The day will come when you want to give arguments to your scripts. These arguments are reflected as the positional parameters inside your script. Most relevant special parameters are described below:
| Parameter(s) | Description |
|---|---|
$0 | the first positional parameter, equivalent to argv[0] in C, see the first argument |
$FUNCNAME | the function name (attention: inside a function, $0 is still the $0 of the shell, not the function name) |
$1 … $9 | the argument list elements from 1 to 9 |
${10} … ${N} | the argument list elements beyond 9 (note the parameter expansion syntax!) |
$* | all positional parameters except $0, see mass usage |
$@ | all positional parameters except $0, see mass usage |
$# | the number of arguments, not counting $0 |
These positional parameters reflect exactly what was given to the
script when it was called. There are no special things interpreted:
Option-switch parsing (-h for displaying help) is not done in this
stage.
See also the dictionary entry for "parameter".
The very first argument you can access is referenced by $0. It
usually is set to the script's name exactly like it was called, and
it's set on shell initialization:
Testscript - it just echos $0:
#!/bin/bash echo "$0"You see,
$0 is always set to however you call that script ($ is the prompt…):
> ./testscript ./testscript
> /usr/bin/testscript /usr/bin/testscript
However, this isn't true for login shells:
> echo "$0" -bash
Also, to be over-exact: $0 is not a positional parameter, it's a
special parameter independent from the real parameter list. Also it
really can be set to anything. In the ideal case it's the pathname
of the script, but since this is set on invocation, the invoking
program can easily influence it (the login program does that for
login shells, by prepending a dash, for example).
Inside a function, $0 still reflects what was described above. To
get the function name, use $FUNCNAME.
The builtin command shift is used to change the positional parameter values:
$1 will be discarded$2 will become $1$3 will become $2$N will become $N-1
The command can take a number as argument: How many positions to shift.
So, a shift 4 will shift $5 to $1.
Enough theory, you want to access your script-arguments. Well, here we go.
One way is to access specific parameters:
#!/bin/bash echo "Total number of arguments: $#" echo "Argument 1: $1" echo "Argument 2: $2" echo "Argument 3: $3" echo "Argument 4: $4" echo "Argument 5: $5"
Well, it might be useful in one or the other situation, but this way is not very flexible. You're fixed in your maximum number of arguments - which is a bad idea if you write a script that takes many filenames as arguments.
⇒ forget that one
There are several ways to loop through the positional parameters.
You can code a C-style for-loop using $#
as end-value. On every iteration, the shift-command is used to
shift the argument list:
numargs=$#
for ((i=1 ; i <= numargs ; i++))
do
echo "$1"
shift
done
Not very stylish, but okay, usable. The numargs variable is used
to store the initial value of $# because it will change due to the
shifting.
Another way to iterate one-by-one is the for-loop without given
wordlist, it will use the positional parameters as wordlist then:
for arg
do
echo "$arg"
done
Advantage: The positional parameters will be preserved and not shifted into nirvana!
The next way is similar to the first example (the for-loop), but
it doesn't test for reaching $#. It shifts and checks if $1
still expands to something, using the test command:
while [ "$1" ]
do
echo "$1"
shift
done
Looks nice, but it has the disadvantage to stop when $1 is empty
(null-string). Let's modify it to run as long as $1 is defined
(but may be empty), using parameter expansion for an alternate value:
while [ "${1+defined}" ]; do
echo "$1"
shift
done
There is a small tutorial dedicated to ''getopts'' (under construction).
Sometimes it's necessary to just "relay" or "hand through" given arguments to another program. It's very inefficient to do that in one of these loops, also you will destroy integrity, most likely (spaces!).
The shell-developers invented $* and $@ for this purpose.
As overview:
| Syntax | Effective result |
|---|---|
$* | $1 $2 $3 … ${N} |
$@ | $1 $2 $3 … ${N} |
"$*" | "$1c$2c$3c…c${N}" |
"$@" | "$1" "$2" "$3" … "${N}" |
You see that without being quoted (double-quoted), both have the same
effect: All positional parameters from $1 to the last used one are
expanded without any specials. A subsequent wordsplitting will
recognize as much words as expanded before (i.e. it "doesn't preserve
words").
When the $* special parameter is doublequoted, it expands to the
equivalent of: "$1c$2c$3c$4c……..$N", where 'c' is the first
character of IFS.
But when the $@ special parameter is used inside doublequotes, it
expands to the equivanent of…
"$1" "$2" "$3" "$4" ….. "$N"
…which exactly reflects all positional parameters like they were
initially set and given to the script or the function. If you want
to re-use your positional parameters to call another program (for
example in a wrapper-script), then this is the choice for you, use the
doublequoted "$@".
Well, let's just say: You almost always want a quoted "$@"!
Another way to mass-expand the positional parameters is similar to what is possible for a range of characters using the substring expansion on normal parameters and the range mass expansion of arrays.
${@:START:COUNT}
${*:START:COUNT}
"${@:START:COUNT}"
"${*:START:COUNT}"
The rules for using @ or * and the quoting are the same as
above. This will expand COUNT number of positional parameters
starting at START. COUNT can be omitted (${@:START}), in
this case all positional parameters beginning at START are
expanded.
If START is negative, the positional parameters are numbered from
the last one backwards.
COUNT may not be negative, so elements are always counted in the
forward direction.
Example: START at the last positional parameter:
echo "${@: -1}"
Attention: Since Bash 4, a START of 0 includes the special parameter $0, i.e. the shell name or whatever it's set to, when the positional parameters are in use. A START of 1 begins at $1. In Bash 3 and older, both 0 and 1 began at $1.
Letting the caller set the positional parameters, by giving parameters on commandline, is not the only way to set them. The set builtin command can be used to "artificially" change the positional parameters from inside the script or function:
set "This is" my new "set of" positional parameters # RESULTS IN # $1: This is # $2: my # $3: new # $4: set of # $5: positional # $6: parameters
It's wise to signal "end of options" when setting positional
parameters this way. If not, dashes might be interpreted as option tag
by set itself:
# both ways work, but behave differently. See the article about the set command! set -- ... set - ...
continue
To make your program accept options like standard command with syntax:
COMMAND [options] <params> # Like 'cat -A file.txt'
See simple option parsing code below. It's not that flexible. It doesn't auto-interpret combined options (-fu USER) but it works and is a good rudimentary way to parse your arguments.
#!/bin/sh
# By keeping options in alphabetical order, it's easy to add more.
while :
do
case "$1" in
-f | --file)
file="$2" # You may want to check validity of $2
shift 2
;;
-h | --help)
display_help # Call your function
# no shifting needed here, we're done.
exit 0
;;
-u | --user)
username="$2" # You may want to check validity of $2
shift 2
;;
-v | --verbose)
# It's better to assign a string, than a number like "verbose=1"
# because if you're debugging script with "bash -x" code like this:
#
# if [ "$verbose" ] ...
#
# You will see:
#
# if [ "verbose" ] ...
#
# Instead of cryptic
#
# if [ "1" ] ...
#
verbose="verbose"
shift
;;
--) # End of all options
shift
break;
-*)
echo "Error: Unknown option: $1" >&2
exit 1
;;
*) # No more options
break
;;
esac
done
# End of file
This simple wrapper allows to filter unwanted options (here: -a
and –all for ls) out of the commandline. It reads the
positional parameters and builds a (filtered) array out of them, then
it calls ls with the new option set. It also respects the –
as "end of options" for ls and doesn't change anything after it:
#!/bin/bash
# simple ls(1) wrapper that doesn't allow the -a option
options=() # the buffer array for the parameters
eoo=0 # end of options reached
while [[ $1 ]]
do
if ! ((eoo)); then
case "$1" in
-a)
shift
;;
--all)
shift
;;
-[^-]*a*|-a?*)
options+=("${1//a}")
shift
;;
--)
eoo=1
options+=("$1")
shift
;;
*)
options+=("$1")
shift
;;
esac
else
options+=("$1")
# Another (worse) way of doing the same thing:
# options=("${options[@]}" "$1")
shift
fi
done
/bin/ls "${options[@]}"
There is a small tutorial dedicated to ''getopts'' (under construction).
Discussion
Without double quotes, $* and $@ are expanding the positional parameters separated by only space, not by IFS.
(Edited: Inserted code tags)
Thank you very much for this finding. I know how
$*works, thus I can't understand why I described it that wrong. I guess it was in some late night session.Thanks again.
#!/bin/bash
OLDIFS="$IFS" IFS='-' #export IFS='-'
#echo -e $* #echo -e $@ #should be echo -e "$*" echo -e "$@" IFS="$OLDIFS"
#should be echo -e "$*"
I would suggext using a different prompt as the $ is confusing to newbies. Otherwise, an excellent treatise on use of positional parameters.
Thanks for the suggestion, I use "> " here now, and I'll change it in whatever text I edit in future (whole wiki). Let's see if "> " is okay.
Here's yet another non-getopts way.
http://bsdpants.blogspot.de/2007/02/option-ize-your-shell-scripts.html
Hi there!
What if I use "$@" in subsequent function calls, but arguments are strings?
I mean, having:
If you use it
But having
#!/bin/bash myfunc() { echo "$@" echo n: $# } ech "$@" echo n: $# myfunc "$@"you get:
As you can see, there is no way to make know the function that a parameter is a string and not a space separated list of arguments.
Any idea of how to solve it? I've test calling functions and doing expansion in almost all ways with no results.
I don't know why it fails for you. It should work if you use
"$@", of course.See the exmaple I used your second script with: