Bash – How does ${!#} work in bash to get the last command-line argument

bash

From the Advanced Bash-Scripting Guide, to get the last command-line argument:

Or: lastarg=${!#}
This is an
indirect reference to the $# variable.
Note that lastarg=${!$#} doesn't work.

Best Answer

As it says it is an indirect reference. Take the following example:

$ var=test
$ test='Hello, world'
$ echo ${!var}
Hello, world

Now if I were to set the following positional parameters:

$ set -- one two three

And $# will represent the total number of positional parameters:

$ echo $#
3

This means when I call it like ${!#} I'm saying give me the value of the parameter named 3:

$ echo ${!#}
three

Related Solutions

Bash Command – Printing and Saving the Last Bash Input Command

You can access the just-executed command line with the history built-in. (I have no idea why history 1 prints the just-executed command line but fc -nl -1 prints the previous commmand, as does fc -nl 0.)

PROMPT_COMMAND='echo -en "\033[38;5;2m"; history 1; echo -en "\033[0m\n"'

This prints a number before the command text. Here's a version that removes the number. (It may be incorrect if you go beyond 99999 history lines, I don't know how bash formats the number then.)

prompt_function () {
  local prompt_history="$(history 1)"
  prompt_history=${prompt_history:7}
  echo -En $'\033[38;5;2m'"$prompt_history"$'\033[0m\n'
}
PROMPT_COMMAND=prompt_function

(Note that echo -en ..."$prompt_history"... would expand backslashes in the command line, so I use echo -E and let the shell expansion generate the control characters with $''.).

Bash – Splitting bash command line argument

Another way would be to use IFS, the shell's built-in method to split strings into fields.

OLDIFS=$IFS
IFS=':'
set -f
for host in $hostlist; do
  set +f
  echo "$host"
done
set +f
IFS=$OLDIFS

set -f turns off filename generation (globbing): without it, wildcards *?\[ would be expanded in each word.

Best Answer

Related Solutions

Bash Command – Printing and Saving the Last Bash Input Command

Bash – Splitting bash command line argument

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