What is the difference between $@ and $* in Bash? When would you use one over the other?

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Accepted Answer

Both $@ and $* represent all the positional parameters passed to a script, but they behave differently when quoted.

Without quotes, they are essentially the same. Both expand to all arguments separated by spaces. The difference becomes important when you put them inside double quotes.

"$*" expands to a single string with all arguments joined by the first character of the IFS variable, which is usually a space. So if you pass three arguments "one", "two", and "three", "$*" becomes one single string: "one two three".

"$@" expands to separate strings, preserving each argument as a distinct word. Using the same example, "$@" gives you three separate strings: "one", "two", and "three". This is crucial when arguments contain spaces.

Here is a practical example:

# script.sh
for arg in "$@"; do
  echo "Argument: $arg"
done

# Running: ./script.sh "hello world" "foo bar"
# With "$@": Argument: hello world
#            Argument: foo bar
# With "$*": Argument: hello world foo bar

The rule of thumb is: almost always use "$@" when you need to pass arguments to another command or iterate over them. This preserves the original argument boundaries. The only time "$*" makes sense is when you intentionally want to join all arguments into a single string.

Common use case: forwarding arguments to another command. Use exec "$@" to pass all arguments correctly to the next program.