Are there lisp-like macros in the shell? - shell

I have a set of shell commands that look like this
if check-some-condition $a then;
do stuff
run-exit-code $a
fi
where check-some-condition and run-exit-code could be replaced by functions taking a single argument $a, while do stuff is a placeholder for possibly several shell commands. Is it possible to emulate the Lisp functionality of a macro where I could just write
(my-macro $a stuff)
and have it replaced by the code above? I am using Bash but I can use any other shell if they have features that make this easier. I thought at first of using functions but I don't think I can pass in a block of commands.

There isn't a macro definition system in the shell, and consequently shell syntax is not walked in order to expand macros. However, the shell has a textual eval command. You can write a function which synthesizes shell syntax, such as by inserting arguments it has been given, into a template. The function can print that syntax, which the caller can capture using $(...) command substitution syntax and pass to eval:
eval $(macro-like foo bar)
The expansion will happen every time that line of code is executed.
I've done something like this on a very small number of occasions. I don't remember all the details, but I remember that the code was also taking advantage of Bash local variables, which have dynamic scope, like ancient Lisp dialects and defvar variables in Common Lisp.
In Bash, eval takes place in a dynamic environment which sees the surrounding local variables, which is something you can exploit; it can help bring about some macro-like semantics. In a Lisp with lexically scoped local variables, eval-ed code has no access to those variables, but macro-substituted code does. Under dynamic scope, evaled code can access and assign surrounding locals.
Here is an example. Note that because eval is a command which has no control over expansions taking place in its argument space because it is called (analogously to eval in Lisp being a function which doesn't control argument evaluation), the client code is encumbered with quoting responsibilities.
# $1 = variable
# $2 = low
# $3 = high
# $4 = body
dofor()
{
cat <<!
$1=$2 ;
while [ \$$1 -lt $3 ] ; do
$4
$1=\$(( $1 + 1 ))
done
!
}
eval "$(dofor i 0 100 'printf "[%d]\n" $i')"
We could make it so that
eval $(dofor i 0 100 'printf "[%d]\n" $i')
works without the quotes, at the cost of more heaps of arcane escapery inside dofor.
Imagine we extended the shell with a built-in command evalcmd, which let us write this instead of the above:
evalcmd dofor i 0 100 'printf "[%d]\n" $i'
Can we write that as a shell function? It turns out, yes:
# run the command specified in the arguments
# capturing its output, which is evaled in quotes
evcmd()
{
eval "$("$#")"
}
evcmd dofor i 0 100 'printf "[%d]\n" $i'
Now, though still monstrously inefficient, it's substantially more ergonomic.
Finally, let's ask: could we split dofor into an a dofor_impl which generates the code, and a dofor command which calls dofor_impl and invokes the evcmd semantics? Also, yes:
dofor_impl()
{
cat <<!
$1=$2 ;
while [ \$$1 -lt $3 ] ; do
$4
$1=\$(( $1 + 1 ))
done
!
}
dofor()
{
# like evcmd, but inserting an operator into the left position
eval "$(dofor_impl "$#")"
}
dofor i 0 100 'printf "[%d]\n" $i'
This is not bad for some simple uses, but what we can't achieve is not having to put the $i into a quote so that the substitution doesn't take place before dofor is invoked.

In Bash you can define functions as follows:
function run_exit_code () {
echo "EXIT $1"
}
function check_some_condition () {
echo "CHECKING $1";
true
}
And your code can execute commands associated with variables:
function my_code () {
var=$1
stuff=$2
if check_some_condition $var; then
echo "OK";
$stuff;
run_exit_code $var
fi
}
So you can write, for example:
$ my_code /tmp 'ls /'
CHECKING /tmp
OK
bin boot cdrom dev etc home lib lib32 lib64 libx32 lost+found media mnt opt proc root run sbin srv swapfile sys tmp usr var
EXIT /tmp
If you want stuff to refer to $var, then you need to add eval:
function my_code () {
var=$1
stuff=$2
if check_some_condition $var; then
echo "OK";
eval $stuff; # <<< eval
run_exit_code $var
fi
}
This allows you to write a quoted bash expression and have it beeing evaluated in the context of your function:
$ my_code / 'ls $var'
CHECKING /
OK
bin boot cdrom dev etc home lib lib32 ...
EXIT /

Related

Evaluate variable at time of function declaration in shell

I'm setting up my shell environments and I want to be able to use some of the same functions/aliases in zsh as in bash. One of these functions opens either .bashrc or .zshrc in an editor (whichever file is relevant), waits for the editor to close, then reloads the rc file.
# a very simplified version of this function
editrc() {
local rcfile=".$(basename $SHELL)rc"
code -w ~/$rcfile
. ~/$rcfile
}
I use the value of rcfile in a few other functions, so I've pulled it out of the function declaration.
_rc=".$(basename $SHELL)rc"
editrc() {
code -w ~/$_rc
. ~/$_rc
}
# ... other functions that use it ...
unset _rc
However, because I'm a neat freak, I want to unset _rc at the end of my script, but I still want my functions to run correctly. Is there a clever way to evaluate $_rc at the time the function is declared?
I know I could use eval and place everything except $_rc instances within single quotes, but that seems like a pain, since the full version of my function uses both single-quotes and double-quotes.
_rc=".$(basename $SHELL)rc"
eval 'editrc() {
echo Here'"'"'s a thing that uses single quotes. As you can see it'"'"'s a pain.
code -w ~/'$_rc'
. ~/'$_rc'
}'
# ... other functions using `_rc`
unset _rc
I'm guessing I could declare my functions, then do some magic with eval "$(declare -f editrc | awk)". It very well be more pain than it's worth, but I'm always interested in learning new things.
Note: I'd love to generalize this into a utility function that does this.
_myvar=foo
anothervar=bar
myfunc() {
echo $_myvar $anothervar
}
# redeclares myfunc with `$_myvar` expanded, but leaves `$anothervar` as-is
expandfunctionvars myfunc '$_myvar'
Is there a clever way to evaluate $_rc at the time the function is declared?
_rc=".$(basename "$SHELL")rc"
# while you could eval here, source lets you work with a stream
source <(
cat <<EOF
editrc() {
local _rc
# first safely trasfer context
$(declare -p _rc)
EOF
# use quoted here string to do anything inside without caring.
cat <<'EOF'
# do anything else
echo "Here's a thing that uses single quotes. As you can see it's not a pain, just choose proper quoting."
code -w "~/$_rc"
. "~/$_rc"
}
EOF
)
unset _rc
Generally first use declare -p to transfer variables as strings to be evaluated. Then after you "import" variables, use a quoted here document to do anything as in a normal script.
References to read:
<<EOF is a here document. Note the difference in parsing when the here delimiter is quoted vs unquoted.
<(..) is a process substitution
The source command reads a pipe created by process substitution. Inside the process subtitution I output the function to be sourced. With the first here document I output the function name definition, with a local of the variable so that it doesn't pollute global namespace. Then with declare -p I output the variable definition as a properly quoted string later to be sourced by source. Then with a quoted here document I output the rest of the function, so that I do not need to care about quoting.
The code is bash specific, I know nothing about zsh and don't use it.
You could do it with eval too:
eval '
editrc() {
local _rc
# first safely trasfer context
'"$(declare -p _rc)"'
# use quoted here string to do anything inside without caring.
# do anything else
echo "Here'\''s a thing that uses single quotes. As you can see it'\''s not a pain, just choose proper quoting."
code -w "~/$_rc"
. "~/$_rc"
}'
But for me using a quoted here document delimiter allows for easier writing.
While KamilCuck was working on their answer, I devised a function that will take in any function name and a set of variable names, expand just those variables, and redeclare the function.
expandFnVars() {
if [[ $# -lt 2 ]]; then
>&2 echo 'expandFnVars requires at least two arguments: the function name and the variable(s) to be expanded'
return 1
fi
local fn="$1"
shift
local vars=("$#")
if [[ -z "$(declare -F $fn 2> /dev/null)" ]]; then
>&2 echo $fn is not a function.
return 1
fi
foundAllVars=true
for v in $vars; do
if [[ -z "$(declare -p $v 2> /dev/null)" ]]; then
>&2 echo $v is not a declared value.
foundAllVars=false
fi
done
[[ $foundAllVars != true ]] && return 1
fn="$(declare -f $fn)"
for v in $vars; do
local val="$(eval 'echo $'$v)" # get the value of the varable represented by $v
val="${val//\"/\\\"}" # escape any double-quotes
val="${val//\\/\\\\\\}" # escape any backslashes
fn="$(echo "$fn" | sed -r 's/"?\$'$v'"?/"'"$val"'"/g')" # replace instances of "$$v" and $$v with $val
done
eval "$fn"
}
Usage:
foo="foo bar"
bar='$foo'
baz=baz
fn() {
echo $bar $baz
}
expandFnVars fn bar
declare -f fn
# prints:
# fn ()
# {
# echo "$foo" $baz
# }
expandFnVars fn foo
declare -f fn
# prints:
# fn ()
# {
# echo "foo bar" $baz
# }
Looking at it now, I see one flaw. Suppose $bar in the original function was in single-quotes. We probably would not want its value to be replaced. This could be fixed by some clever regex lookbehinds to count the number of unescaped 's, but I'm happy with it as-is.

Using an environment variable to pass arguments to a command

I'm trying to write a bash script that takes an environment variable and passes it along to a command.
So if I had something like:
export OUT="-a=arg1 -b=\"arg2.0 arg2.1\""
I want in my bash script to do something like:
<command> -a=arg1 '-b=arg2.0 arg2.1'
I have one approach that seems to do this, but it involves using eval:
eval <command> ${OUT}
If I include set -x right about the command, I will see:
+ eval <command> a=arg1 'b="arg2.0' 'arg2.1"'
++ <command> -a=arg1 '-b=arg2.0 arg.1'
However, I've poked around the dangers of using eval and since this will be taking the arguments from user input, it's less than ideal.
Since this is bash, I've also considered using arrays to store my arguments and simply put: <command> "$ARRAY[#]" to do what I want. I've been trying to use IFS, but I'm not sure what I should be splitting on.
If you're not completely inflexible about the format of $OUT, one possibility would be to repeat the option= string to allow for concatenation. Then you'd write:
export OUT="a=arg1 b=arg2.0 b=arg2.1"
If that is acceptable, the following script will work
#!/bin/bash
# Parse $OUT into an associative array.
# Instead of using $OUT, it would be cleaner to use "$#".
declare -A args
for arg in $OUT; do
if [[ "$arg" =~ ^([[:alnum:]]+)=(.*)$ ]]; then
key=${BASH_REMATCH[1]}
val=${BASH_REMATCH[2]}
if [[ -z ${args[$key]} ]]; then
args[$key]=-$key="$val"
else
args[$key]+=" $val"
fi
fi
done
# Test, approximately as specified
command() { :; }
set -x
command "${args[#]}"
set +x
I can't say I like it much, but it's the closest I've been able to come.
Here's a sample run:
$ export OUT="a=foo b=bar b=glitch s9= s9=* "
./command-runner
+ command -a=foo '-b=bar glitch' '-s9= *'
+ :
+ set +x
If you import a bash function (for example, in your bash startup file), you can make much better use of arrays. Here's one approach:
# This goes into your bash startup file:
declare -a SAVED_ARGS
save_args() {
SAVED_ARGS=("$#")
}
do_script() {
/path/to/script.sh "${SAVED_ARGS[#]}" "$#"
}
For expository purposes, script.sh:
#!/bin/bash
command() { :; }
set -x
command "${#/#/-}"
set +x
Example:
$ save_args x=3 y="a few words from our sponsor"
$ do_script a=3 b="arg2.0 arg2.1"
+ command -x=3 '-y=a few words from our sponsor' -a=3 '-b=arg2.0 arg2.1'
+ :
+ set +x
$ do_script a=42
+ command -x=3 '-y=a few words from our sponsor' -a=42
+ :
+ set +x
In case it's not obvious:
command() { :; }
defines a bash function called command which does almost nothing (except invoke the builtin : which does nothing), and
"${#/#/-}"
expands to the positional parameters, inserting a dash at the beginning of each one use a find-and-replace substitution. The pattern # is actually an empty pattern which only matches at the beginning of the string.
For the simplified problem described in the answer above; i.e., turning the following environment variable into three arguments inside a bash script:
export OPTS="a=arg1 b=arg2.0 b=arg2.1"
Just do the following:
#!/bin/bash
opts=( $OPTS )
my-command "${opts[#]}"
# Use this for debugging:
echo "number of opts = ${#opts[#]}; opts are: ${opts[#]}"
set your env varivbale as:
export abc=123
while execution of any script where abc need to pass as an argument pass as below:
./testing.sh "$abc"

Set a parent shell's variable from a subshell

How do I set a variable in the parent shell, from a subshell?
a=3
(a=4)
echo $a
The whole point of a subshell is that it doesn't affect the calling session. In bash a subshell is a child process, other shells differ but even then a variable setting in a subshell does not affect the caller. By definition.
Do you need a subshell? If you just need a group then use braces:
a=3
{ a=4;}
echo $a
gives 4 (be careful of the spaces in that one). Alternatively, write the variable value to stdout and capture it in the caller:
a=3
a=$(a=4;echo $a)
echo $a
avoid using back-ticks ``, they are deprecated and can be difficult to read.
There is the gdb-bash-variable hack:
gdb --batch-silent -ex "attach $$" -ex 'set bind_variable("a", "4", 0)';
although that always sets a variable in the global scope, not just the parent scope
You don't. The subshell doesn't have access to its parent's environment. (At least within the abstraction that Bash provides. You could potentially try to use gdb, or smash the stack, or whatnot, to gain such access clandestinely. I wouldn't recommend that, though.)
One alternative is for the subshell to write assignment statements to a temporary file for its parent to read:
a=3
(echo 'a=4' > tmp)
. tmp
rm tmp
echo "$a"
If the problem is related to a while loop, one way to fix this is by using Process Substitution:
var=0
while read i;
do
# perform computations on $i
((var++))
done < <(find . -type f -name "*.bin" -maxdepth 1)
as shown here: https://stackoverflow.com/a/13727116/2547445
To change variables in a script called from a parent script, you can call the script preceded with a "."
(EDIT - for explanation)
In most shells "." is an alias for "source". the source command just inserts the text of another file at that position in the executing script. In the context of this question this answer avoids a sub-shell
a=3
echo $a
. ./calledScript.sh
echo $a
in calledScript.sh
a=4
Expected output
3
4
By reading the answer from #ruakh (thank you) with a temporary file approach and the comments asking for a file descriptors solution, I got the following idea:
a=3
. <(echo a=4; echo b=5)
echo $a
echo $b
It allows returning different variables at once (which could be an issue in the subshell variant of the accepted answer).
No iteration is needed,
No temporary file to take care of.
Close to the syntax proposed by the OP.
Result:
4
5
With xtrace enabled is visible that we are sourcing from the file descriptor created for the output of the subshell:
+ a=3
+ . /dev/fd/63 # <-- the file descriptor ;)
++ echo a=4
++ echo b=5
++ a=4
++ b=5
+ echo 4
4
+ echo 5
5
You can output the value in the subshell and assign the subshell output to a variable in the caller script:
# subshell.sh
echo Value
# caller
myvar=$(subshell.sh)
If the subshell has more to output you can separate the variable value and other messages by redirecting them into different output streams:
# subshell.sh
echo "Writing value" 1>&2
echo Value
# caller
myvar=$(subshell.sh 2>/dev/null) # or to somewhere else
echo $myvar
Alternatively, you can output variable assignments in the subshell, evaluate them in the caller script and avoid using files to exchange information:
# subshell.sh
echo "a=4"
# caller
# export $(subshell.sh) would be more secure, since export accepts name=value only.
eval $(subshell.sh)
echo $a
The last way I can think of is to use exit codes but this covers the integer values exchange only (and in a limited range) and breaks the convention for interpreting exit codes (0 for success non-0 for everything else).
Instead of accessing the variable from the parent shell, change the order of the commands and use the process substitution:
a=3
echo 5 | (read a)
echo $a
prints 3
a=3
read a < <(echo 5)
echo $a
prints 5
Another example:
let i=0
seq $RANDOM | while read r
do
let i=r
done
echo $i
vs
let i=0
while read r
do
let i=r
done < <(seq $RANDOM)
echo $i
Alternatively, when job control is inactive (e.g. in scripts) you can use the lastpipe shell option to achieve the same result without changing the order of the commands:
#!/bin/bash
shopt -s lastpipe
let i=0
seq $RANDOM | while read r
do
let i=r
done
echo $i
Unless you can apply all io to pipes and use file handles, basic variable updating is impossible within $(command) and any other sub-process.
Regular files, however, are bash's global variables for normal sequential processing. Note: Due to race conditions, this simple approach is not good for parallel processing.
Create an set/get/default function like this:
globalVariable() { # NEW-VALUE
# set/get/default globalVariable
if [ 0 = "$#" ]; then
# new value not given -- echo the value
[ -e "$aRam/globalVariable" ] \
&& cat "$aRam/globalVariable" \
|| printf "default-value-here"
else
# new value given -- set the value
printf "%s" "$1" > "$aRam/globalVariable"
fi
}
"$aRam" is the directory where values are stored. I like it to be a ram disk for speed and volatility:
aRam="$(mktemp -td $(basename "$0").XXX)" # temporary directory
mount -t tmpfs ramdisk "$aRam" # mount the ram disk there
trap "umount "$aRam" && rm -rf "$aRam"" EXIT # auto-eject
To read the value:
v="$(globalVariable)" # or part of any command
To set the value:
globalVariable newValue # newValue will be written to file
To unset the value:
rm -f "$aRam/globalVariable"
The only real reason for the access function is to apply a default value because cat will error given a non-existent file. It is also useful to apply other get/set logic. Otherwise, it would not be needed at all.
An ugly read method avoiding cat's non-existent file error:
v="$(cat "$aRam/globalVariable 2>/dev/null")"
A cool feature of this mess is that you can open another terminal and examine the contents of the files while the program is running.
While it's harder to get multiple variables out of a subshell, you can set multiple variables inside a function without using globals.
You can pass the name of a variable into a function that uses local -n to turn it into a special variable called a nameref:
myfunc() {
local -n OUT=$1
local -n SIDEEFFECT=$2
OUT='foo'
SIDEEFFECT='bar'
}
myfunc A B
echo $A
> foo
echo $B
> bar
This is the technique I ended up using instead of getting subshell FOO=$(myfunc) working setting multiple variables.
A very simple and practical method that allows multiple variables is as follows, eventually may add parameters to the call:
function ComplexReturn(){
# do your processing...
a=123
b=456
echo -n "AAA=${a}; BBB=${b};"
}
# ... this can be internal function or any subshell command
eval $(ComplexReturn)
echo $AAA $BBB

Quoting parameters with spaces for later execution

I have this (test) script:
#!/bin/bash
my_cmd_bad_ ( ) {
cmd="$#"
$cmd
}
my_cmd_good_ ( ) {
"$#"
}
my_cmd_bad_ ls -l "file with space"
my_cmd_good_ ls -l "file with space"
The output is (the file does not exist, which is not the point of this question):
ยป ~/test.sh
ls: cannot access file: No such file or directory
ls: cannot access with: No such file or directory
ls: cannot access space: No such file or directory
ls: cannot access file with space: No such file or directory
I am surprised that the first version does not work as expected: the parameter is not quoted, and instead of processing one file, it processes three. Why?
How can I save the command that I want to execute, properly quoted? I need to execute it later, where I do not have "$#" anymore.
A simple rework of this test script would be appreciated.
See similar question: How to pass command line parameters with quotes stored in single variable?
Use those utility functions ho save a command to a string for later execution:
bash_escape() {
# backtick indirection strictly necessary here: we use it to strip the
# trailing newline from sed's output, which Solaris/BSD sed *always* output
# (unlike GNU sed, which outputs "test": printf %s test | sed -e s/dummy//)
out=`echo "$1" | sed -e s/\\'/\\''\\\\'\\'\\'/g`
printf \'%s\' "$out"
}
append_bash_escape() {
printf "%s " "$1"
bash_escape "$2"
}
your_cmd_fixed_ ( ) {
cmd="$#"
while [ $# -gt 0 ] ; do
cmd=`append_bash_escape "$cmd" "$1"` ; shift
done
$cmd
}
You can quote any single parameter and evaluate it later:
my_cmd_bad_ ( ) {
j=0
for i in "$#"; do
cmd["$j"]=\"$"$i"\"
j=$(( $j + 1 ))
done;
eval ${cmd[*]}
}
You are combining three space-delimited strings "ls", "-l", and "file with space" into a single space-delimited string cmd. There's no way to know which spaces were originally quoted (in "file with space") and which spaces were introduced during the assignment to cmd.
Typically, it is not a good idea to try to build up command lines into a single string. Use functions, or isolate the actual command and leave the arguments in $#.
Rewrite the command like this:
my_cmd_bad_ () {
cmd=$1; shift
$cmd "$#"
}
See http://mywiki.wooledge.org/BashFAQ/050
Note that your second version is greatly preferred most of the time. The only exceptions are if you need to do something special. For example, you can't bundle an assignment or redirect or compound command into a parameter list.
The correct way to handle the quoting issue requires non-standard features. Semi-realistic example involving a template:
function myWrapper {
typeset x IFS=$' \t\n'
{ eval "$(</dev/fd/0)"; } <<-EOF
for x in $(printf '%q ' "$#"); do
echo "\$x"
done
EOF
}
myWrapper 'foo bar' $'baz\nbork'
Make sure you understand exactly what's going on here and that you really have a good reason for doing this. It requires ensuring side-effects can't affect the arguments. This specific example doesn't demonstrate a very good use case because everything is hard-coded so you're able to correctly escape things in advance and expand the arguments quoted if you wanted.

How to access command line arguments of the caller inside a function?

I'm attempting to write a function in bash that will access the scripts command line arguments, but they are replaced with the positional arguments to the function. Is there any way for the function to access the command line arguments if they aren't passed in explicitly?
# Demo function
function stuff {
echo $0 $*
}
# Echo's the name of the script, but no command line arguments
stuff
# Echo's everything I want, but trying to avoid
stuff $*
If you want to have your arguments C style (array of arguments + number of arguments) you can use $# and $#.
$# gives you the number of arguments.
$# gives you all arguments. You can turn this into an array by args=("$#").
So for example:
args=("$#")
echo $# arguments passed
echo ${args[0]} ${args[1]} ${args[2]}
Note that here ${args[0]} actually is the 1st argument and not the name of your script.
My reading of the Bash Reference Manual says this stuff is captured in BASH_ARGV,
although it talks about "the stack" a lot.
#!/bin/bash
shopt -s extdebug
function argv {
for a in ${BASH_ARGV[*]} ; do
echo -n "$a "
done
echo
}
function f {
echo f $1 $2 $3
echo -n f ; argv
}
function g {
echo g $1 $2 $3
echo -n g; argv
f
}
f boo bar baz
g goo gar gaz
Save in f.sh
$ ./f.sh arg0 arg1 arg2
f boo bar baz
fbaz bar boo arg2 arg1 arg0
g goo gar gaz
ggaz gar goo arg2 arg1 arg0
f
fgaz gar goo arg2 arg1 arg0
#!/usr/bin/env bash
echo name of script is $0
echo first argument is $1
echo second argument is $2
echo seventeenth argument is $17
echo number of arguments is $#
Edit: please see my comment on question
Ravi's comment is essentially the answer. Functions take their own arguments. If you want them to be the same as the command-line arguments, you must pass them in. Otherwise, you're clearly calling a function without arguments.
That said, you could if you like store the command-line arguments in a global array to use within other functions:
my_function() {
echo "stored arguments:"
for arg in "${commandline_args[#]}"; do
echo " $arg"
done
}
commandline_args=("$#")
my_function
You have to access the command-line arguments through the commandline_args variable, not $#, $1, $2, etc., but they're available. I'm unaware of any way to assign directly to the argument array, but if someone knows one, please enlighten me!
Also, note the way I've used and quoted $# - this is how you ensure special characters (whitespace) don't get mucked up.
# Save the script arguments
SCRIPT_NAME=$0
ARG_1=$1
ARGS_ALL=$*
function stuff {
# use script args via the variables you saved
# or the function args via $
echo $0 $*
}
# Call the function with arguments
stuff 1 2 3 4
One can do it like this as well
#!/bin/bash
# script_name function_test.sh
function argument(){
for i in $#;do
echo $i
done;
}
argument $#
Now call your script like
./function_test.sh argument1 argument2
This is #mcarifio response with several comments incorporated:
#!/bin/bash
shopt -s extdebug
function stuff() {
local argIndex="${#BASH_ARGV[#]}"
while [[ argIndex -gt 0 ]] ; do
argIndex=$((argIndex - 1))
echo -n "${BASH_ARGV[$argIndex]} "
done
echo
}
stuff
I want to highlight:
The shopt -s extdebug is important. Without this the BASH_ARGV array will be empty unless you use it in top level part of the script (it means outside of the stuff function). Details here: Why does the variable BASH_ARGV have a different value in a function, depending on whether it is used before calling the function
BASH_ARGV is a stack so arguments are stored there in backward order. That's the reason why I decrement the index inside loop so we get arguments in the right order.
Double quotes around the ${BASH_ARGV[#]} and the # as an index instead of * are needed so arguments with spaces are handled properly. Details here: bash arrays - what is difference between ${#array_name[*]} and ${#array_name[#]}
You can use the shift keyword (operator?) to iterate through them.
Example:
#!/bin/bash
function print()
{
while [ $# -gt 0 ]
do
echo "$1"
shift 1
done
}
print "$#"
I do it like this:
#! /bin/bash
ORIGARGS="$#"
function init(){
ORIGOPT= "- $ORIGARGS -" # tacs are for sed -E
echo "$ORIGOPT"
}
The simplest and likely the best way to get arguments passed from the command line to a particular function is to include the arguments directly in the function call.
# first you define your function
function func_ImportantPrints() {
printf '%s\n' "$1"
printf '%s\n' "$2"
printf '%s\n' "$3"
}
# then when you make your function call you do this:
func_ImportantPrints "$#"
This is useful no matter if you are sending the arguments to main or some function like func_parseArguments (a function containing a case statement as seen in previous examples) or any function in the script.

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