Bash: Passing Arguments with Quotes

Today I needed to write a bash script that eventually calls another program, passing all the arguments to it. It seemed like a simple enough task: just use $@. However, the problem was that the program accepts arguments surrounded with quotes, and the $@ stripped the quotes away. I was really surprised that google didn't immediately find an answer.

Finally I reached this forum:
http://www.linuxforums.org/forum/linux-programming-scripting/62564-bash-script-problem-preserving-quotes-arguments.html

And to save you the trouble, here's a script that preserves the quotes:

#!/bin/bash

# blah blah blah

./foo "$@"

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