Assigning “$@” to a variable in Bash and preserving escaping

Question

If I save the following script as 'args.bash' and run bash args.bash -a 'foo bar'

#!/bin/bash

main() {
    set -x
    which "$@"

    local args="$@"
    which "$args"
    which $args

    local args=$(printf " %q" "$@")
    which "$args"
    which $args
}
main "$@"

I'm surprised to see that these print different results. The first one prints:

which -a 'foo bar'

Which I expect! It groups the arguments properly.

The rest of them print, in order:

which '-a foo bar'
which -a foo bar
which ' -a foo\ bar'
which -a 'foo\' bar

Which are all incorrect to one degree or another. How can I assign "$@" to a variable, and get it to echo the same output as the first command?

Answer 1

Use an array variable, not a scalar:

items=( "$@" )
which -a "${items[@]}"

However, if you must use a scalar _[1] , you had the right approach with printf '%q ' ; note, however, that printf %q creates a string which will evaluate back to itself when run through the full parser (thus requiring eval or an equivalent to ensure that the full parser is invoked):

printf -v items_str '%q ' "$@"
eval "which -a $items_str"

---

_[1] The number of circumstances where one has bash extensions such as printf -v and %q but can't use an array is limited, but they do exist; particularly, passing values to a subprocess through the environment comes to mind. However, in the bulk of cases where arrays are inappropriate, I'd expect either using a series of distinct environment variables (one per value), or passing a NUL-delimited stream (as created by printf '%s\\n' "$@" ) to be more appropriate, even if (as for the environment) it might need to be escaped (for instance, via base64-encoding).