Make a Bash alias that takes a parameter?

Question

I used to use CShell (csh), which lets you make an alias that takes a parameter. The notation was something like

alias junk=\"mv \\\\!* ~/.Trash\"
         


        

Answer 1

TL;DR: Do this instead

Its far easier and more readable to use a function than an alias to put arguments in the middle of a command.

$ wrap_args() { echo "before $@ after"; }
$ wrap_args 1 2 3
before 1 2 3 after

If you read on, you'll learn things that you don't need to know about shell argument processing. Knowledge is dangerous. Just get the outcome you want, before the dark side forever controls your destiny.

Clarification

bash aliases do accept arguments, but only at the end:

$ alias speak=echo
$ speak hello world
hello world

Putting arguments into the middle of command via alias is indeed possible but it gets ugly.

Don't try this at home, kiddies!

If you like circumventing limitations and doing what others say is impossible, here's the recipe. Just don't blame me if your hair gets frazzled and your face ends up covered in soot mad-scientist-style.

The workaround is to pass the arguments that alias accepts only at the end to a wrapper that will insert them in the middle and then execute your command.

Solution 1

If you're really against using a function per se, you can use:

$ alias wrap_args='f(){ echo before "$@" after;  unset -f f; }; f'
$ wrap_args x y z
before x y z after

You can replace $@ with $1 if you only want the first argument.

Explanation 1

This creates a temporary function f, which is passed the arguments (note that f is called at the very end). The unset -f removes the function definition as the alias is executed so it doesn't hang around afterwards.

Solution 2

You can also use a subshell:

$ alias wrap_args='sh -c '\''echo before "$@" after'\'' _'

Explanation 2

The alias builds a command like:

sh -c 'echo before "$@" after' _

Comments:

• The placeholder _ is required, but it could be anything. It gets set to sh's $0, and is required so that the first of the user-given arguments don't get consumed. Demonstration:

  sh -c 'echo Consumed: "$0" Printing: "$@"' alcohol drunken babble
  Consumed: alcohol Printing: drunken babble
  

• The single-quotes inside single-quotes are required. Here's an example of it not working with double quotes:

  $ sh -c "echo Consumed: $0 Printing: $@" alcohol drunken babble
  Consumed: -bash Printing:
  

  Here the values of the interactive shell's $0 and $@ are replaced into the double quoted before it is passed to sh. Here's proof:

  echo "Consumed: $0 Printing: $@"
  Consumed: -bash Printing:
  

  The single quotes ensure that these variables are not interpreted by interactive shell, and are passed literally to sh -c.

  You could use double-quotes and \$@, but best practice is to quote your arguments (as they may contain spaces), and \"\$@\" looks even uglier, but may help you win an obfuscation contest where frazzled hair is a prerequisite for entry.