How to get the second last argument from shell script?

Question

I want to get the second last item given to a shell program. Currently I do it like this:

file1_tmp="${@: -2}"
oldIFS=$IFS
IFS=" "
count=0
for value in $file1; do
  if [[ count -e 0 ]]; then
    file1=$value
  fi
    count=1
done
oldIFS=$IFS 

I'm sure that there is a much easier way to do this. So how can I get the second last argument from a shell script input in as few lines as possible?

Answer 1

set -- "first argument" "second argument" \
       "third argument" "fourth argument" \
       "fifth argument"
second_to_last="${@:(-2):1}"
echo "$second_to_last"

Note the quoting, which ensures that arguments with whitespace stick together -- which your original solution doesn't do.

Answer 2

There are some options for all bash versions:

$ set -- aa bb cc dd ee ff
$ echo "${@: -2:1}   ${@:(-2):1}   ${@:(~1):1}   ${@:~1:1}   ${@:$#-1:1}"
ee   ee   ee   ee   ee

The (~) is the bitwise negation operator (search in the ARITHMETIC EVALUATION section).
It means flip all bits.

The selection even could be done with (integer) variables:

$ a=1  ; b=-a; echo "${@:b-1:1}   ${@:(b-1):1}   ${@:(~a):1}   ${@:~a:1}   ${@:$#-a:1}"
ee   ee   ee   ee   ee

$ a=2  ; b=-a; echo "${@:b-1:1}   ${@:(b-1):1}   ${@:(~a):1}   ${@:~a:1}   ${@:$#-a:1}"
dd   dd   dd   dd   dd

For really old shells, you must use eval:

eval "printf \"%s\n\" \"\$$(($#-1))\""