How to pass bash script arguments to a subshell

I have a wrapper script that does some work and then passes the original parameters on to another tool:

#!/bin/bash
# ...
other_tool -a -b "$@"

This works fine, unless the "other tool" is run in a subshell:

#!/bin/bash
# ...
bash -c "other_tool -a -b $@"

If I call my wrapper script like this:

wrapper.sh -x "blah blup"

then, only the first orginal argument (-x) is handed to "other_tool". In reality, I do not create a subshell, but pass the original arguments to a shell on an Android phone, which shouldn't make any difference:

#!/bin/bash
# ...
adb sh -c "other_tool -a -b $@"

Answer

Bash's printf command has a feature that'll quote/escape/whatever a string, so as long as both the parent and subshell are actually bash, this should work:

[Edit: as siegi pointed out in a comment, if you do this the obvious way there's a problem when no arguments are supplied, where it acts like there actually was a single empty argument. I've added a workaround below, wrapping the format string with ${1+}, which only includes the format string if the first argument is defined. It's a bit of a kluge, but it does work.]

#!/bin/bash

quoted_args="$(printf "${1+ %q}" "$@")" # Note: this will have a leading space before the first arg
# echo "Quoted args:$quoted_args" # Uncomment this to see what it's doing
bash -c "other_tool -a -b$quoted_args"

Note that you can also do it in a single line: bash -c "other_tool -a -b$(printf "${1+ %q}" "$@")"