Accessing bash command line args $@ vs $*

Question

In many SO questions and bash tutorials I see that I can access command line args in bash scripts in two ways:

$ ~ >cat testargs.sh  #!/bin/bash  echo "you passed me" $* echo "you passed me" $@ 

Which results in:

$ ~> bash testargs.sh arg1 arg2 you passed me arg1 arg2 you passed me arg1 arg2 

What is the difference between $* and $@?
When should one use the former and when shall one use the latter?

Answer 1

The difference appears when the special parameters are quoted. Let me illustrate the differences:

$ set -- "arg  1" "arg  2" "arg  3"  $ for word in $*; do echo "$word"; done arg 1 arg 2 arg 3  $ for word in $@; do echo "$word"; done arg 1 arg 2 arg 3  $ for word in "$*"; do echo "$word"; done arg  1 arg  2 arg  3  $ for word in "$@"; do echo "$word"; done arg  1 arg  2 arg  3 

---

one further example on the importance of quoting: note there are 2 spaces between "arg" and the number, but if I fail to quote $word:

$ for word in "$@"; do echo $word; done arg 1 arg 2 arg 3 

and in bash, "$@" is the "default" list to iterate over:

$ for word; do echo "$word"; done arg  1 arg  2 arg  3 

Answer 2

A nice handy overview table from the Bash Hackers Wiki:

Syntax	Effective result
$*	$1 $2 $3 … ${N}
$@	$1 $2 $3 … ${N}
"$*"	"$1c$2c$3c…c${N}"
"$@"	"$1" "$2" "$3" … "${N}"

where c in the third row is the first character of $IFS, the Input Field Separator, a shell variable.

If the arguments are to be stored in a script variable and the arguments are expected to contain spaces, I wholeheartedly recommend employing a "$*" trick with the input field separator set to tab IFS=$'\t'.