How to prevent "source" in a bash script from passing the script's arguments?

These questions address the functionality of source: What is the difference between executing a Bash script vs sourcing it? What does 'source' do?

but I'm confused as to why calling source in a script passes the script's arguments. E.g. I have these 2 scripts:

caller.sh

source sourced.sh
source sourced.sh ""

sourced.sh

echo [$*]

When I do

./caller.sh arg1 arg2

I get

[arg1 arg2]
[]

"arg1 arg2" are passed to sourced.sh even though I didn't specify source sourced.sh $*.

Why?

I found that appending "" prevents the arguments from being passed. Is this the recommended way to prevent arguments passed?

Solution 1:

source allows you to execute a command in the current context (arguments $* are part of context).

The second source call overwrites these arguments. Note it overwrites them only for the call, they are restored right after.