Getting the last argument passed to a shell script

Question

$1 is the first argument.
$@ is all of them.

How can I find the last argument passed to a shell script?

Answer 1

echo $argv[$#argv]

Now I just need to add some text because my answer was too short to post. I need to add more text to edit.

Answer 2

$ set quick brown fox jumps

$ echo ${*: -1:1} # last argument
jumps

$ echo ${*: -1} # or simply
jumps

$ echo ${*: -2:1} # next to last
fox

The space is necessary so that it doesnt get interpreted as a default value.

Answer 3

For tcsh:

set X = `echo $* | awk -F " " '{print $NF}'`
somecommand "$X"

I'm quite sure this would be a portable solution, except for the assignment.

Answer 4

I found @AgileZebra's answer (plus @starfry's comment) the most useful, but it sets heads to a scalar. An array is probably more useful:

heads=( "${@:1:$(($# - 1))}" )
tail=${@:${#@}}

Answer 5

If you are using Bash >= 3.0

echo ${BASH_ARGV[0]}

Answer 6

This is part of my copy function:

eval echo $(echo '$'"$#")

To use in scripts, do this:

a=$(eval echo $(echo '$'"$#"))

Explanation (most nested first):

1. $(echo '$'"$#") returns $[nr] where [nr] is the number of parameters. E.g. the string $123 (unexpanded).
2. echo $123 returns the value of 123rd parameter, when evaluated.
3. eval just expands $123 to the value of the parameter, e.g. last_arg. This is interpreted as a string and returned.

Works with Bash as of mid 2015.