Run bash script as source without source command

Bash forks and starts 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.

When you are running a shell, there are two ways to invoke a shell script:

• Executing a script spawns a new process inside which the script is running. This is done by typing the script name, if it is made executable and starts with a

  #!/bin/bash

  line, or directly invoking

  /bin/bash mycmd.sh

• Sourcing a script runs it inside its parent shell (i.e. the one you are typing commands into). This is done by typing

  source mycmd.sh

  or

  . mycmd.sh

So the cd inside a shell script that isn't sourced is never going to propagate to its parent shell, as this would violate process isolation.

If the cd is all you are interested about, you can get rid of the script by using cd "shortcuts"... Take a look into the bash doc, at the CDPATH env var.

Otherwise, you can use an alias to be able to type a single command, instead of source or .:

alias mycmd="source mycmd.sh"

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.