Ubuntu – What does !$ means in Bash scripting

bashcommand line

When reading the content of a script, I found something like:

echo "An Example" > !$

So I wonder what does !$ mean in bash .

Best Answer

From man bash (somewhere after the line 3813):

!      Start  a  history substitution, except when followed by a blank,
          newline, carriage return, = or ( (when the extglob shell  option
          is enabled using the shopt builtin).

[...]

$      The  last  word.   This  is  usually the last argument, but will
          expand to the zeroth word if there is only one word in the line.

So, the !$ will recall the last argument from the last command from history.

Here are some examples and equivalents:

$ echo foo bar
foo bar
$ echo !$ # recall the last argument from the last command from history
echo bar
bar

$ echo foo bar
foo bar
$ echo !:$ # the same like `!$'
echo bar
bar

$ echo foo bar
foo bar
$ echo !:2  # recall the second (in this case the last) argument from the last command from history
echo bar
bar

$ echo foo bar
foo bar
$ echo $_ # gives the last argument from the last command from history
bar

$ echo foo bar
foo bar
$ echo Alt+. # pressing Alt and . (dot) in the same time will automatically insert the last argument from the last command from history
bar

$ echo foo bar
foo bar
$ echo Alt+_ # the same like when you press Alt+.
bar

$ echo foo bar
foo bar
$ echo Esc. # the same like when you press Alt+.
bar

All of these will work only inside of an interactive shell. But if you use that command from your question inside of a script as you said that you sow it, then the things are different: when you run the script, this will start in a new shell, a non-interactive shell, so the history expansion doesn't has any effect in this case and so the command:

echo "An Example" > !$

creates the file !$ if not present (otherwise overwrites it) and write An Example into it.

Related Solutions

Ubuntu – How to escape `$` and `#` symbols in bash

Because you already have single quotes in the string, use double quotes, and escape only the $ sign:

echo "\$(document).ready(function() {\$('#cf-footer-paragraph').append('<p>Revision 12345</p>');});"

Ubuntu – What does the “-” in “bash -” mean

When in doubt, read the source code. =)

Bash 4.3, shell.c line 830, in function parse_shell_options():

  /* A single `-' signals the end of options.  From the 4.3 BSD sh.
     An option `--' means the same thing; this is the standard
     getopt(3) meaning. */
  if (arg_string[0] == '-' &&
       (arg_string[1] == '\0' ||
         (arg_string[1] == '-' && arg_string[2] == '\0')))
    return (next_arg);

In other words, the - is saying that there are no more options. If there were any more words on the command line, they would be treated as a filename, even if the word started with a -.

In your example, of course, that - is completely redundant, as there is nothing following it anyway. In other words, bash - is exactly equivalent to bash.

Bash takes its commands

from a script file if it's provided on the command line, or
non-interactively from its stdin if its stdin is not a TTY (such as in your example: stdin is a pipe, so Bash will execute the contents of that URL as a script), or
interactively if its stdin is a TTY.

It is a misconception that bash - tells Bash to read its commands from its standard input. While it is true that in your example, Bash will read its commands from stdin, it would have done so regardless of whether there was a - on the command line, because, as stated above, bash - is identical to bash.

To further illustrate that - does not mean stdin, consider:

The cat command is designed to interpret a - as stdin. For example:

$ echo xxx | cat /etc/hosts - /etc/shells
127.0.0.1 localhost
xxx
# /etc/shells: valid login shells
/bin/sh
/bin/dash
/bin/bash
/bin/rbash
/bin/zsh
/usr/bin/zsh
/usr/bin/screen
/bin/tcsh
/usr/bin/tcsh
/usr/bin/tmux
/bin/ksh93

In contrast, you can't get Bash to execute /bin/date then /bin/hostname by trying this:
```
$ echo date | bash - hostname
/bin/hostname: /bin/hostname: cannot execute binary file
```
Rather, it tries to interpret /bin/hostname as a shell script file, which fails because it's a bunch of binary gobbledygook.

You can't execute date +%s using bash - either.

$ date +%s
1448696965
$ echo date | bash -
Sat Nov 28 07:49:31 UTC 2015
$ echo date | bash - +%s
bash: +%s: No such file or directory

Can you write xargs bash instead? No. curl | xargs bash would invoke bash with the contents of the script as command-line arguments. The first word of the contents would be the first argument, and it would likely be misinterpreted as a script filename.

Best Answer

Related Solutions

Ubuntu – How to escape `$` and `#` symbols in bash

Ubuntu – What does the “-” in “bash -” mean

Related Question