Difference between executing a script with 'bash cd.sh' and 'source cd.sh'?

Question

Explain the difference between executing a script with bash cd.sh and source cd.sh

cd.sh contains:

#!/bin/sh
cd /tmp

Answer 1

bash execute the script in a child shell that cannot modify the environment of the invoking shell while source executes the script in the current shell:

test.sh

#!/bin/sh
export MY_NAME=chucksmash
echo $MY_NAME

Running test.sh:

chuck@precision:~$ bash test.sh
chucksmash
chuck@precision:~$ echo $MY_NAME

chuck@precision:~$ source test.sh
chucksmash
chuck@precision:~$ echo $MY_NAME
chucksmash
chuck@precision:~$ 

Answer 2

In bash, commands that look like source script.sh (or . script.sh) run the script in the current shell, regardless of the #! line.

Therefore, if you have a script (named script.sh in this example):

#!/bin/bash
VALUE=1
cd /tmp

This would print nothing (because VALUE is null) and not change your directory (because the commands were executed in another instance of bash):

bash script.sh
echo $VALUE

This would print 1 and change your directory to /tmp:

source script.sh
echo $VALUE

If you instead had this script (named script.py in this example):

#!/usr/bin/env python
print 'Hello, world"

This would give a WEIRD bash error (because it tries to interpret it as a bash script):

source shell.py

This would *also *give a WEIRD bash error (because it tries to interpret it as a bash script):

bash shell.py

This would print Hello, world:

./shell.py # assuming the execute bit it set