Run bash script as source without source command

Question

Is there any way to mark a script to be "run as source" so you don't have to add the source or "." command to it every time? i.e., if I write a script called "sup", I'd like to call it as

sup Argument

rather than

source sup Argument

or

. sup Argument

Basically, I'm trying to use cd within a script.

Answer 1

Create an alias for it:

alias sup=". ~/bin/sup"

Or along those lines.

See also: Why doesn't cd work in a bash shell script?

---

Answering comment by counter-example: experimentation with Korn Shell on Solaris 10 shows that I can do:

$ pwd
/work1/jleffler
$ echo "cd /work5/atria" > $HOME/bin/yyy
$ alias yyy=". ~/bin/yyy"
$ yyy
$ pwd
/work5/atria
$

Experimentation with Bash (3.00.16) on Solaris 10 also shows the same behaviour.

---

Answer 2

Bash forks and stars a subshell way before it or your kernel even considers what it's supposed to do in there. It's not something you can "undo". So no, it's impossible.

Thankfully.

Look into bash functions instead:

sup() {
    ...
}

Put that in your ~/.bashrc.

Answer 3

what about when you run the script

./scriptname.sh

I tried running my script using this command but it does not recognize the script. Works fine if I do

scriptname.sh