After reading ilkkachu's answer to this question I learned on the existence of the declare (with argument -n) shell built in. help declare brings: > Set variable values and attributes. > > Declare variables and give them attributes. If no NAMEs are given, > display the attributes and values of all variables. > -n ... make NAME a reference to the variable named by its value I ask for a general explanation with an example regarding declare because I don't understand the man. I know what is a variable and expanding it but I still miss the man on declare (variable attribute?). Maybe you'd like to explain this based on the code by ilkkachu in the answer: #!/bin/bash function read_and_verify { read -p "Please enter value for '$1': " tmp1 read -p "Please repeat the value to verify: " tmp2 if [ "$tmp1" != "$tmp2" ]; then echo "Values unmatched. Please try again."; return 2 else declare -n ref="$1" ref=$tmp1 fi }

**How do I get the FZF Preview Window to Display Functions from my Current Bash Environment?** I want to list my custom bash functions using FZF, and view the code of a selected function in the FZF Preview Window. However, it does not appear that the bash enviroment used by FZF to execute my command can see the functions in my terminal bash environment. For example:

$ declare -F | fzf --preview="type {3}"

/bin/bash: line 1: type: g: not found

However, the following works:

$ declare -F

declare -f fcd
declare -f fz
declare -f g

$ type g
g is a function
g ()
{
    search="";
    for term in $@;
    do
        search="$search%20$term";
    done;
    nohup google-chrome --app-url "http://www.google.com/search?q=$search " > /dev/null 2>&1 &
}

declare -F | fzf --preview="echo {3}"

g # my function g()

One reason I suspect that the FZF Preview Window environment may not be able to see my terminal environment is because they have different process ID's.

$ echo $BASHPID

1129439

$ declare -F | fzf --preview="echo $BASHPID"

1208203

**How do I get the FZF Preview Window to Display Functions from my Current Bash Environment?**

I've read what's listed in Bibliography regarding unset, declare, local and "Shell Functions". My version of Bash is

GNU bash, version 5.1.16(1)-release (x86_64-pc-linux-gnu)

var='outer'
declare -p var
unset var
declare -p var
function foo {
  echo \""I'm in"\"
  local var='inner'
  declare -p var
  unset var
  declare -p var
}
foo

(the strange quoting around I'm in is there just to preserve syntax highlithing in the following block quote.) This prints

declare -- var="outer"
bash: declare: var: not found
"I'm in"
declare -- var="inner"
declare -- var

There is a difference in unsetting a global variable and unsetting a local variable inside a function. In the former case, the variable is removed. In the latter, the variable stays declared (but unset). Is there a way to completely remove a local variable inside a function (before the function returns)? That is that the output would be

declare -- var="outer"
bash: declare: var: not found
"I'm in"
declare -- var="inner"
bash: declare: var: not found

That would be useful to test if a variable is non existing inside a function, like in

function foo {
  local var
  while
    declare -p var
  do
    echo "$var"
    ((var++))
    [[ "$var" -eq 4 ]] \
    && unset var
  done
}

This code loops indefinitely, printing

declare -- var="1"
1
declare -- var="2"
2
declare -- var="3"
3
declare -- var

declare -- var="1"
1
declare -- var="2"
2
declare -- var="3"
3
declare -- var
[omissis]

Is there anything wrong in checking if a variable exists using declare -p? The only mention I found in the Bash reference manual about the difference between unsetting a global variable and a local one is > If a variable at the current local scope is unset, it will remain so (appearing as unset) until it is reset in that scope or until the function returns. (ref. Bash reference manual - sec. "Shell Functions") where the word _appearing_ is the only clue. **Bibliography** + [Bash reference manual] sec. "4 Shell Builtin Commands" + [What does unset do?]

Our RHEL8 servers don't have nc available, but at some time in the past someone using this shared account defined an nc function:

-shellsession
$ declare -f nc | head -1
nc ()
$

I am trying to determine where this function is defined:

-shellsession
$ (shopt -s extdebug; declare -F nc)
nc 0 environment
$

I was expecting a line number and a filename there, and as you can see I got 0 environment instead. Can somebody tell me what this means, please?

Main question: how would one get the delta for declare -F, between that in the current shell, and that as if the shell just started (first two commands below). $(declare -F) does not solve the problem because the subshell is a copy of the shell process . Subsidiary: why does the third command below output nothing? $ exec env -i bash $ declare -F declare -f ShowInstallerIsoInfo declare -f __expand_tilde_by_ref declare -f __get_cword_at_cursor_by_ref declare -f __load_completion declare -f __ltrim_colon_completions declare -f __parse_options declare -f __reassemble_comp_words_by_ref declare -f _allowed_groups declare -f _allowed_users declare -f _available_interfaces declare -f _bashcomp_try_faketty declare -f _bq_completer declare -f _cd declare -f _cd_devices declare -f _command declare -f _command_offset declare -f _complete_as_root declare -f _completer declare -f _completion_loader declare -f _configured_interfaces declare -f _count_args declare -f _dvd_devices declare -f _expand declare -f _filedir declare -f _filedir_xspec declare -f _fstypes declare -f _get_comp_words_by_ref declare -f _get_cword declare -f _get_first_arg declare -f _get_pword declare -f _gids declare -f _have declare -f _included_ssh_config_files declare -f _init_completion declare -f _installed_modules declare -f _ip_addresses declare -f _kernel_versions declare -f _known_hosts declare -f _known_hosts_real declare -f _longopt declare -f _mac_addresses declare -f _minimal declare -f _modules declare -f _ncpus declare -f _open_files_for_editing declare -f _parse_help declare -f _parse_usage declare -f _pci_ids declare -f _pgids declare -f _pids declare -f _pnames declare -f _python_argcomplete declare -f _quote_readline_by_ref declare -f _realcommand declare -f _rl_enabled declare -f _root_command declare -f _service declare -f _services declare -f _shells declare -f _signals declare -f _split_longopt declare -f _sysvdirs declare -f _terms declare -f _tilde declare -f _uids declare -f _upvar declare -f _upvars declare -f _usb_ids declare -f _user_at_host declare -f _usergroup declare -f _userland declare -f _variables declare -f _xfunc declare -f _xinetd_services declare -f dequote declare -f quote declare -f quote_readline $ "$SHELL" -c 'declare -F' Other: $ uname -a Linux elitebook 6.7.3-arch1-2 #1 SMP PREEMPT_DYNAMIC Fri, 02 Feb 2024 17:03:55 +0000 x86_64 GNU/Linux $ bash --version GNU bash, version 5.2.26(1)-release (x86_64-pc-linux-gnu) $ echo $SHELL /bin/bash ## Update: > The clean state is an empty list of functions, as seen in the output of bash -c 'declare -F' Based on the first command I expected the second to output two functions rather than zero. $ grep -F -f I expected the second to output two functions rather than zero. Here might be the reason for sourcing not taking effect: $ cat ~/.bashrc # # ~/.bashrc # # If not running interactively, don't do anything [[ $- != *i* ]] && return Workaround does output two functions. $ bash -c 'source /dev/null

This is specifically about bash's declare - the general case is pretty exhaustively dealt with in this answer (which mentions "the typeset/declare/export -p output of ksh93, mksh, zsh" but _not_ that of bash). Given a local/exported/array/assocative-array (but maybe not nameref) variable foo, is the output of declare -p foo in bash guaranteed to be reusable by bash? The official documentation doesn't mention anything like that: > The -p option will display the attributes and values of each name. > When -p is used with name arguments, additional options, other > than -f and -F, are ignored. And I looked through the CHANGES , and saw this about _functions_:

This document details the changes between this version, bash-2.05-beta1,
and the previous version, bash-2.05-alpha1.
...
b.  When `set' is called without options, it prints function definitions in a
    way that allows them to be reused as input.  This affects `declare' and
    `declare -p' as well.

And for a couple of other commands, -p is meant to produce reusable output:

s.  The shopt' -p' option now causes output to be displayed in a reusable
    format.
...
u.  umask' now has a -p' option to print output in a reusable format.

And Chet Ramey's Bash FAQ has:

Bash-2.0 contained extensive changes and new features from bash-1.14.7.
Here's a short list:
...
most builtins use -p option to display output in a reusable form
	(for consistency)

But nothing I can find about declare -p for variables.

I'm attempting to write a function that writes arrays with a name that's passed in. Given the following bash function: function writeToArray { local name="$1" echo "$name" declare -a "$name" ${name}="does this work?" } Running like this: writeToArray $("test") I get two errors: bash: declare: `': not a valid identifier =does this work?: command not found I am expecting to be able to do this: writeToArray $("test") for item in "${test[@]}"; do echo "item" echo "$item" done This should print: item does this work? How could I properly configure this to write the array (named test in the example, such that this array named test is readable outside the function)?

With bash >5, I'm trying to assign a different value to variables depending on the architecture specified in a variable. I use a function to do so. This works perfectly: # arguments: variable name to assign, value for mac arch, value for pi arch create_variable_for_arch() { if [ "$_run_for_arch" = "mac" ]; then eval $1=\$2 else eval $1=\$3 fi } However, this breaks my script for some reason: create_variable_for_arch() { if [ "$_run_for_arch" = "mac" ]; then declare "$1"="$2" else declare "$1"="$3" fi } Here is a snippet to demonstrate how I use create_variable_from_arch()

declare _moonlight_opt_audio
declare _arch_specific_stream_command

#
while getopts "b:fahdr:s" options; do
  case $options in
    a)
      create_variable_for_arch "_moonlight_opt_audio" \
        "--audio-on-host" "-localaudio"
      ;;
  esac
done

create_variable_for_arch "_moonlight_opt_fps" "--fps 60" "-fps 60"

start_streaming() {
  _arch_specific_options="$_moonlight_opt_resolution $_moonlight_opt_fps $_moonlight_opt_audio $_moonlight_opt_display_type $_moonlight_opt_bitrate"
  create_variable_for_arch "_arch_specific_stream_command" "$_arch_specific_options stream $_target_computer_ip $_moonlight_opt_app_name" "stream $_arch_specific_options -app $_moonlight_opt_app_name $_target_computer_ip"

  moonlight $_arch_specific_stream_command
}

The trace looks like this with eval()

+ start_streaming
+ _arch_specific_options='--resolution 1920x1080 --fps 60   --bitrate 5000'
+ create_variable_for_arch _arch_specific_stream_command '--resolution 1920x1080 --fps 60   --bitrate 5000 stream 192.168.1.30 StreamMouse' 'stream --resolution 1920x1080 --fps 60   --bitrate 5000 -app StreamMouse 192.168.1.30'
+ '[' mac = mac ']'
+ eval '_arch_specific_stream_command=$2'
++ _arch_specific_stream_command='--resolution 1920x1080 --fps 60   --bitrate 5000 stream 192.168.1.30 StreamMouse'
+ moonlight --resolution 1920x1080 --fps 60 --bitrate 5000 stream 192.168.1.30 StreamMouse
moonlight --resolution 1920x1080 --fps 60 --bitrate 5000 stream 192.168.1.30 StreamMouse

But with declare it looks like this:

+ start_streaming
    + _arch_specific_options=
    + create_variable_for_arch _arch_specific_stream_command ' stream 192.168.1.30 ' 'stream  -app  192.168.1.30'
    + '[' mac = mac ']'
    + declare '_arch_specific_stream_command= stream 192.168.1.30 '
    + echo moonlight
    moonlight

$_arch_specific_options ends up with no value. What is going on? I've tried a few different ways of quoting or not quoting variables, but I don't really understand what's doing what in terms of quotations.

I downloaded the CLI Client habash for the habit/routine gameification project habatica.com . In the fandom wiki for habash it is written that, I need to set environment variables. Additonally I want to make the program habash an ordinary CLI-program, so that I dont have to invoke it via the fullpath. I describe now, what I have been doing. I did the following. - After declare -x HABITICA_UUID=[myUserID] I did not find any entries in ~/.bashrc. Any one knows why? - Therfore I added HABITICA_UUID and HABITICA_TOKEN at the top of ~/.bashrc (and made a comment for myself) - chmod 600 ~/.bashrc because UUID and TOKEN are regarded as PWs. - sudo mv ./habash /opt - sudo ln -s /opt/habash/habash /usr/local/bin Is this the best way how to do it? (storing in /opt and linkint to /usr/local/bin ; variables in .bashrc and chmoding it with 600)

I learned from the following sources: - **curl -O**: [Download a file with curl on Linux / Unix command line](https://www.cyberciti.biz/faq/download-a-file-with-curl-on-linux-unix-command-line/) - **jq:** [How to urlencode data for curl command?](https://stackoverflow.com/questions/296536/how-to-urlencode-data-for-curl-command/34407620#34407620) - **Multiple files and curl -J:** https://unix.stackexchange.com/questions/91798/download-pdf-files-from-using-curl - **Condition for loop:** https://unix.stackexchange.com/questions/377446/shell-how-to-use-2-variables-with-for-condition and https://unix.stackexchange.com/questions/291049/unable-to-download-data-using-curl-for-loop Description of the script: 1. Variables, required by GitLab's [Repository files API](https://docs.gitlab.com/ee/api/repository_files.html) :

branch="master"
      repo="my-dotfiles"
      private_token="XXY_wwwwwx-qQQQRRSSS"
      username="gusbemacbe"

2. I used a declaration for multiple files:

declare -a context_dirs=(
        "home/.config/Code - Insiders/Preferences"
        "home/.config/Code - Insiders/languagepacks.json"
        "home/.config/Code - Insiders/rapid_render.json"
        "home/.config/Code - Insiders/storage.json"
      )

3. I used the condition for loop with jq to convert all files from the declaration context_dirs to encoded URLs:

for urlencode in "${context_dirs[@]}"; do
        paths=$(jq -nr --arg v "$urlencode" '$v|@uri')
      done

4. I used the condition for loop to download with curl multiple files taken from paths converted by jq. It is important that I used -0 and -J to output the file name, and -H for "PRIVATE-TOKEN: $private_token":

for file in "${paths[@]}"; do 
          curl -sLOJH "PRIVATE-TOKEN: $private_token" "https://gitlab.com/api/v4/projects/$username%2F$repo/repository/files/$file/raw?ref=$branch "
      done

Complete source code:

branch="master"
id="1911000X"
repo="my-dotfiles"
private_token="XXY_wwwwwx-qQQQRRSSS"
username="gusbemacbe"

declare -a context_dirs=(
  "home/.config/Code - Insiders/Preferences"
  "home/.config/Code - Insiders/languagepacks.json"
  "home/.config/Code - Insiders/rapid_render.json"
  "home/.config/Code - Insiders/storage.json"
)

for urlencode in "${context_dirs[@]}"; do
  paths=$(jq -nr --arg v "$urlencode" '$v|@uri')
done

for file in "${paths[@]}"; do 
    curl -sLOJH "PRIVATE-TOKEN: $private_token" "https://gitlab.com/api/v4/projects/$username%2F$repo/repository/files/$file/raw?ref=$branch "
done

But the two conditions for loop output only an encoded path and downloaded only a file.

I stumbled by accident on the following bash behaviour, which is for me kind of unexpected. # The following works $ declare bar=Hello # Line 1 $ declare -p bar # Line 2 declare -- bar="Hello" $ foo=bar # Line 3 $ declare ${foo}=Bye # Line 4 $ declare -p bar # Line 5 declare -- bar="Bye" # The following fails, though $ declare -a array=( A B C ) # Line 6 $ declare -p array # Line 7 declare -a array=(="A" ="B" ="C") $ foo=array # Line 8 $ declare -a ${foo}=(="A" ="XXX" ="C") # Line 9 bash: syntax error near unexpected token (' # Quoting the assignment fixes the problem $ declare -a "${foo}=(A YYY C)" # Line 10 $ declare -p array # Line 11 declare -a array=(="A" ="YYY" ="C") Since shell expansion 1. Brace expansion 2. * Tilde expansion * Parameter and variable expansion * Arithmetic expansion * Process substitution * Command substitution 3. Word splitting 4. Filename expansion is performed on the command line after it has been split into tokens (followed by quote removal) but before the final command is executed, I would not have expected line 9 to fail. **Which is the rationale behind it, that makes bash not accept line 9?** Or, said differently, **what am I missing in the way line 9 is processed by bash that makes it fail but makes line 10 succeed?** In any case, quoting is not always going to straightforwardly work and it would require extra attention in case the array elements are strings with e.g. spaces.

I'm new to bash. I'm trying to write a script that will read data from a text find and declare some variables. In the example below we read from a tab delimited file "ab.txt" that looks like this: a->AA b->BB Where -> denotes a tab. I'm reading this data with this code:

#!/bin/bash
while read line
do
	tmp=(${line///})
	fieldName=${tmp}
	
	case $fieldName in
	"a")
		a=${tmp}
		;;
	"b")
		b=${tmp}
		;;
	esac
		
done < "ab.txt"


echo "a:"
echo $a	

echo "b:"
echo $b

echo "concat a,b"
echo $a$b

echo "concat b,a"
echo $b$a

This gives me "a" and "b" nicely, but will not concatenate a and b! The output is this:

a:
AA
b:
BB
concat a,b
BB
concat b,a
AA

What am I doing wrong?

I have this very simple script: #!/bin/bash read local _test echo "_test: $_test" This is the output. $ ./jltest.sh sdfsdfs _test: I want the variable _test to be local only. Is this possible?

I aim to understand the general concept of "variable attributes" hoping it will help me understand what is declare in Bash . What is a variable attribute? Why would someone want to give an attribute to a variable? Why isn't just creating an variables and expanding them in execution be "enough" when working with variables?

Does declare -a A create an empty array A in bash, or does it just set an attribute in case A is assigned to later? Consider this code: set -u declare -a A echo ${#A[*]} echo ${A[*]} A=() echo ${#A[*]} echo ${A[*]} A=(1 2) echo ${#A[*]} echo ${A[*]} What should be the expected output? In Bash 4.3.48(1) I get bash: A: unbound variable when querying the number of elements after declare. I also get that error when accessing all the elements. I know that later versions of Bash treat this differently. Still I'd like to know whether declare actually *defines* a variable (to be empty).

If I had a script which sets variables read-only to some odd values, and sets errexit because of other unsafe operations: #!/bin/bash set -e declare -r NOTIFY=$(case "$OS" in (macosx) echo macos_notify ;; (linux) echo linux_notify ;; (*) echo : ;; esac) declare -r SAY=_say # _say is a function declare -r VERSION=0.99 set +e And I source it to get the definitions, the second time because it's in development: $ . s.bash $ . s.bash bash: declare: NOTIFY: readonly variable Exited Normally declare -r EXISTING_VAR would neither stop the script nor remove the old, working definition of EXISTING_VAR. But with errexit, assigning to an existing variable is understandably a failure. The easy options are to remove -r or use set +e for that part of the script. Barring those, **is it possible to write a Bash function to take the place of declare -r but not re-assign if the name already exists**? I tried: # arg #1: var name, #2: value set_var_once () { # test whether the variable with the # name stored in $1 exists if [[ -z "${!1}" ]] then # if it doesn't, set it declare -r $1=$2 fi } I also tried things along the lines of eval "declare -r $1=$(eval $2)", it feels like eval is required somewhere here but I'm not sure where. All of the versions of set_var_once result in not setting the variable they should.

I have this: > cd 31 > /Users/alexamil/WebstormProjects/oresoftware/botch/botch-shell-overrides.sh the above style output, is given by this command: declare -F my_bash_func How can I grab just the file name from that result? something like: file=$(declare -F my_bash_func | grab_3rd_entry) _______________ I have to use: shopt -s extdebug declare -F my_bash_func shopt -u extdebug but this doesn't work on MacOS: shopt -s extdebug declare -pf my_bash_func shopt -u extdebug the latter yields a weird error: > declare: my_bash_func: not found but using declare -F can find the function, so not sure why the -pf option doesn't work.