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Accessing bash command line args $@ vs $*
5 answers
There are 4 bash snippets below. I call them with ./script.sh a b c
for arg in $@; do
echo "$arg"
done ## output "a\nb\nc"
for arg in "$@"; do
echo "$arg"
done ## output "a\nb\nc" -- I don't know why
for arg in $*; do
echo "$arg"
done ## output "a\nb\nc"
for arg in "$*"; do
echo "$arg"
done ## output "abc"
I don't know what is the exact difference between $@
and $*
,
and I think "$@"
and "$*"
should be the same, but they are not. Why?
If you have a script foo.sh
:
asterisk "$*"
at-sign "$@"
and call it with:
./foo.sh "a a" "b b" "c c"
it's equivalent to:
asterisk "a a b b c c"
at-sign "a a" "b b" "c c"
Without the quotes, they're the same:
asterisk $*
at-sign $@
would be equivalent to:
asterisk "a" "a" "b" "b" "c" "c"
at-sign "a" "a" "b" "b" "c" "c"
Difference between $* and $@ is::
"$*" All the positional parameters (as a single word) *
"$@" All the positional parameters (as separate strings)
If you pass three command-line arguments given to a bash script to a C program using ./my_c $@,
you get the result ARG[1] == "par1" ARG[2] == "par2" ARG[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 ARG[1] == "par1 par2 par3"
This matters in shell scripts: for example the script testargs.sh
#! /bin/bash -p
echo $#
for i in $(seq 1 $#)
do
echo "$i: ${!i}"
done
for val in "$@"; do
echo "in quote @, $val"
done
for val in "$*"; do
echo "in quote *, $val"
done
for val in $@; do
echo "not in quote @, $val"
done
for val in $*; do
echo "not in quote *, $val"
done
If this script is executed as /tmp/testargs.sh a b c 'd e'
,
the results are:
4
1: a
2: b
3: c
4: d e
in quote @, a
in quote @, b
in quote @, c
in quote @, d e
in quote *, a b c d e
not in quote @, a
not in quote @, b
not in quote @, c
not in quote @, d
not in quote @, e
not in quote *, a
not in quote *, b
not in quote *, c
not in quote *, d
not in quote *, e
Thus, if number of arguments are to be preserved, always use "$@" or iterate through each argument using the for i in $(seq 1 $#)
loop. Without quotes, both are same.