What is the difference between $* and $@?
Consider the following code:
foo () {
echo $*
}
bar () {
echo $@
}
foo 1 2 3 4
bar 1 2 3 4
It outputs:
1 2 3 4
1 2 3 4
I am using Ksh88, but I am interested in other common shells as well. If you happen to know any particularity for specific shells, please do mention them.
I found the follwing in the Ksh man page on Solaris:
The meaning of $* and $@ is
identical when not quoted or when used as a parameter
assignment value or as a file name. However, when used as a
command argument, $* is equivalent to ``$1d$2d...'', where d
is the first character of the IFS variable, whereas $@ is
equivalent to $1 $2 ....
I tried modifying the IFS
variable, but it doesn't modify the output. Maybe I'm doing something wrong?
shell quoting ksh arguments
add a comment |
Consider the following code:
foo () {
echo $*
}
bar () {
echo $@
}
foo 1 2 3 4
bar 1 2 3 4
It outputs:
1 2 3 4
1 2 3 4
I am using Ksh88, but I am interested in other common shells as well. If you happen to know any particularity for specific shells, please do mention them.
I found the follwing in the Ksh man page on Solaris:
The meaning of $* and $@ is
identical when not quoted or when used as a parameter
assignment value or as a file name. However, when used as a
command argument, $* is equivalent to ``$1d$2d...'', where d
is the first character of the IFS variable, whereas $@ is
equivalent to $1 $2 ....
I tried modifying the IFS
variable, but it doesn't modify the output. Maybe I'm doing something wrong?
shell quoting ksh arguments
add a comment |
Consider the following code:
foo () {
echo $*
}
bar () {
echo $@
}
foo 1 2 3 4
bar 1 2 3 4
It outputs:
1 2 3 4
1 2 3 4
I am using Ksh88, but I am interested in other common shells as well. If you happen to know any particularity for specific shells, please do mention them.
I found the follwing in the Ksh man page on Solaris:
The meaning of $* and $@ is
identical when not quoted or when used as a parameter
assignment value or as a file name. However, when used as a
command argument, $* is equivalent to ``$1d$2d...'', where d
is the first character of the IFS variable, whereas $@ is
equivalent to $1 $2 ....
I tried modifying the IFS
variable, but it doesn't modify the output. Maybe I'm doing something wrong?
shell quoting ksh arguments
Consider the following code:
foo () {
echo $*
}
bar () {
echo $@
}
foo 1 2 3 4
bar 1 2 3 4
It outputs:
1 2 3 4
1 2 3 4
I am using Ksh88, but I am interested in other common shells as well. If you happen to know any particularity for specific shells, please do mention them.
I found the follwing in the Ksh man page on Solaris:
The meaning of $* and $@ is
identical when not quoted or when used as a parameter
assignment value or as a file name. However, when used as a
command argument, $* is equivalent to ``$1d$2d...'', where d
is the first character of the IFS variable, whereas $@ is
equivalent to $1 $2 ....
I tried modifying the IFS
variable, but it doesn't modify the output. Maybe I'm doing something wrong?
shell quoting ksh arguments
shell quoting ksh arguments
edited May 24 '14 at 2:07
Gilles
543k12811001617
543k12811001617
asked Jun 25 '12 at 9:51
rahmurahmu
10.5k2070112
10.5k2070112
add a comment |
add a comment |
8 Answers
8
active
oldest
votes
When they are not quoted, $*
and $@
are the same. You shouldn't use either of these, because they can break unexpectedly as soon as you have arguments containing spaces or wildcards.
"$*"
expands to a single word "$1c$2c..."
. Usually c
is a space, but it's actually the first character of IFS
, so it can be anything you choose.
The only good use I've ever found for it is:
join arguments with comma (simple version)
join1() {
typeset IFS=,
echo "$*"
}
join1 a b c # => a,b,c
join arguments with the specified delimiter (better version)
join2() {
typeset IFS=$1 # typeset makes a local variable in ksh (see footnote)
shift
echo "$*"
}
join2 + a b c # => a+b+c
"$@"
expands to separate words: "$1"
"$2"
...
This is almost always what you want. It expands each positional parameter to a separate word, which makes it perfect for taking command line or function arguments in and then passing them on to another command or function. And because it expands using double quotes, it means things don't break if, say, "$1"
contains a space or an asterisk (*
).
Let's write a script called svim
that runs vim
with sudo
. We'll do three versions to illustrate the difference.
svim1
#!/bin/sh
sudo vim $*
svim2
#!/bin/sh
sudo vim "$*"
svim3
#!/bin/sh
sudo vim "$@"
All of them will be fine for simple cases, e.g. a single file name that doesn't contain spaces:
svim1 foo.txt # == sudo vim foo.txt
svim2 foo.txt # == sudo vim "foo.txt"
svim2 foo.txt # == sudo vim "foo.txt"
But only $*
and "$@"
work properly if you have multiple arguments.
svim1 foo.txt bar.txt # == sudo vim foo.txt bar.txt
svim2 foo.txt bar.txt # == sudo vim "foo.txt bar.txt" # one file name!
svim3 foo.txt bar.txt # == sudo vim "foo.txt" "bar.txt"
And only "$*"
and "$@"
work properly if you have arguments containing spaces.
svim1 "shopping list.txt" # == sudo vim shopping list.txt # two file names!
svim2 "shopping list.txt" # == sudo vim "shopping list.txt"
svim3 "shopping list.txt" # == sudo vim "shopping list.txt"
So only "$@"
will work properly all the time.
typeset
is how to make a local variable in ksh
(bash
and ash
use local
instead). It means IFS
will be restored to its previous value when the function returns. This is important, because the commands you run afterward might not work properly if IFS
is set to something non-standard.
1
Wonderful explanation, thank you very much.
– rahmu
Jun 26 '12 at 15:29
Thank you for an example of using$*
. I was always considering it to be completely useless... join with delimiter is a good use case.
– anishsane
Mar 31 '16 at 5:01
add a comment |
Short answer: use "$@"
(note the double quotes). The other forms are very rarely useful.
"$@"
is a rather strange syntax. It is replaced by all the positional parameters, as separate fields. If there are no positional parameters ($#
is 0), then "$@"
expands to nothing (not an empty string, but a list with 0 elements), if there is one positional parameter then "$@"
is equivalent to "$1"
, if there are two positional parameters then "$@"
is equivalent to "$1" "$2"
, etc.
"$@"
allows you to pass down the arguments of a script or function to another command. It is very useful for wrappers that do things like setting environment variables, preparing data files, etc. before calling a command with the same arguments and options that the wrapper was called with.
For example, the following function filters the output of cvs -nq update
. Apart from the output filtering and the return status (which is that of grep
rather than that of cvs
), calling cvssm
on some arguments behaves like calling cvs -nq update
with these arguments.
cvssm () { cvs -nq update "$@" | egrep -v '^[?A]'; }
"$@"
expands to the list of positional parameters. In shells that support arrays, there is a similar syntax to expand to the list of elements of the array: "${array[@]}"
(the braces are mandatory except in zsh). Again, the double quotes are somewhat misleading: they protect against field splitting and pattern generation of the array elements, but each array element ends up in its own field.
Some ancient shells had what is arguably a bug: when there were no positional arguments, "$@"
expanded to a single field containing an empty string, rather than into no field. This led to the workaround ${1+"$@"}
(made famous via the Perl documentation). Only older versions of the actual Bourne shell and the OSF1 implementation are affected, none of its modern compatible replacements (ash, ksh, bash, …) are. /bin/sh
is not affected on any system that was released in the 21st century that I know of (unless you count Tru64 maintenance release, and even there /usr/xpg4/bin/sh
is safe so only #!/bin/sh
script are affected, not #!/usr/bin/env sh
scripts as long as your PATH is set up for POSIX compliance). In short, this is a historical anecdote that you don't need to worry about.
"$*"
always expands to one word. This word contains the positional parameters, concatenated with a space in between. (More generally, the separator is the first character of the value of the IFS
variable. If the value of IFS
is the empty string, the separator is the empty string.) If there are no positional parameters then "$*"
is the empty string, if there are two positional parameters and IFS
has its default value then "$*"
is equivalent to "$1 $2"
, etc.
$@
and $*
outside quotes are equivalent. They expand to the list of positional parameters, as separate fields, like "$@"
; but each resulting field is then split into separate fields which are treated as file name wildcard patterns, as usual with unquoted variable expansions.
For example, if the current directory contains three files bar
, baz
and foo
, then:
set -- # no positional parameters
for x in "$@"; do echo "$x"; done # prints nothing
for x in "$*"; do echo "$x"; done # prints 1 empty line
for x in $*; do echo "$x"; done # prints nothing
set -- "b* c*" "qux"
echo "$@" # prints `b* c* qux`
echo "$*" # prints `b* c* qux`
echo $* # prints `bar baz c* qux`
for x in "$@"; do echo "$x"; done # prints 2 lines: `b* c*` and `qux`
for x in "$*"; do echo "$x"; done # prints 1 lines: `b* c* qux`
for x in $*; do echo "$x"; done # prints 4 lines: `bar`, `baz`, `c*` and `qux`
1
Historical note: on some ancient Bourne shells,"$@"
did indeed expand to a list consisiting of the empty string: unix.stackexchange.com/questions/68484/…
– ninjalj
Oct 9 '13 at 13:01
add a comment |
Here is a simple script to demonstrates the difference between $*
and $@
:
#!/bin/bash
test_param() {
echo "Receive $# parameters"
echo Using '$*'
echo
for param in $*; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$*"'
for param in "$*"; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '$@'
for param in $@; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$@"';
for param in "$@"; do
printf '==>%s<==n' "$param"
done
}
IFS="^${IFS}"
test_param 1 2 3 "a b c"
Output:
% cuonglm at ~
% bash test.sh
Receive 4 parameters
Using $*
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$*"
==>1^2^3^a b c<==
Using $@
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$@"
==>1<==
==>2<==
==>3<==
==>a b c<==
In array syntax, there is no difference when using $*
or $@
. It only make sense when you use them with double quotes "$*"
and "$@"
.
Excellent example! Can you explain the use ofIFS="^${IFS}"
, though?
– Russ
Sep 16 '16 at 12:47
@Russ: It show you how value is concat with the first character inIFS
.
– cuonglm
Sep 16 '16 at 13:03
In the same way thatIFS="^xxxxx"
would do? The trailing${IFS}
suffix made me think you were doing something trickier, like somehow automatically recovering the original IFS at the end (eg: first char automatically shifted out or something).
– Russ
Sep 16 '16 at 14:48
add a comment |
The code you provided will give the same result. To understand it better, try this:
foo () {
for i in "$*"; do
echo "$i"
done
}
bar () {
for i in "$@"; do
echo "$i"
done
}
The output should now be different. Here's what I get:
$ foo() 1 2 3 4
1 2 3 4
$ bar() 1 2 3 4
1
2
3
4
This worked for me on bash
. As far as I know, ksh should not differ much. Essentially, quoting $*
will treat everything as one word, and quoting $@
will treat the list as separate words, as can be seen in the example above.
As an example of using the IFS
variable with $*
, consider this
fooifs () {
IFS="c"
for i in "$*"; do
echo "$i"
done
unset IFS # reset to the original value
}
I get this as a result:
$ fooifs 1 2 3 4
1c2c3c4
Also, I've just confirmed it works the same in ksh
. Both bash
and ksh
tested here were under OSX but I can't see how that would matter much.
"changed within a function - doesn't affect the global". Did you test that?
– Mikel
Jun 25 '12 at 15:11
Yes, I've checked that. For the peace of mind, I'll add theunset IFS
at the end to reset it to the original, but it worked for me no problem and doing anecho $IFS
resulted in the standard output I get from it. Setting theIFS
withing the curly brackets introduces a new scope, so unless you export it, it will not affect the outsideIFS
.
– Wojtek Rzepala
Jun 25 '12 at 16:00
echo $IFS
doesn't prove anything, because the shell sees the,
, but then does word splitting usingIFS
! Tryecho "$IFS"
.
– Mikel
Jun 25 '12 at 16:43
good point. unsettingIFS
should solve that.
– Wojtek Rzepala
Jun 25 '12 at 17:39
UnlessIFS
had a different custom value before calling the function. But yes, most of the time unsetting IFS will work.
– Mikel
Jun 25 '12 at 20:51
add a comment |
The difference is important when writing scripts that should use the positional parameters in the right way...
Imagine the following call:
$ myuseradd -m -c "Carlos Campderrós" ccampderros
Here there are just 4 parameters:
$1 => -m
$2 => -c
$3 => Carlos Campderrós
$4 => ccampderros
In my case, myuseradd
is just a wrapper for useradd
that accepts the same parameters, but adds a quota for the user:
#!/bin/bash -e
useradd "$@"
setquota -u "${!#}" 10000 11000 1000 1100
Notice the call to useradd "$@"
, with $@
quoted. This will respect the parameters and send them as-they-are to useradd
. If you were to unquote $@
(or to use $*
also unquoted), useradd would see 5 parameters, as the 3rd parameter which contained a space would be split in two:
$1 => -m
$2 => -c
$3 => Carlos
$4 => Campderrós
$5 => ccampderros
(and conversely, if you were to use "$*"
, useradd would only see one parameter: -m -c Carlos Campderrós ccampderros
)
So, in short, if you need to work with parameters respecting multi-word parameters, use "$@"
.
add a comment |
* Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, it expands to a sin‐
gle word with the value of each parameter separated by the first
character of the IFS special variable. That is, "$*" is equiva‐
lent to "$1c$2c...", where c is the first character of the value
of the IFS variable. If IFS is unset, the parameters are sepa‐
rated by spaces. If IFS is null, the parameters are joined
without intervening separators.
@ Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, each parameter
expands to a separate word. That is, "$@" is equivalent to "$1"
"$2" ... If the double-quoted expansion occurs within a word,
the expansion of the first parameter is joined with the begin‐
ning part of the original word, and the expansion of the last
parameter is joined with the last part of the original word.
When there are no positional parameters, "$@" and $@ expand to
nothing (i.e., they are removed).
// man bash . is ksh, afair, similar behaviour.
add a comment |
Talking about differences between zsh
and bash
:
With quotes around $@
and $*
, zsh
and bash
behave the same, and I guess the result is quite standard among all shells:
$ f () { for i in "$@"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
++
$ f () { for i in "$*"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a b +
Without quotes, results are the same for $*
and $@
, but different in bash
and in zsh
. In this case zsh
shows some odd behaviour:
bash$ f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a+
+a+
+b+
zsh% f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
(Zsh usually don't split textual data using IFS, unless explicitly requested, but notice that here the empty argument is unexpectedly missing in the list.)
1
In zsh,$@
isn't special in this respect:$x
expands to at most one word, but empty variables expand to nothing (not an empty word). Tryprint -l a $foo b
withfoo
empty or undefined.
– Gilles
Jun 25 '12 at 23:27
add a comment |
One of the answers says $*
(which I think of as a "splat") is rarely useful.
I search google with G() { IFS='+' ; w3m "https://encrypted.google.com/search?q=$*" ; }
Since URL’s are often split with a +
, but my keyboard makes easier to reach than
+
, $*
+ $IFS
feel worthwhile.
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
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When they are not quoted, $*
and $@
are the same. You shouldn't use either of these, because they can break unexpectedly as soon as you have arguments containing spaces or wildcards.
"$*"
expands to a single word "$1c$2c..."
. Usually c
is a space, but it's actually the first character of IFS
, so it can be anything you choose.
The only good use I've ever found for it is:
join arguments with comma (simple version)
join1() {
typeset IFS=,
echo "$*"
}
join1 a b c # => a,b,c
join arguments with the specified delimiter (better version)
join2() {
typeset IFS=$1 # typeset makes a local variable in ksh (see footnote)
shift
echo "$*"
}
join2 + a b c # => a+b+c
"$@"
expands to separate words: "$1"
"$2"
...
This is almost always what you want. It expands each positional parameter to a separate word, which makes it perfect for taking command line or function arguments in and then passing them on to another command or function. And because it expands using double quotes, it means things don't break if, say, "$1"
contains a space or an asterisk (*
).
Let's write a script called svim
that runs vim
with sudo
. We'll do three versions to illustrate the difference.
svim1
#!/bin/sh
sudo vim $*
svim2
#!/bin/sh
sudo vim "$*"
svim3
#!/bin/sh
sudo vim "$@"
All of them will be fine for simple cases, e.g. a single file name that doesn't contain spaces:
svim1 foo.txt # == sudo vim foo.txt
svim2 foo.txt # == sudo vim "foo.txt"
svim2 foo.txt # == sudo vim "foo.txt"
But only $*
and "$@"
work properly if you have multiple arguments.
svim1 foo.txt bar.txt # == sudo vim foo.txt bar.txt
svim2 foo.txt bar.txt # == sudo vim "foo.txt bar.txt" # one file name!
svim3 foo.txt bar.txt # == sudo vim "foo.txt" "bar.txt"
And only "$*"
and "$@"
work properly if you have arguments containing spaces.
svim1 "shopping list.txt" # == sudo vim shopping list.txt # two file names!
svim2 "shopping list.txt" # == sudo vim "shopping list.txt"
svim3 "shopping list.txt" # == sudo vim "shopping list.txt"
So only "$@"
will work properly all the time.
typeset
is how to make a local variable in ksh
(bash
and ash
use local
instead). It means IFS
will be restored to its previous value when the function returns. This is important, because the commands you run afterward might not work properly if IFS
is set to something non-standard.
1
Wonderful explanation, thank you very much.
– rahmu
Jun 26 '12 at 15:29
Thank you for an example of using$*
. I was always considering it to be completely useless... join with delimiter is a good use case.
– anishsane
Mar 31 '16 at 5:01
add a comment |
When they are not quoted, $*
and $@
are the same. You shouldn't use either of these, because they can break unexpectedly as soon as you have arguments containing spaces or wildcards.
"$*"
expands to a single word "$1c$2c..."
. Usually c
is a space, but it's actually the first character of IFS
, so it can be anything you choose.
The only good use I've ever found for it is:
join arguments with comma (simple version)
join1() {
typeset IFS=,
echo "$*"
}
join1 a b c # => a,b,c
join arguments with the specified delimiter (better version)
join2() {
typeset IFS=$1 # typeset makes a local variable in ksh (see footnote)
shift
echo "$*"
}
join2 + a b c # => a+b+c
"$@"
expands to separate words: "$1"
"$2"
...
This is almost always what you want. It expands each positional parameter to a separate word, which makes it perfect for taking command line or function arguments in and then passing them on to another command or function. And because it expands using double quotes, it means things don't break if, say, "$1"
contains a space or an asterisk (*
).
Let's write a script called svim
that runs vim
with sudo
. We'll do three versions to illustrate the difference.
svim1
#!/bin/sh
sudo vim $*
svim2
#!/bin/sh
sudo vim "$*"
svim3
#!/bin/sh
sudo vim "$@"
All of them will be fine for simple cases, e.g. a single file name that doesn't contain spaces:
svim1 foo.txt # == sudo vim foo.txt
svim2 foo.txt # == sudo vim "foo.txt"
svim2 foo.txt # == sudo vim "foo.txt"
But only $*
and "$@"
work properly if you have multiple arguments.
svim1 foo.txt bar.txt # == sudo vim foo.txt bar.txt
svim2 foo.txt bar.txt # == sudo vim "foo.txt bar.txt" # one file name!
svim3 foo.txt bar.txt # == sudo vim "foo.txt" "bar.txt"
And only "$*"
and "$@"
work properly if you have arguments containing spaces.
svim1 "shopping list.txt" # == sudo vim shopping list.txt # two file names!
svim2 "shopping list.txt" # == sudo vim "shopping list.txt"
svim3 "shopping list.txt" # == sudo vim "shopping list.txt"
So only "$@"
will work properly all the time.
typeset
is how to make a local variable in ksh
(bash
and ash
use local
instead). It means IFS
will be restored to its previous value when the function returns. This is important, because the commands you run afterward might not work properly if IFS
is set to something non-standard.
1
Wonderful explanation, thank you very much.
– rahmu
Jun 26 '12 at 15:29
Thank you for an example of using$*
. I was always considering it to be completely useless... join with delimiter is a good use case.
– anishsane
Mar 31 '16 at 5:01
add a comment |
When they are not quoted, $*
and $@
are the same. You shouldn't use either of these, because they can break unexpectedly as soon as you have arguments containing spaces or wildcards.
"$*"
expands to a single word "$1c$2c..."
. Usually c
is a space, but it's actually the first character of IFS
, so it can be anything you choose.
The only good use I've ever found for it is:
join arguments with comma (simple version)
join1() {
typeset IFS=,
echo "$*"
}
join1 a b c # => a,b,c
join arguments with the specified delimiter (better version)
join2() {
typeset IFS=$1 # typeset makes a local variable in ksh (see footnote)
shift
echo "$*"
}
join2 + a b c # => a+b+c
"$@"
expands to separate words: "$1"
"$2"
...
This is almost always what you want. It expands each positional parameter to a separate word, which makes it perfect for taking command line or function arguments in and then passing them on to another command or function. And because it expands using double quotes, it means things don't break if, say, "$1"
contains a space or an asterisk (*
).
Let's write a script called svim
that runs vim
with sudo
. We'll do three versions to illustrate the difference.
svim1
#!/bin/sh
sudo vim $*
svim2
#!/bin/sh
sudo vim "$*"
svim3
#!/bin/sh
sudo vim "$@"
All of them will be fine for simple cases, e.g. a single file name that doesn't contain spaces:
svim1 foo.txt # == sudo vim foo.txt
svim2 foo.txt # == sudo vim "foo.txt"
svim2 foo.txt # == sudo vim "foo.txt"
But only $*
and "$@"
work properly if you have multiple arguments.
svim1 foo.txt bar.txt # == sudo vim foo.txt bar.txt
svim2 foo.txt bar.txt # == sudo vim "foo.txt bar.txt" # one file name!
svim3 foo.txt bar.txt # == sudo vim "foo.txt" "bar.txt"
And only "$*"
and "$@"
work properly if you have arguments containing spaces.
svim1 "shopping list.txt" # == sudo vim shopping list.txt # two file names!
svim2 "shopping list.txt" # == sudo vim "shopping list.txt"
svim3 "shopping list.txt" # == sudo vim "shopping list.txt"
So only "$@"
will work properly all the time.
typeset
is how to make a local variable in ksh
(bash
and ash
use local
instead). It means IFS
will be restored to its previous value when the function returns. This is important, because the commands you run afterward might not work properly if IFS
is set to something non-standard.
When they are not quoted, $*
and $@
are the same. You shouldn't use either of these, because they can break unexpectedly as soon as you have arguments containing spaces or wildcards.
"$*"
expands to a single word "$1c$2c..."
. Usually c
is a space, but it's actually the first character of IFS
, so it can be anything you choose.
The only good use I've ever found for it is:
join arguments with comma (simple version)
join1() {
typeset IFS=,
echo "$*"
}
join1 a b c # => a,b,c
join arguments with the specified delimiter (better version)
join2() {
typeset IFS=$1 # typeset makes a local variable in ksh (see footnote)
shift
echo "$*"
}
join2 + a b c # => a+b+c
"$@"
expands to separate words: "$1"
"$2"
...
This is almost always what you want. It expands each positional parameter to a separate word, which makes it perfect for taking command line or function arguments in and then passing them on to another command or function. And because it expands using double quotes, it means things don't break if, say, "$1"
contains a space or an asterisk (*
).
Let's write a script called svim
that runs vim
with sudo
. We'll do three versions to illustrate the difference.
svim1
#!/bin/sh
sudo vim $*
svim2
#!/bin/sh
sudo vim "$*"
svim3
#!/bin/sh
sudo vim "$@"
All of them will be fine for simple cases, e.g. a single file name that doesn't contain spaces:
svim1 foo.txt # == sudo vim foo.txt
svim2 foo.txt # == sudo vim "foo.txt"
svim2 foo.txt # == sudo vim "foo.txt"
But only $*
and "$@"
work properly if you have multiple arguments.
svim1 foo.txt bar.txt # == sudo vim foo.txt bar.txt
svim2 foo.txt bar.txt # == sudo vim "foo.txt bar.txt" # one file name!
svim3 foo.txt bar.txt # == sudo vim "foo.txt" "bar.txt"
And only "$*"
and "$@"
work properly if you have arguments containing spaces.
svim1 "shopping list.txt" # == sudo vim shopping list.txt # two file names!
svim2 "shopping list.txt" # == sudo vim "shopping list.txt"
svim3 "shopping list.txt" # == sudo vim "shopping list.txt"
So only "$@"
will work properly all the time.
typeset
is how to make a local variable in ksh
(bash
and ash
use local
instead). It means IFS
will be restored to its previous value when the function returns. This is important, because the commands you run afterward might not work properly if IFS
is set to something non-standard.
edited Feb 27 at 16:10
glenn jackman
52.4k573113
52.4k573113
answered Jun 25 '12 at 15:25
MikelMikel
39.9k10103127
39.9k10103127
1
Wonderful explanation, thank you very much.
– rahmu
Jun 26 '12 at 15:29
Thank you for an example of using$*
. I was always considering it to be completely useless... join with delimiter is a good use case.
– anishsane
Mar 31 '16 at 5:01
add a comment |
1
Wonderful explanation, thank you very much.
– rahmu
Jun 26 '12 at 15:29
Thank you for an example of using$*
. I was always considering it to be completely useless... join with delimiter is a good use case.
– anishsane
Mar 31 '16 at 5:01
1
1
Wonderful explanation, thank you very much.
– rahmu
Jun 26 '12 at 15:29
Wonderful explanation, thank you very much.
– rahmu
Jun 26 '12 at 15:29
Thank you for an example of using
$*
. I was always considering it to be completely useless... join with delimiter is a good use case.– anishsane
Mar 31 '16 at 5:01
Thank you for an example of using
$*
. I was always considering it to be completely useless... join with delimiter is a good use case.– anishsane
Mar 31 '16 at 5:01
add a comment |
Short answer: use "$@"
(note the double quotes). The other forms are very rarely useful.
"$@"
is a rather strange syntax. It is replaced by all the positional parameters, as separate fields. If there are no positional parameters ($#
is 0), then "$@"
expands to nothing (not an empty string, but a list with 0 elements), if there is one positional parameter then "$@"
is equivalent to "$1"
, if there are two positional parameters then "$@"
is equivalent to "$1" "$2"
, etc.
"$@"
allows you to pass down the arguments of a script or function to another command. It is very useful for wrappers that do things like setting environment variables, preparing data files, etc. before calling a command with the same arguments and options that the wrapper was called with.
For example, the following function filters the output of cvs -nq update
. Apart from the output filtering and the return status (which is that of grep
rather than that of cvs
), calling cvssm
on some arguments behaves like calling cvs -nq update
with these arguments.
cvssm () { cvs -nq update "$@" | egrep -v '^[?A]'; }
"$@"
expands to the list of positional parameters. In shells that support arrays, there is a similar syntax to expand to the list of elements of the array: "${array[@]}"
(the braces are mandatory except in zsh). Again, the double quotes are somewhat misleading: they protect against field splitting and pattern generation of the array elements, but each array element ends up in its own field.
Some ancient shells had what is arguably a bug: when there were no positional arguments, "$@"
expanded to a single field containing an empty string, rather than into no field. This led to the workaround ${1+"$@"}
(made famous via the Perl documentation). Only older versions of the actual Bourne shell and the OSF1 implementation are affected, none of its modern compatible replacements (ash, ksh, bash, …) are. /bin/sh
is not affected on any system that was released in the 21st century that I know of (unless you count Tru64 maintenance release, and even there /usr/xpg4/bin/sh
is safe so only #!/bin/sh
script are affected, not #!/usr/bin/env sh
scripts as long as your PATH is set up for POSIX compliance). In short, this is a historical anecdote that you don't need to worry about.
"$*"
always expands to one word. This word contains the positional parameters, concatenated with a space in between. (More generally, the separator is the first character of the value of the IFS
variable. If the value of IFS
is the empty string, the separator is the empty string.) If there are no positional parameters then "$*"
is the empty string, if there are two positional parameters and IFS
has its default value then "$*"
is equivalent to "$1 $2"
, etc.
$@
and $*
outside quotes are equivalent. They expand to the list of positional parameters, as separate fields, like "$@"
; but each resulting field is then split into separate fields which are treated as file name wildcard patterns, as usual with unquoted variable expansions.
For example, if the current directory contains three files bar
, baz
and foo
, then:
set -- # no positional parameters
for x in "$@"; do echo "$x"; done # prints nothing
for x in "$*"; do echo "$x"; done # prints 1 empty line
for x in $*; do echo "$x"; done # prints nothing
set -- "b* c*" "qux"
echo "$@" # prints `b* c* qux`
echo "$*" # prints `b* c* qux`
echo $* # prints `bar baz c* qux`
for x in "$@"; do echo "$x"; done # prints 2 lines: `b* c*` and `qux`
for x in "$*"; do echo "$x"; done # prints 1 lines: `b* c* qux`
for x in $*; do echo "$x"; done # prints 4 lines: `bar`, `baz`, `c*` and `qux`
1
Historical note: on some ancient Bourne shells,"$@"
did indeed expand to a list consisiting of the empty string: unix.stackexchange.com/questions/68484/…
– ninjalj
Oct 9 '13 at 13:01
add a comment |
Short answer: use "$@"
(note the double quotes). The other forms are very rarely useful.
"$@"
is a rather strange syntax. It is replaced by all the positional parameters, as separate fields. If there are no positional parameters ($#
is 0), then "$@"
expands to nothing (not an empty string, but a list with 0 elements), if there is one positional parameter then "$@"
is equivalent to "$1"
, if there are two positional parameters then "$@"
is equivalent to "$1" "$2"
, etc.
"$@"
allows you to pass down the arguments of a script or function to another command. It is very useful for wrappers that do things like setting environment variables, preparing data files, etc. before calling a command with the same arguments and options that the wrapper was called with.
For example, the following function filters the output of cvs -nq update
. Apart from the output filtering and the return status (which is that of grep
rather than that of cvs
), calling cvssm
on some arguments behaves like calling cvs -nq update
with these arguments.
cvssm () { cvs -nq update "$@" | egrep -v '^[?A]'; }
"$@"
expands to the list of positional parameters. In shells that support arrays, there is a similar syntax to expand to the list of elements of the array: "${array[@]}"
(the braces are mandatory except in zsh). Again, the double quotes are somewhat misleading: they protect against field splitting and pattern generation of the array elements, but each array element ends up in its own field.
Some ancient shells had what is arguably a bug: when there were no positional arguments, "$@"
expanded to a single field containing an empty string, rather than into no field. This led to the workaround ${1+"$@"}
(made famous via the Perl documentation). Only older versions of the actual Bourne shell and the OSF1 implementation are affected, none of its modern compatible replacements (ash, ksh, bash, …) are. /bin/sh
is not affected on any system that was released in the 21st century that I know of (unless you count Tru64 maintenance release, and even there /usr/xpg4/bin/sh
is safe so only #!/bin/sh
script are affected, not #!/usr/bin/env sh
scripts as long as your PATH is set up for POSIX compliance). In short, this is a historical anecdote that you don't need to worry about.
"$*"
always expands to one word. This word contains the positional parameters, concatenated with a space in between. (More generally, the separator is the first character of the value of the IFS
variable. If the value of IFS
is the empty string, the separator is the empty string.) If there are no positional parameters then "$*"
is the empty string, if there are two positional parameters and IFS
has its default value then "$*"
is equivalent to "$1 $2"
, etc.
$@
and $*
outside quotes are equivalent. They expand to the list of positional parameters, as separate fields, like "$@"
; but each resulting field is then split into separate fields which are treated as file name wildcard patterns, as usual with unquoted variable expansions.
For example, if the current directory contains three files bar
, baz
and foo
, then:
set -- # no positional parameters
for x in "$@"; do echo "$x"; done # prints nothing
for x in "$*"; do echo "$x"; done # prints 1 empty line
for x in $*; do echo "$x"; done # prints nothing
set -- "b* c*" "qux"
echo "$@" # prints `b* c* qux`
echo "$*" # prints `b* c* qux`
echo $* # prints `bar baz c* qux`
for x in "$@"; do echo "$x"; done # prints 2 lines: `b* c*` and `qux`
for x in "$*"; do echo "$x"; done # prints 1 lines: `b* c* qux`
for x in $*; do echo "$x"; done # prints 4 lines: `bar`, `baz`, `c*` and `qux`
1
Historical note: on some ancient Bourne shells,"$@"
did indeed expand to a list consisiting of the empty string: unix.stackexchange.com/questions/68484/…
– ninjalj
Oct 9 '13 at 13:01
add a comment |
Short answer: use "$@"
(note the double quotes). The other forms are very rarely useful.
"$@"
is a rather strange syntax. It is replaced by all the positional parameters, as separate fields. If there are no positional parameters ($#
is 0), then "$@"
expands to nothing (not an empty string, but a list with 0 elements), if there is one positional parameter then "$@"
is equivalent to "$1"
, if there are two positional parameters then "$@"
is equivalent to "$1" "$2"
, etc.
"$@"
allows you to pass down the arguments of a script or function to another command. It is very useful for wrappers that do things like setting environment variables, preparing data files, etc. before calling a command with the same arguments and options that the wrapper was called with.
For example, the following function filters the output of cvs -nq update
. Apart from the output filtering and the return status (which is that of grep
rather than that of cvs
), calling cvssm
on some arguments behaves like calling cvs -nq update
with these arguments.
cvssm () { cvs -nq update "$@" | egrep -v '^[?A]'; }
"$@"
expands to the list of positional parameters. In shells that support arrays, there is a similar syntax to expand to the list of elements of the array: "${array[@]}"
(the braces are mandatory except in zsh). Again, the double quotes are somewhat misleading: they protect against field splitting and pattern generation of the array elements, but each array element ends up in its own field.
Some ancient shells had what is arguably a bug: when there were no positional arguments, "$@"
expanded to a single field containing an empty string, rather than into no field. This led to the workaround ${1+"$@"}
(made famous via the Perl documentation). Only older versions of the actual Bourne shell and the OSF1 implementation are affected, none of its modern compatible replacements (ash, ksh, bash, …) are. /bin/sh
is not affected on any system that was released in the 21st century that I know of (unless you count Tru64 maintenance release, and even there /usr/xpg4/bin/sh
is safe so only #!/bin/sh
script are affected, not #!/usr/bin/env sh
scripts as long as your PATH is set up for POSIX compliance). In short, this is a historical anecdote that you don't need to worry about.
"$*"
always expands to one word. This word contains the positional parameters, concatenated with a space in between. (More generally, the separator is the first character of the value of the IFS
variable. If the value of IFS
is the empty string, the separator is the empty string.) If there are no positional parameters then "$*"
is the empty string, if there are two positional parameters and IFS
has its default value then "$*"
is equivalent to "$1 $2"
, etc.
$@
and $*
outside quotes are equivalent. They expand to the list of positional parameters, as separate fields, like "$@"
; but each resulting field is then split into separate fields which are treated as file name wildcard patterns, as usual with unquoted variable expansions.
For example, if the current directory contains three files bar
, baz
and foo
, then:
set -- # no positional parameters
for x in "$@"; do echo "$x"; done # prints nothing
for x in "$*"; do echo "$x"; done # prints 1 empty line
for x in $*; do echo "$x"; done # prints nothing
set -- "b* c*" "qux"
echo "$@" # prints `b* c* qux`
echo "$*" # prints `b* c* qux`
echo $* # prints `bar baz c* qux`
for x in "$@"; do echo "$x"; done # prints 2 lines: `b* c*` and `qux`
for x in "$*"; do echo "$x"; done # prints 1 lines: `b* c* qux`
for x in $*; do echo "$x"; done # prints 4 lines: `bar`, `baz`, `c*` and `qux`
Short answer: use "$@"
(note the double quotes). The other forms are very rarely useful.
"$@"
is a rather strange syntax. It is replaced by all the positional parameters, as separate fields. If there are no positional parameters ($#
is 0), then "$@"
expands to nothing (not an empty string, but a list with 0 elements), if there is one positional parameter then "$@"
is equivalent to "$1"
, if there are two positional parameters then "$@"
is equivalent to "$1" "$2"
, etc.
"$@"
allows you to pass down the arguments of a script or function to another command. It is very useful for wrappers that do things like setting environment variables, preparing data files, etc. before calling a command with the same arguments and options that the wrapper was called with.
For example, the following function filters the output of cvs -nq update
. Apart from the output filtering and the return status (which is that of grep
rather than that of cvs
), calling cvssm
on some arguments behaves like calling cvs -nq update
with these arguments.
cvssm () { cvs -nq update "$@" | egrep -v '^[?A]'; }
"$@"
expands to the list of positional parameters. In shells that support arrays, there is a similar syntax to expand to the list of elements of the array: "${array[@]}"
(the braces are mandatory except in zsh). Again, the double quotes are somewhat misleading: they protect against field splitting and pattern generation of the array elements, but each array element ends up in its own field.
Some ancient shells had what is arguably a bug: when there were no positional arguments, "$@"
expanded to a single field containing an empty string, rather than into no field. This led to the workaround ${1+"$@"}
(made famous via the Perl documentation). Only older versions of the actual Bourne shell and the OSF1 implementation are affected, none of its modern compatible replacements (ash, ksh, bash, …) are. /bin/sh
is not affected on any system that was released in the 21st century that I know of (unless you count Tru64 maintenance release, and even there /usr/xpg4/bin/sh
is safe so only #!/bin/sh
script are affected, not #!/usr/bin/env sh
scripts as long as your PATH is set up for POSIX compliance). In short, this is a historical anecdote that you don't need to worry about.
"$*"
always expands to one word. This word contains the positional parameters, concatenated with a space in between. (More generally, the separator is the first character of the value of the IFS
variable. If the value of IFS
is the empty string, the separator is the empty string.) If there are no positional parameters then "$*"
is the empty string, if there are two positional parameters and IFS
has its default value then "$*"
is equivalent to "$1 $2"
, etc.
$@
and $*
outside quotes are equivalent. They expand to the list of positional parameters, as separate fields, like "$@"
; but each resulting field is then split into separate fields which are treated as file name wildcard patterns, as usual with unquoted variable expansions.
For example, if the current directory contains three files bar
, baz
and foo
, then:
set -- # no positional parameters
for x in "$@"; do echo "$x"; done # prints nothing
for x in "$*"; do echo "$x"; done # prints 1 empty line
for x in $*; do echo "$x"; done # prints nothing
set -- "b* c*" "qux"
echo "$@" # prints `b* c* qux`
echo "$*" # prints `b* c* qux`
echo $* # prints `bar baz c* qux`
for x in "$@"; do echo "$x"; done # prints 2 lines: `b* c*` and `qux`
for x in "$*"; do echo "$x"; done # prints 1 lines: `b* c* qux`
for x in $*; do echo "$x"; done # prints 4 lines: `bar`, `baz`, `c*` and `qux`
edited Apr 13 '17 at 12:36
Community♦
1
1
answered Oct 9 '13 at 1:20
GillesGilles
543k12811001617
543k12811001617
1
Historical note: on some ancient Bourne shells,"$@"
did indeed expand to a list consisiting of the empty string: unix.stackexchange.com/questions/68484/…
– ninjalj
Oct 9 '13 at 13:01
add a comment |
1
Historical note: on some ancient Bourne shells,"$@"
did indeed expand to a list consisiting of the empty string: unix.stackexchange.com/questions/68484/…
– ninjalj
Oct 9 '13 at 13:01
1
1
Historical note: on some ancient Bourne shells,
"$@"
did indeed expand to a list consisiting of the empty string: unix.stackexchange.com/questions/68484/…– ninjalj
Oct 9 '13 at 13:01
Historical note: on some ancient Bourne shells,
"$@"
did indeed expand to a list consisiting of the empty string: unix.stackexchange.com/questions/68484/…– ninjalj
Oct 9 '13 at 13:01
add a comment |
Here is a simple script to demonstrates the difference between $*
and $@
:
#!/bin/bash
test_param() {
echo "Receive $# parameters"
echo Using '$*'
echo
for param in $*; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$*"'
for param in "$*"; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '$@'
for param in $@; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$@"';
for param in "$@"; do
printf '==>%s<==n' "$param"
done
}
IFS="^${IFS}"
test_param 1 2 3 "a b c"
Output:
% cuonglm at ~
% bash test.sh
Receive 4 parameters
Using $*
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$*"
==>1^2^3^a b c<==
Using $@
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$@"
==>1<==
==>2<==
==>3<==
==>a b c<==
In array syntax, there is no difference when using $*
or $@
. It only make sense when you use them with double quotes "$*"
and "$@"
.
Excellent example! Can you explain the use ofIFS="^${IFS}"
, though?
– Russ
Sep 16 '16 at 12:47
@Russ: It show you how value is concat with the first character inIFS
.
– cuonglm
Sep 16 '16 at 13:03
In the same way thatIFS="^xxxxx"
would do? The trailing${IFS}
suffix made me think you were doing something trickier, like somehow automatically recovering the original IFS at the end (eg: first char automatically shifted out or something).
– Russ
Sep 16 '16 at 14:48
add a comment |
Here is a simple script to demonstrates the difference between $*
and $@
:
#!/bin/bash
test_param() {
echo "Receive $# parameters"
echo Using '$*'
echo
for param in $*; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$*"'
for param in "$*"; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '$@'
for param in $@; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$@"';
for param in "$@"; do
printf '==>%s<==n' "$param"
done
}
IFS="^${IFS}"
test_param 1 2 3 "a b c"
Output:
% cuonglm at ~
% bash test.sh
Receive 4 parameters
Using $*
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$*"
==>1^2^3^a b c<==
Using $@
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$@"
==>1<==
==>2<==
==>3<==
==>a b c<==
In array syntax, there is no difference when using $*
or $@
. It only make sense when you use them with double quotes "$*"
and "$@"
.
Excellent example! Can you explain the use ofIFS="^${IFS}"
, though?
– Russ
Sep 16 '16 at 12:47
@Russ: It show you how value is concat with the first character inIFS
.
– cuonglm
Sep 16 '16 at 13:03
In the same way thatIFS="^xxxxx"
would do? The trailing${IFS}
suffix made me think you were doing something trickier, like somehow automatically recovering the original IFS at the end (eg: first char automatically shifted out or something).
– Russ
Sep 16 '16 at 14:48
add a comment |
Here is a simple script to demonstrates the difference between $*
and $@
:
#!/bin/bash
test_param() {
echo "Receive $# parameters"
echo Using '$*'
echo
for param in $*; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$*"'
for param in "$*"; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '$@'
for param in $@; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$@"';
for param in "$@"; do
printf '==>%s<==n' "$param"
done
}
IFS="^${IFS}"
test_param 1 2 3 "a b c"
Output:
% cuonglm at ~
% bash test.sh
Receive 4 parameters
Using $*
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$*"
==>1^2^3^a b c<==
Using $@
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$@"
==>1<==
==>2<==
==>3<==
==>a b c<==
In array syntax, there is no difference when using $*
or $@
. It only make sense when you use them with double quotes "$*"
and "$@"
.
Here is a simple script to demonstrates the difference between $*
and $@
:
#!/bin/bash
test_param() {
echo "Receive $# parameters"
echo Using '$*'
echo
for param in $*; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$*"'
for param in "$*"; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '$@'
for param in $@; do
printf '==>%s<==n' "$param"
done;
echo
echo Using '"$@"';
for param in "$@"; do
printf '==>%s<==n' "$param"
done
}
IFS="^${IFS}"
test_param 1 2 3 "a b c"
Output:
% cuonglm at ~
% bash test.sh
Receive 4 parameters
Using $*
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$*"
==>1^2^3^a b c<==
Using $@
==>1<==
==>2<==
==>3<==
==>a<==
==>b<==
==>c<==
Using "$@"
==>1<==
==>2<==
==>3<==
==>a b c<==
In array syntax, there is no difference when using $*
or $@
. It only make sense when you use them with double quotes "$*"
and "$@"
.
edited Jan 29 '18 at 0:21
Jeff Schaller
43.7k1161141
43.7k1161141
answered Oct 8 '13 at 17:18
cuonglmcuonglm
105k25209307
105k25209307
Excellent example! Can you explain the use ofIFS="^${IFS}"
, though?
– Russ
Sep 16 '16 at 12:47
@Russ: It show you how value is concat with the first character inIFS
.
– cuonglm
Sep 16 '16 at 13:03
In the same way thatIFS="^xxxxx"
would do? The trailing${IFS}
suffix made me think you were doing something trickier, like somehow automatically recovering the original IFS at the end (eg: first char automatically shifted out or something).
– Russ
Sep 16 '16 at 14:48
add a comment |
Excellent example! Can you explain the use ofIFS="^${IFS}"
, though?
– Russ
Sep 16 '16 at 12:47
@Russ: It show you how value is concat with the first character inIFS
.
– cuonglm
Sep 16 '16 at 13:03
In the same way thatIFS="^xxxxx"
would do? The trailing${IFS}
suffix made me think you were doing something trickier, like somehow automatically recovering the original IFS at the end (eg: first char automatically shifted out or something).
– Russ
Sep 16 '16 at 14:48
Excellent example! Can you explain the use of
IFS="^${IFS}"
, though?– Russ
Sep 16 '16 at 12:47
Excellent example! Can you explain the use of
IFS="^${IFS}"
, though?– Russ
Sep 16 '16 at 12:47
@Russ: It show you how value is concat with the first character in
IFS
.– cuonglm
Sep 16 '16 at 13:03
@Russ: It show you how value is concat with the first character in
IFS
.– cuonglm
Sep 16 '16 at 13:03
In the same way that
IFS="^xxxxx"
would do? The trailing ${IFS}
suffix made me think you were doing something trickier, like somehow automatically recovering the original IFS at the end (eg: first char automatically shifted out or something).– Russ
Sep 16 '16 at 14:48
In the same way that
IFS="^xxxxx"
would do? The trailing ${IFS}
suffix made me think you were doing something trickier, like somehow automatically recovering the original IFS at the end (eg: first char automatically shifted out or something).– Russ
Sep 16 '16 at 14:48
add a comment |
The code you provided will give the same result. To understand it better, try this:
foo () {
for i in "$*"; do
echo "$i"
done
}
bar () {
for i in "$@"; do
echo "$i"
done
}
The output should now be different. Here's what I get:
$ foo() 1 2 3 4
1 2 3 4
$ bar() 1 2 3 4
1
2
3
4
This worked for me on bash
. As far as I know, ksh should not differ much. Essentially, quoting $*
will treat everything as one word, and quoting $@
will treat the list as separate words, as can be seen in the example above.
As an example of using the IFS
variable with $*
, consider this
fooifs () {
IFS="c"
for i in "$*"; do
echo "$i"
done
unset IFS # reset to the original value
}
I get this as a result:
$ fooifs 1 2 3 4
1c2c3c4
Also, I've just confirmed it works the same in ksh
. Both bash
and ksh
tested here were under OSX but I can't see how that would matter much.
"changed within a function - doesn't affect the global". Did you test that?
– Mikel
Jun 25 '12 at 15:11
Yes, I've checked that. For the peace of mind, I'll add theunset IFS
at the end to reset it to the original, but it worked for me no problem and doing anecho $IFS
resulted in the standard output I get from it. Setting theIFS
withing the curly brackets introduces a new scope, so unless you export it, it will not affect the outsideIFS
.
– Wojtek Rzepala
Jun 25 '12 at 16:00
echo $IFS
doesn't prove anything, because the shell sees the,
, but then does word splitting usingIFS
! Tryecho "$IFS"
.
– Mikel
Jun 25 '12 at 16:43
good point. unsettingIFS
should solve that.
– Wojtek Rzepala
Jun 25 '12 at 17:39
UnlessIFS
had a different custom value before calling the function. But yes, most of the time unsetting IFS will work.
– Mikel
Jun 25 '12 at 20:51
add a comment |
The code you provided will give the same result. To understand it better, try this:
foo () {
for i in "$*"; do
echo "$i"
done
}
bar () {
for i in "$@"; do
echo "$i"
done
}
The output should now be different. Here's what I get:
$ foo() 1 2 3 4
1 2 3 4
$ bar() 1 2 3 4
1
2
3
4
This worked for me on bash
. As far as I know, ksh should not differ much. Essentially, quoting $*
will treat everything as one word, and quoting $@
will treat the list as separate words, as can be seen in the example above.
As an example of using the IFS
variable with $*
, consider this
fooifs () {
IFS="c"
for i in "$*"; do
echo "$i"
done
unset IFS # reset to the original value
}
I get this as a result:
$ fooifs 1 2 3 4
1c2c3c4
Also, I've just confirmed it works the same in ksh
. Both bash
and ksh
tested here were under OSX but I can't see how that would matter much.
"changed within a function - doesn't affect the global". Did you test that?
– Mikel
Jun 25 '12 at 15:11
Yes, I've checked that. For the peace of mind, I'll add theunset IFS
at the end to reset it to the original, but it worked for me no problem and doing anecho $IFS
resulted in the standard output I get from it. Setting theIFS
withing the curly brackets introduces a new scope, so unless you export it, it will not affect the outsideIFS
.
– Wojtek Rzepala
Jun 25 '12 at 16:00
echo $IFS
doesn't prove anything, because the shell sees the,
, but then does word splitting usingIFS
! Tryecho "$IFS"
.
– Mikel
Jun 25 '12 at 16:43
good point. unsettingIFS
should solve that.
– Wojtek Rzepala
Jun 25 '12 at 17:39
UnlessIFS
had a different custom value before calling the function. But yes, most of the time unsetting IFS will work.
– Mikel
Jun 25 '12 at 20:51
add a comment |
The code you provided will give the same result. To understand it better, try this:
foo () {
for i in "$*"; do
echo "$i"
done
}
bar () {
for i in "$@"; do
echo "$i"
done
}
The output should now be different. Here's what I get:
$ foo() 1 2 3 4
1 2 3 4
$ bar() 1 2 3 4
1
2
3
4
This worked for me on bash
. As far as I know, ksh should not differ much. Essentially, quoting $*
will treat everything as one word, and quoting $@
will treat the list as separate words, as can be seen in the example above.
As an example of using the IFS
variable with $*
, consider this
fooifs () {
IFS="c"
for i in "$*"; do
echo "$i"
done
unset IFS # reset to the original value
}
I get this as a result:
$ fooifs 1 2 3 4
1c2c3c4
Also, I've just confirmed it works the same in ksh
. Both bash
and ksh
tested here were under OSX but I can't see how that would matter much.
The code you provided will give the same result. To understand it better, try this:
foo () {
for i in "$*"; do
echo "$i"
done
}
bar () {
for i in "$@"; do
echo "$i"
done
}
The output should now be different. Here's what I get:
$ foo() 1 2 3 4
1 2 3 4
$ bar() 1 2 3 4
1
2
3
4
This worked for me on bash
. As far as I know, ksh should not differ much. Essentially, quoting $*
will treat everything as one word, and quoting $@
will treat the list as separate words, as can be seen in the example above.
As an example of using the IFS
variable with $*
, consider this
fooifs () {
IFS="c"
for i in "$*"; do
echo "$i"
done
unset IFS # reset to the original value
}
I get this as a result:
$ fooifs 1 2 3 4
1c2c3c4
Also, I've just confirmed it works the same in ksh
. Both bash
and ksh
tested here were under OSX but I can't see how that would matter much.
edited Jun 25 '12 at 21:09
answered Jun 25 '12 at 12:26
Wojtek RzepalaWojtek Rzepala
1,84411222
1,84411222
"changed within a function - doesn't affect the global". Did you test that?
– Mikel
Jun 25 '12 at 15:11
Yes, I've checked that. For the peace of mind, I'll add theunset IFS
at the end to reset it to the original, but it worked for me no problem and doing anecho $IFS
resulted in the standard output I get from it. Setting theIFS
withing the curly brackets introduces a new scope, so unless you export it, it will not affect the outsideIFS
.
– Wojtek Rzepala
Jun 25 '12 at 16:00
echo $IFS
doesn't prove anything, because the shell sees the,
, but then does word splitting usingIFS
! Tryecho "$IFS"
.
– Mikel
Jun 25 '12 at 16:43
good point. unsettingIFS
should solve that.
– Wojtek Rzepala
Jun 25 '12 at 17:39
UnlessIFS
had a different custom value before calling the function. But yes, most of the time unsetting IFS will work.
– Mikel
Jun 25 '12 at 20:51
add a comment |
"changed within a function - doesn't affect the global". Did you test that?
– Mikel
Jun 25 '12 at 15:11
Yes, I've checked that. For the peace of mind, I'll add theunset IFS
at the end to reset it to the original, but it worked for me no problem and doing anecho $IFS
resulted in the standard output I get from it. Setting theIFS
withing the curly brackets introduces a new scope, so unless you export it, it will not affect the outsideIFS
.
– Wojtek Rzepala
Jun 25 '12 at 16:00
echo $IFS
doesn't prove anything, because the shell sees the,
, but then does word splitting usingIFS
! Tryecho "$IFS"
.
– Mikel
Jun 25 '12 at 16:43
good point. unsettingIFS
should solve that.
– Wojtek Rzepala
Jun 25 '12 at 17:39
UnlessIFS
had a different custom value before calling the function. But yes, most of the time unsetting IFS will work.
– Mikel
Jun 25 '12 at 20:51
"changed within a function - doesn't affect the global". Did you test that?
– Mikel
Jun 25 '12 at 15:11
"changed within a function - doesn't affect the global". Did you test that?
– Mikel
Jun 25 '12 at 15:11
Yes, I've checked that. For the peace of mind, I'll add the
unset IFS
at the end to reset it to the original, but it worked for me no problem and doing an echo $IFS
resulted in the standard output I get from it. Setting the IFS
withing the curly brackets introduces a new scope, so unless you export it, it will not affect the outside IFS
.– Wojtek Rzepala
Jun 25 '12 at 16:00
Yes, I've checked that. For the peace of mind, I'll add the
unset IFS
at the end to reset it to the original, but it worked for me no problem and doing an echo $IFS
resulted in the standard output I get from it. Setting the IFS
withing the curly brackets introduces a new scope, so unless you export it, it will not affect the outside IFS
.– Wojtek Rzepala
Jun 25 '12 at 16:00
echo $IFS
doesn't prove anything, because the shell sees the ,
, but then does word splitting using IFS
! Try echo "$IFS"
.– Mikel
Jun 25 '12 at 16:43
echo $IFS
doesn't prove anything, because the shell sees the ,
, but then does word splitting using IFS
! Try echo "$IFS"
.– Mikel
Jun 25 '12 at 16:43
good point. unsetting
IFS
should solve that.– Wojtek Rzepala
Jun 25 '12 at 17:39
good point. unsetting
IFS
should solve that.– Wojtek Rzepala
Jun 25 '12 at 17:39
Unless
IFS
had a different custom value before calling the function. But yes, most of the time unsetting IFS will work.– Mikel
Jun 25 '12 at 20:51
Unless
IFS
had a different custom value before calling the function. But yes, most of the time unsetting IFS will work.– Mikel
Jun 25 '12 at 20:51
add a comment |
The difference is important when writing scripts that should use the positional parameters in the right way...
Imagine the following call:
$ myuseradd -m -c "Carlos Campderrós" ccampderros
Here there are just 4 parameters:
$1 => -m
$2 => -c
$3 => Carlos Campderrós
$4 => ccampderros
In my case, myuseradd
is just a wrapper for useradd
that accepts the same parameters, but adds a quota for the user:
#!/bin/bash -e
useradd "$@"
setquota -u "${!#}" 10000 11000 1000 1100
Notice the call to useradd "$@"
, with $@
quoted. This will respect the parameters and send them as-they-are to useradd
. If you were to unquote $@
(or to use $*
also unquoted), useradd would see 5 parameters, as the 3rd parameter which contained a space would be split in two:
$1 => -m
$2 => -c
$3 => Carlos
$4 => Campderrós
$5 => ccampderros
(and conversely, if you were to use "$*"
, useradd would only see one parameter: -m -c Carlos Campderrós ccampderros
)
So, in short, if you need to work with parameters respecting multi-word parameters, use "$@"
.
add a comment |
The difference is important when writing scripts that should use the positional parameters in the right way...
Imagine the following call:
$ myuseradd -m -c "Carlos Campderrós" ccampderros
Here there are just 4 parameters:
$1 => -m
$2 => -c
$3 => Carlos Campderrós
$4 => ccampderros
In my case, myuseradd
is just a wrapper for useradd
that accepts the same parameters, but adds a quota for the user:
#!/bin/bash -e
useradd "$@"
setquota -u "${!#}" 10000 11000 1000 1100
Notice the call to useradd "$@"
, with $@
quoted. This will respect the parameters and send them as-they-are to useradd
. If you were to unquote $@
(or to use $*
also unquoted), useradd would see 5 parameters, as the 3rd parameter which contained a space would be split in two:
$1 => -m
$2 => -c
$3 => Carlos
$4 => Campderrós
$5 => ccampderros
(and conversely, if you were to use "$*"
, useradd would only see one parameter: -m -c Carlos Campderrós ccampderros
)
So, in short, if you need to work with parameters respecting multi-word parameters, use "$@"
.
add a comment |
The difference is important when writing scripts that should use the positional parameters in the right way...
Imagine the following call:
$ myuseradd -m -c "Carlos Campderrós" ccampderros
Here there are just 4 parameters:
$1 => -m
$2 => -c
$3 => Carlos Campderrós
$4 => ccampderros
In my case, myuseradd
is just a wrapper for useradd
that accepts the same parameters, but adds a quota for the user:
#!/bin/bash -e
useradd "$@"
setquota -u "${!#}" 10000 11000 1000 1100
Notice the call to useradd "$@"
, with $@
quoted. This will respect the parameters and send them as-they-are to useradd
. If you were to unquote $@
(or to use $*
also unquoted), useradd would see 5 parameters, as the 3rd parameter which contained a space would be split in two:
$1 => -m
$2 => -c
$3 => Carlos
$4 => Campderrós
$5 => ccampderros
(and conversely, if you were to use "$*"
, useradd would only see one parameter: -m -c Carlos Campderrós ccampderros
)
So, in short, if you need to work with parameters respecting multi-word parameters, use "$@"
.
The difference is important when writing scripts that should use the positional parameters in the right way...
Imagine the following call:
$ myuseradd -m -c "Carlos Campderrós" ccampderros
Here there are just 4 parameters:
$1 => -m
$2 => -c
$3 => Carlos Campderrós
$4 => ccampderros
In my case, myuseradd
is just a wrapper for useradd
that accepts the same parameters, but adds a quota for the user:
#!/bin/bash -e
useradd "$@"
setquota -u "${!#}" 10000 11000 1000 1100
Notice the call to useradd "$@"
, with $@
quoted. This will respect the parameters and send them as-they-are to useradd
. If you were to unquote $@
(or to use $*
also unquoted), useradd would see 5 parameters, as the 3rd parameter which contained a space would be split in two:
$1 => -m
$2 => -c
$3 => Carlos
$4 => Campderrós
$5 => ccampderros
(and conversely, if you were to use "$*"
, useradd would only see one parameter: -m -c Carlos Campderrós ccampderros
)
So, in short, if you need to work with parameters respecting multi-word parameters, use "$@"
.
answered Jun 25 '12 at 14:55
Carlos CampderrósCarlos Campderrós
8711716
8711716
add a comment |
add a comment |
* Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, it expands to a sin‐
gle word with the value of each parameter separated by the first
character of the IFS special variable. That is, "$*" is equiva‐
lent to "$1c$2c...", where c is the first character of the value
of the IFS variable. If IFS is unset, the parameters are sepa‐
rated by spaces. If IFS is null, the parameters are joined
without intervening separators.
@ Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, each parameter
expands to a separate word. That is, "$@" is equivalent to "$1"
"$2" ... If the double-quoted expansion occurs within a word,
the expansion of the first parameter is joined with the begin‐
ning part of the original word, and the expansion of the last
parameter is joined with the last part of the original word.
When there are no positional parameters, "$@" and $@ expand to
nothing (i.e., they are removed).
// man bash . is ksh, afair, similar behaviour.
add a comment |
* Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, it expands to a sin‐
gle word with the value of each parameter separated by the first
character of the IFS special variable. That is, "$*" is equiva‐
lent to "$1c$2c...", where c is the first character of the value
of the IFS variable. If IFS is unset, the parameters are sepa‐
rated by spaces. If IFS is null, the parameters are joined
without intervening separators.
@ Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, each parameter
expands to a separate word. That is, "$@" is equivalent to "$1"
"$2" ... If the double-quoted expansion occurs within a word,
the expansion of the first parameter is joined with the begin‐
ning part of the original word, and the expansion of the last
parameter is joined with the last part of the original word.
When there are no positional parameters, "$@" and $@ expand to
nothing (i.e., they are removed).
// man bash . is ksh, afair, similar behaviour.
add a comment |
* Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, it expands to a sin‐
gle word with the value of each parameter separated by the first
character of the IFS special variable. That is, "$*" is equiva‐
lent to "$1c$2c...", where c is the first character of the value
of the IFS variable. If IFS is unset, the parameters are sepa‐
rated by spaces. If IFS is null, the parameters are joined
without intervening separators.
@ Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, each parameter
expands to a separate word. That is, "$@" is equivalent to "$1"
"$2" ... If the double-quoted expansion occurs within a word,
the expansion of the first parameter is joined with the begin‐
ning part of the original word, and the expansion of the last
parameter is joined with the last part of the original word.
When there are no positional parameters, "$@" and $@ expand to
nothing (i.e., they are removed).
// man bash . is ksh, afair, similar behaviour.
* Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, it expands to a sin‐
gle word with the value of each parameter separated by the first
character of the IFS special variable. That is, "$*" is equiva‐
lent to "$1c$2c...", where c is the first character of the value
of the IFS variable. If IFS is unset, the parameters are sepa‐
rated by spaces. If IFS is null, the parameters are joined
without intervening separators.
@ Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, each parameter
expands to a separate word. That is, "$@" is equivalent to "$1"
"$2" ... If the double-quoted expansion occurs within a word,
the expansion of the first parameter is joined with the begin‐
ning part of the original word, and the expansion of the last
parameter is joined with the last part of the original word.
When there are no positional parameters, "$@" and $@ expand to
nothing (i.e., they are removed).
// man bash . is ksh, afair, similar behaviour.
answered Jun 25 '12 at 10:13
rushrush
19.3k46596
19.3k46596
add a comment |
add a comment |
Talking about differences between zsh
and bash
:
With quotes around $@
and $*
, zsh
and bash
behave the same, and I guess the result is quite standard among all shells:
$ f () { for i in "$@"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
++
$ f () { for i in "$*"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a b +
Without quotes, results are the same for $*
and $@
, but different in bash
and in zsh
. In this case zsh
shows some odd behaviour:
bash$ f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a+
+a+
+b+
zsh% f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
(Zsh usually don't split textual data using IFS, unless explicitly requested, but notice that here the empty argument is unexpectedly missing in the list.)
1
In zsh,$@
isn't special in this respect:$x
expands to at most one word, but empty variables expand to nothing (not an empty word). Tryprint -l a $foo b
withfoo
empty or undefined.
– Gilles
Jun 25 '12 at 23:27
add a comment |
Talking about differences between zsh
and bash
:
With quotes around $@
and $*
, zsh
and bash
behave the same, and I guess the result is quite standard among all shells:
$ f () { for i in "$@"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
++
$ f () { for i in "$*"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a b +
Without quotes, results are the same for $*
and $@
, but different in bash
and in zsh
. In this case zsh
shows some odd behaviour:
bash$ f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a+
+a+
+b+
zsh% f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
(Zsh usually don't split textual data using IFS, unless explicitly requested, but notice that here the empty argument is unexpectedly missing in the list.)
1
In zsh,$@
isn't special in this respect:$x
expands to at most one word, but empty variables expand to nothing (not an empty word). Tryprint -l a $foo b
withfoo
empty or undefined.
– Gilles
Jun 25 '12 at 23:27
add a comment |
Talking about differences between zsh
and bash
:
With quotes around $@
and $*
, zsh
and bash
behave the same, and I guess the result is quite standard among all shells:
$ f () { for i in "$@"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
++
$ f () { for i in "$*"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a b +
Without quotes, results are the same for $*
and $@
, but different in bash
and in zsh
. In this case zsh
shows some odd behaviour:
bash$ f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a+
+a+
+b+
zsh% f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
(Zsh usually don't split textual data using IFS, unless explicitly requested, but notice that here the empty argument is unexpectedly missing in the list.)
Talking about differences between zsh
and bash
:
With quotes around $@
and $*
, zsh
and bash
behave the same, and I guess the result is quite standard among all shells:
$ f () { for i in "$@"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
++
$ f () { for i in "$*"; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a b +
Without quotes, results are the same for $*
and $@
, but different in bash
and in zsh
. In this case zsh
shows some odd behaviour:
bash$ f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a+
+a+
+b+
zsh% f () { for i in $*; do echo +"$i"+; done; }; f 'a a' 'b' ''
+a a+
+b+
(Zsh usually don't split textual data using IFS, unless explicitly requested, but notice that here the empty argument is unexpectedly missing in the list.)
edited Jun 25 '12 at 13:02
answered Jun 25 '12 at 12:56
Stéphane GimenezStéphane Gimenez
19.6k25174
19.6k25174
1
In zsh,$@
isn't special in this respect:$x
expands to at most one word, but empty variables expand to nothing (not an empty word). Tryprint -l a $foo b
withfoo
empty or undefined.
– Gilles
Jun 25 '12 at 23:27
add a comment |
1
In zsh,$@
isn't special in this respect:$x
expands to at most one word, but empty variables expand to nothing (not an empty word). Tryprint -l a $foo b
withfoo
empty or undefined.
– Gilles
Jun 25 '12 at 23:27
1
1
In zsh,
$@
isn't special in this respect: $x
expands to at most one word, but empty variables expand to nothing (not an empty word). Try print -l a $foo b
with foo
empty or undefined.– Gilles
Jun 25 '12 at 23:27
In zsh,
$@
isn't special in this respect: $x
expands to at most one word, but empty variables expand to nothing (not an empty word). Try print -l a $foo b
with foo
empty or undefined.– Gilles
Jun 25 '12 at 23:27
add a comment |
One of the answers says $*
(which I think of as a "splat") is rarely useful.
I search google with G() { IFS='+' ; w3m "https://encrypted.google.com/search?q=$*" ; }
Since URL’s are often split with a +
, but my keyboard makes easier to reach than
+
, $*
+ $IFS
feel worthwhile.
add a comment |
One of the answers says $*
(which I think of as a "splat") is rarely useful.
I search google with G() { IFS='+' ; w3m "https://encrypted.google.com/search?q=$*" ; }
Since URL’s are often split with a +
, but my keyboard makes easier to reach than
+
, $*
+ $IFS
feel worthwhile.
add a comment |
One of the answers says $*
(which I think of as a "splat") is rarely useful.
I search google with G() { IFS='+' ; w3m "https://encrypted.google.com/search?q=$*" ; }
Since URL’s are often split with a +
, but my keyboard makes easier to reach than
+
, $*
+ $IFS
feel worthwhile.
One of the answers says $*
(which I think of as a "splat") is rarely useful.
I search google with G() { IFS='+' ; w3m "https://encrypted.google.com/search?q=$*" ; }
Since URL’s are often split with a +
, but my keyboard makes easier to reach than
+
, $*
+ $IFS
feel worthwhile.
answered Mar 9 at 2:01
isomorphismesisomorphismes
437412
437412
add a comment |
add a comment |
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