Difference between $@ and $* in bash scripting [duplicate]

Answer 1

They may appear the same when you are using echo but this is due to them being treated the same by echo and not being equivalent.

If pass three command-line arguments given to a bash script to a C program using ./my_c $@,

you get the result ARGV[1] == "par1" ARGV[2] == "par2" ARGV[3] == "par3".

If you pass three command-line arguments given to a bash script to a C program using ./my_c $*,

you get the result ARGV[1] == "par1 par2 par3".

(ARGV is the array of supplied arguments in C, the first element is always the command-name the program was invoked with)

It's to allow greater flexibility with what you do with the given parameters later in the script.

Answer 2

From http://tldp.org/LDP/abs/html/refcards.html:

"$*" All the positional parameters (as a single word) *

"$@" All the positional parameters (as separate strings)

This code shows it: given a string with items separated by spaces, $@ considers every word as a new item, while $* considers them all together the same parameter.

echo "Params for: \$@"
for item in "${@}"
do
        echo $item --
done

echo "Params for : \$*"
for item in "${*}"
do
        echo $item --
done

Test:

$ ./a par1 par2 par3
Your command line contains 3 arguments
Params for: $@
par1 --
par2 --
par3 --
Params for : $*
par1 par2 par3 --