CODEWITHSUNDEEP
bashcommand-line-arguments
oz123
oz123

Reputation: 28878

Accessing bash command line args $@ vs $*

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?

Upvotes: 447

Views: 266446

Answers (5)

stefansson
stefansson

Reputation: 579

This example may highlight the difference between @ and * when we use them. I declared two arrays fruits and vegetables

fruits=(apple pear plum peach melon)            
vegetables=(carrot tomato cucumber potato onion)

printf "Fruits:\t%s\n" "${fruits[*]}"            
printf "Fruits:\t%s\n" "${fruits[@]}"            
echo + --------------------------------------------- +      
printf "Vegetables:\t%s\n" "${vegetables[*]}"    
printf "Vegetables:\t%s\n" "${vegetables[@]}"

See the following result the code above:

Fruits: apple pear plum peach melon
Fruits: apple
Fruits: pear
Fruits: plum
Fruits: peach
Fruits: melon
+ --------------------------------------------- +
Vegetables: carrot tomato cucumber potato onion
Vegetables: carrot
Vegetables: tomato
Vegetables: cucumber
Vegetables: potato
Vegetables: onion

Upvotes: 3

Serge Stroobandt
Serge Stroobandt

Reputation: 31618

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, load them in an array variable.

Upvotes: 367

glenn jackman
glenn jackman

Reputation: 247182

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

Upvotes: 546

Muffo
Muffo

Reputation: 1743

$*

Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.

$@

Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" ... If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed).

Source: Bash man

Upvotes: 53

rkosegi
rkosegi

Reputation: 14678

$@ is same as $*, but each parameter is a quoted string, that is, the parameters are passed on intact, without interpretation or expansion. This means, among other things, that each parameter in the argument list is seen as a separate word.

Of course, "$@" should be quoted.

http://tldp.org/LDP/abs/html/internalvariables.html#ARGLIST

Upvotes: 24

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