p }

That replaces the start of each arg with -p<space>.

Unfortunately this does not work properly if you have spaces in your arguments. Spaces are preserved properly with ARGS=("$@"), but not when you do the ${ARGS[@]/#/-p } expansion. You can put double quotes around that expansion, but then you get -p arg1 as a single argument, not two arguments.

If you don't need -p<space> prepended, just -p then putting double quotes around the expansion should work fine. Experiment with and without double quotes around the expansion to see what works best for you.

Answer 3

You can preserve the spaces in the ${ARGS[@]/#/-p } expansion if you set the IFS variable to a null string!

# sample code
(
set -- 1 2 3 'arg with spaces' $'bar\n\tbaz bay'
printf 'oldIFS: %q\n' "$IFS"
IFS=""                       
#IFS=" "                       
printf 'newIFS: %q\n' "$IFS"
ARGS=("$@")
ARGS=( ${ARGS[@]/#/-p } )
for ((i=0; i < ${#ARGS[@]}; i++)); do
  echo "$i: ${ARGS[i]}"
done
)

Inside a function you may limit the scope of a modified IFS variable to this function by using declare IFS="".