The difference between $* and $@ [duplicate]

Question

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What the difference between “$@” and “$*” in bash?

For years and on dozens of occasions, I have hesitated between the use of $* and $@ in shell scripts. Having read the applicable section of Bash's manpage over and over again, having tried both $* and $@, I more or less completely fail to understand the practical difference of application between the two variables. Can you enlighten me, please?

I have been using $* recently, but don't ask me why. I don't know why, because I don't know why $@ even exists, except as an almost exact synonym for $*.

Is there any practical difference?

(I personally tend to use Bash, but remain agnostic regarding the choice of shell. My question is not specific to Bash as far as I know.)

Answer 1

Unquoted, there is no difference -- they're expanded to all the arguments and they're split accordingly. The difference comes when quoting. "$@" expands to properly quoted arguments and "$*" makes all arguments into a single argument. Take this for example:

#!/bin/bash  function print_args_at {     printf "%s\n" "$@" }  function print_args_star {     printf "%s\n" "$*" }  print_args_at "one" "two three" "four" print_args_star "one" "two three" "four" 

Then:

$ ./printf.sh   one two three four  one two three four 

Answer 2

Consider:

foo() { mv "$@"; }  bar() { mv "$*"; } foo a b bar a b 

The call to foo will attempt to mv file a to b. The call to bar will fail since it calls mv with only one argument.