Passing alias as function argument linux bash

Question

Hi everyone I'm learning how to use the .bashrc file in linux and as my title states I'm wondering how to make a function recognize an argument as an alias

I have an alias called home defined as: alias home=$HOME

and a function go defined as

function go(){
cd $1
ls $1
}

but when I do go home i get

bash: cd: home: No such file or directory ls: cannot access home: No such file or directory

when I want it to do go $HOME

how would i go about implementing this?

Answer 1

An alias is not a word substitution but a small newly created command:

$ alias bla=ls
$ bla
file1
file2
file3
…

So, it cannot be used in the way you assumed.

You might want to use variable substitution for this:

$ home=$HOME
$ function go() {
  cd "$(eval echo \$"$1")"
}
$ go home

In case you want to use an alias despite that this is an abuse, try this:

$ alias home=$HOME
$ function go() {
  cd "$(type "$1" | sed -e 's/.*is aliased to .//' -e 's/.$//')"
}
$ go home

Answer 2

I'm wondering how to make a function recognize an argument as an alias

To answer your question literally, you could use type -t inside your function to determine whether its argument is an alias. But that's probably not what you want.

AFAIK, what you actually want isn't possible. From the Bash manual (emphasis mine):

Aliases allow a string to be substituted for a word when it is used as the first word of a simple command.

The closest you could get is:

go `home`