Print bash arguments in reverse order

Question

I have to write a script, which will take all arguments and print them in reverse.

I\'ve made a solution, but find it very bad. Do you have a smarter idea?

Answer 1

Portably and POSIXly, without arrays and working with spaces and newlines:

Reverse the positional parameters:

flag=''; c=1; for a in "$@"; do set -- "$a" ${flag-"$@"}; unset flag; done

Print them:

printf '<%s>' "$@"; echo