Related
I'm working with this:
GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)
I have a script like below:
#!/bin/bash
e=2
function test1() {
e=4
echo "hello"
}
test1
echo "$e"
Which returns:
hello
4
But if I assign the result of the function to a variable, the global variable e is not modified:
#!/bin/bash
e=2
function test1() {
e=4
echo "hello"
}
ret=$(test1)
echo "$ret"
echo "$e"
Returns:
hello
2
I've heard of the use of eval in this case, so I did this in test1:
eval 'e=4'
But the same result.
Could you explain me why it is not modified? How could I save the echo of the test1 function in ret and modify the global variable too?
When you use a command substitution (i.e., the $(...) construct), you are creating a subshell. Subshells inherit variables from their parent shells, but this only works one way: A subshell cannot modify the environment of its parent shell.
Your variable e is set within a subshell, but not the parent shell. There are two ways to pass values from a subshell to its parent. First, you can output something to stdout, then capture it with a command substitution:
myfunc() {
echo "Hello"
}
var="$(myfunc)"
echo "$var"
The above outputs:
Hello
For a numerical value in the range of 0 through 255, you can use return to pass the number as the exit status:
mysecondfunc() {
echo "Hello"
return 4
}
var="$(mysecondfunc)"
num_var=$?
echo "$var - num is $num_var"
This outputs:
Hello - num is 4
This needs bash 4.1 if you use {fd} or local -n.
The rest should work in bash 3.x I hope. I am not completely sure due to printf %q - this might be a bash 4 feature.
Summary
Your example can be modified as follows to archive the desired effect:
# Add following 4 lines:
_passback() { while [ 1 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; return $1; }
passback() { _passback "$#" "$?"; }
_capture() { { out="$("${#:2}" 3<&-; "$2_" >&3)"; ret=$?; printf "%q=%q;" "$1" "$out"; } 3>&1; echo "(exit $ret)"; }
capture() { eval "$(_capture "$#")"; }
e=2
# Add following line, called "Annotation"
function test1_() { passback e; }
function test1() {
e=4
echo "hello"
}
# Change following line to:
capture ret test1
echo "$ret"
echo "$e"
prints as desired:
hello
4
Note that this solution:
Works for e=1000, too.
Preserves $? if you need $?
The only bad sideffects are:
It needs a modern bash.
It forks quite more often.
It needs the annotation (named after your function, with an added _)
It sacrifices file descriptor 3.
You can change it to another FD if you need that.
In _capture just replace all occurances of 3 with another (higher) number.
The following (which is quite long, sorry for that) hopefully explains, how to adpot this recipe to other scripts, too.
The problem
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
d1=$(d)
d2=$(d)
d3=$(d)
d4=$(d)
echo $x $d1 $d2 $d3 $d4
outputs
0 20171129-123521 20171129-123521 20171129-123521 20171129-123521
while the wanted output is
4 20171129-123521 20171129-123521 20171129-123521 20171129-123521
The cause of the problem
Shell variables (or generally speaking, the environment) is passed from parental processes to child processes, but not vice versa.
If you do output capturing, this usually is run in a subshell, so passing back variables is difficult.
Some even tell you, that it is impossible to fix. This is wrong, but it is a long known difficult to solve problem.
There are several ways on how to solve it best, this depends on your needs.
Here is a step by step guide on how to do it.
Passing back variables into the parental shell
There is a way to pass back variables to a parental shell. However this is a dangerous path, because this uses eval. If done improperly, you risk many evil things. But if done properly, this is perfectly safe, provided that there is no bug in bash.
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
d() { let x++; d=$(date +%Y%m%d-%H%M%S); _passback x d; }
x=0
eval `d`
d1=$d
eval `d`
d2=$d
eval `d`
d3=$d
eval `d`
d4=$d
echo $x $d1 $d2 $d3 $d4
prints
4 20171129-124945 20171129-124945 20171129-124945 20171129-124945
Note that this works for dangerous things, too:
danger() { danger="$*"; passback danger; }
eval `danger '; /bin/echo *'`
echo "$danger"
prints
; /bin/echo *
This is due to printf '%q', which quotes everything such, that you can re-use it in a shell context safely.
But this is a pain in the a..
This does not only look ugly, it also is much to type, so it is error prone. Just one single mistake and you are doomed, right?
Well, we are at shell level, so you can improve it. Just think about an interface you want to see, and then you can implement it.
Augment, how the shell processes things
Let's go a step back and think about some API which allows us to easily express, what we want to do.
Well, what do we want do do with the d() function?
We want to capture the output into a variable.
OK, then let's implement an API for exactly this:
# This needs a modern bash 4.3 (see "help declare" if "-n" is present,
# we get rid of it below anyway).
: capture VARIABLE command args..
capture()
{
local -n output="$1"
shift
output="$("$#")"
}
Now, instead of writing
d1=$(d)
we can write
capture d1 d
Well, this looks like we haven't changed much, as, again, the variables are not passed back from d into the parent shell, and we need to type a bit more.
However now we can throw the full power of the shell at it, as it is nicely wrapped in a function.
Think about an easy to reuse interface
A second thing is, that we want to be DRY (Don't Repeat Yourself).
So we definitively do not want to type something like
x=0
capture1 x d1 d
capture1 x d2 d
capture1 x d3 d
capture1 x d4 d
echo $x $d1 $d2 $d3 $d4
The x here is not only redundant, it's error prone to always repeate in the correct context. What if you use it 1000 times in a script and then add a variable? You definitively do not want to alter all the 1000 locations where a call to d is involved.
So leave the x away, so we can write:
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
d() { let x++; output=$(date +%Y%m%d-%H%M%S); _passback output x; }
xcapture() { local -n output="$1"; eval "$("${#:2}")"; }
x=0
xcapture d1 d
xcapture d2 d
xcapture d3 d
xcapture d4 d
echo $x $d1 $d2 $d3 $d4
outputs
4 20171129-132414 20171129-132414 20171129-132414 20171129-132414
This already looks very good. (But there still is the local -n which does not work in oder common bash 3.x)
Avoid changing d()
The last solution has some big flaws:
d() needs to be altered
It needs to use some internal details of xcapture to pass the output.
Note that this shadows (burns) one variable named output,
so we can never pass this one back.
It needs to cooperate with _passback
Can we get rid of this, too?
Of course, we can! We are in a shell, so there is everything we need to get this done.
If you look a bit closer to the call to eval you can see, that we have 100% control at this location. "Inside" the eval we are in a subshell,
so we can do everything we want without fear of doing something bad to the parental shell.
Yeah, nice, so let's add another wrapper, now directly inside the eval:
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
# !DO NOT USE!
_xcapture() { "${#:2}" > >(printf "%q=%q;" "$1" "$(cat)"); _passback x; } # !DO NOT USE!
# !DO NOT USE!
xcapture() { eval "$(_xcapture "$#")"; }
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
xcapture d1 d
xcapture d2 d
xcapture d3 d
xcapture d4 d
echo $x $d1 $d2 $d3 $d4
prints
4 20171129-132414 20171129-132414 20171129-132414 20171129-132414
However, this, again, has some major drawback:
The !DO NOT USE! markers are there,
because there is a very bad race condition in this,
which you cannot see easily:
The >(printf ..) is a background job. So it might still
execute while the _passback x is running.
You can see this yourself if you add a sleep 1; before printf or _passback.
_xcapture a d; echo then outputs x or a first, respectively.
The _passback x should not be part of _xcapture,
because this makes it difficult to reuse that recipe.
Also we have some unneded fork here (the $(cat)),
but as this solution is !DO NOT USE! I took the shortest route.
However, this shows, that we can do it, without modification to d() (and without local -n)!
Please note that we not neccessarily need _xcapture at all,
as we could have written everyting right in the eval.
However doing this usually isn't very readable.
And if you come back to your script in a few years,
you probably want to be able to read it again without much trouble.
Fix the race
Now let's fix the race condition.
The trick could be to wait until printf has closed it's STDOUT, and then output x.
There are many ways to archive this:
You cannot use shell pipes, because pipes run in different processes.
One can use temporary files,
or something like a lock file or a fifo. This allows to wait for the lock or fifo,
or different channels, to output the information, and then assemble the output in some correct sequence.
Following the last path could look like (note that it does the printf last because this works better here):
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
_xcapture() { { printf "%q=%q;" "$1" "$("${#:2}" 3<&-; _passback x >&3)"; } 3>&1; }
xcapture() { eval "$(_xcapture "$#")"; }
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
xcapture d1 d
xcapture d2 d
xcapture d3 d
xcapture d4 d
echo $x $d1 $d2 $d3 $d4
outputs
4 20171129-144845 20171129-144845 20171129-144845 20171129-144845
Why is this correct?
_passback x directly talks to STDOUT.
However, as STDOUT needs to be captured in the inner command,
we first "save" it into FD3 (you can use others, of course) with '3>&1'
and then reuse it with >&3.
The $("${#:2}" 3<&-; _passback x >&3) finishes after the _passback,
when the subshell closes STDOUT.
So the printf cannot happen before the _passback,
regardless how long _passback takes.
Note that the printf command is not executed before the complete
commandline is assembled, so we cannot see artefacts from printf,
independently how printf is implemented.
Hence first _passback executes, then the printf.
This resolves the race, sacrificing one fixed file descriptor 3.
You can, of course, choose another file descriptor in the case,
that FD3 is not free in your shellscript.
Please also note the 3<&- which protects FD3 to be passed to the function.
Make it more generic
_capture contains parts, which belong to d(), which is bad,
from a reusability perspective. How to solve this?
Well, do it the desparate way by introducing one more thing,
an additional function, which must return the right things,
which is named after the original function with _ attached.
This function is called after the real function, and can augment things.
This way, this can be read as some annotation, so it is very readable:
_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }
_capture() { { printf "%q=%q;" "$1" "$("${#:2}" 3<&-; "$2_" >&3)"; } 3>&1; }
capture() { eval "$(_capture "$#")"; }
d_() { _passback x; }
d() { let x++; date +%Y%m%d-%H%M%S; }
x=0
capture d1 d
capture d2 d
capture d3 d
capture d4 d
echo $x $d1 $d2 $d3 $d4
still prints
4 20171129-151954 20171129-151954 20171129-151954 20171129-151954
Allow access to the return-code
There is only on bit missing:
v=$(fn) sets $? to what fn returned. So you probably want this, too.
It needs some bigger tweaking, though:
# This is all the interface you need.
# Remember, that this burns FD=3!
_passback() { while [ 1 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; return $1; }
passback() { _passback "$#" "$?"; }
_capture() { { out="$("${#:2}" 3<&-; "$2_" >&3)"; ret=$?; printf "%q=%q;" "$1" "$out"; } 3>&1; echo "(exit $ret)"; }
capture() { eval "$(_capture "$#")"; }
# Here is your function, annotated with which sideffects it has.
fails_() { passback x y; }
fails() { x=$1; y=69; echo FAIL; return 23; }
# And now the code which uses it all
x=0
y=0
capture wtf fails 42
echo $? $x $y $wtf
prints
23 42 69 FAIL
There is still a lot room for improvement
_passback() can be elmininated with passback() { set -- "$#" "$?"; while [ 1 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; return $1; }
_capture() can be eliminated with capture() { eval "$({ out="$("${#:2}" 3<&-; "$2_" >&3)"; ret=$?; printf "%q=%q;" "$1" "$out"; } 3>&1; echo "(exit $ret)")"; }
The solution pollutes a file descriptor (here 3) by using it internally.
You need to keep that in mind if you happen to pass FDs.
Note thatbash 4.1 and above has {fd} to use some unused FD.
(Perhaps I will add a solution here when I come around.)
Note that this is why I use to put it in separate functions like _capture, because stuffing this all into one line is possible, but makes it increasingly harder to read and understand
Perhaps you want to capture STDERR of the called function, too.
Or you want to even pass in and out more than one filedescriptor
from and to variables.
I have no solution yet, however here is a way to catch more than one FD, so we can probably pass back the variables this way, too.
Also do not forget:
This must call a shell function, not an external command.
There is no easy way to pass environment variables out of external commands.
(With LD_PRELOAD= it should be possible, though!)
But this then is something completely different.
Last words
This is not the only possible solution. It is one example to a solution.
As always you have many ways to express things in the shell.
So feel free to improve and find something better.
The solution presented here is quite far from being perfect:
It was nearly not tested at all, so please forgive typos.
There is a lot of room for improvement, see above.
It uses many features from modern bash, so probably is hard to port to other shells.
And there might be some quirks I haven't thought about.
However I think it is quite easy to use:
Add just 4 lines of "library".
Add just 1 line of "annotation" for your shell function.
Sacrifices just one file descriptor temporarily.
And each step should be easy to understand even years later.
Maybe you can use a file, write to file inside function, read from file after it. I have changed e to an array. In this example blanks are used as separator when reading back the array.
#!/bin/bash
declare -a e
e[0]="first"
e[1]="secondddd"
function test1 () {
e[2]="third"
e[1]="second"
echo "${e[#]}" > /tmp/tempout
echo hi
}
ret=$(test1)
echo "$ret"
read -r -a e < /tmp/tempout
echo "${e[#]}"
echo "${e[0]}"
echo "${e[1]}"
echo "${e[2]}"
Output:
hi
first second third
first
second
third
What you are doing, you are executing test1
$(test1)
in a sub-shell( child shell ) and Child shells cannot modify anything in parent.
You can find it in bash manual
Please Check: Things results in a subshell here
I had a similar problem when I wanted to remove temporary files I had created automatically. The solution I came up with was not to use command substitution, but rather to pass the name of the variable, that should take the final result, into the function. E.g.
#!/usr/bin/env bash
# array that keeps track of tmp-files
remove_later=()
# function that manages tmp-files
new_tmp_file() {
file=$(mktemp)
remove_later+=( "$file" )
# assign value (safe form of `eval "$1=$file"`)
printf -v "$1" -- "$file"
}
# function to remove all tmp-files
remove_tmp_files() { rm -- "${remove_later[#]}"; }
# define trap to remove all tmp-files upon EXIT
trap remove_tmp_files EXIT
# generate tmp-files
new_tmp_file tmpfile1
new_tmp_file tmpfile2
So, adapting this to the OP, it would be:
#!/usr/bin/env bash
e=2
function test1() {
e=4
printf -v "$1" -- "hello"
}
test1 ret
echo "$ret"
echo "$e"
Works and has no restrictions on the "return value".
Assuming that local -n is available, the following script lets the function test1 modify a global variable:
#!/bin/bash
e=2
function test1() {
local -n var=$1
var=4
echo "hello"
}
test1 e
echo "$e"
Which gives the following output:
hello
4
I'm not sure if this works on your terminal, but I found out that if you don't provide any outputs whatsoever it gets naturally treated as a void function, and can make global variable changes.
Here's the code I used:
let ran1=$(( (1<<63)-1)/3 ))
let ran2=$(( (1<<63)-1)/5 ))
let c=0
function randomize {
c=$(( ran1+ran2 ))
ran2=$ran1
ran1=$c
c=$(( c > 0 ))
}
It's a simple randomizer for games that effectively modifies the needed variables.
It's because command substitution is performed in a subshell, so while the subshell inherits the variables, changes to them are lost when the subshell ends.
Reference:
Command substitution, commands grouped with parentheses, and asynchronous commands are invoked in a subshell environment that is a duplicate of the shell environment
A solution to this problem, without having to introduce complex functions and heavily modify the original one, is to store the value in a temporary file and read / write it when needed.
This approach helped me greatly when I had to mock a bash function called multiple times in a bats test case.
For example, you could have:
# Usage read_value path_to_tmp_file
function read_value {
cat "${1}"
}
# Usage: set_value path_to_tmp_file the_value
function set_value {
echo "${2}" > "${1}"
}
#----
# Original code:
function test1() {
e=4
set_value "${tmp_file}" "${e}"
echo "hello"
}
# Create the temp file
# Note that tmp_file is available in test1 as well
tmp_file=$(mktemp)
# Your logic
e=2
# Store the value
set_value "${tmp_file}" "${e}"
# Run test1
test1
# Read the value modified by test1
e=$(read_value "${tmp_file}")
echo "$e"
The drawback is that you might need multiple temp files for different variables. And also you might need to issue a sync command to persist the contents on the disk between one write and read operations.
You can always use an alias:
alias next='printf "blah_%02d" $count;count=$((count+1))'
I have a bash script that looks as such:
#!/bin/bash
function one {
echo "I am function one!!"
}
function two {
echo "I am function two!!"
}
one
two
If I simply do bash test.sh both functions are being executed.
What I'd like to do is to call the script from the terminal while also specifying one of the two functions, and executing only it.
Maybe something like: bash test.sh$one() and it should only print out
I am function one!!
Is this possible and if so, how will I go about achieving it?
Thanks!
=========================
EDIT: As per #Waqas suggestion I ended up implementing the below which did the trick for me:
function main {
if [ -z "$1" ]
then
some commands
# else run the given function only
else
$1
fi
}
main "$#"
Thanks!!!
There are many ways to write the code in order to fulfill your requirement. The way I will write the code for this, is the following:
#!/bin/bash
function main {
# If the argument is empty then run both functions else only run provided function as argument $1.
[ -z "$1" ] && { one; two; } || $1
}
function one {
echo "I am function one!!"
}
function two {
echo "I am function two!!"
}
main "$#"
If you only execute the script without passing argument then both functions will run and with passing argument only single function will work.
Example1 (Both functions will run): bash script_name
Example2 (Only function one will run): bash script_name one
Example3 (Only function two will run): bash script_name two
You better separate the files: Move the function definitions in a separate file, say ~/lib/testlib.src. Your test.sh then becomes
#!/bin/bash
. ~/lib/testlib.src
one
two
If you need the definitions in your interactive shell, either do there a . ~/lib/testlib.src manually, or if you want to have them always available, put this statement into your ~/.bashrc.
You could do this by writing a case statement after defining the functions but before any other lines of code.
#!/bin/bash
function one {
echo "I am function one!!"
}
function two {
echo "I am function two!!"
}
case $1 in
one)
one
;;
two)
two
;;
*)
one
two
;;
esac
Which could then be used as:
$ ./test.sh one
# I am function one!!
$ ./test.sh two
# I am function two!!
In the above example I put the body of your script under the *) option, but if it better suits your needs, you could instead have the one) and two) options "exit" after calling their single function:
case $1 in
one)
one
exit 0
;;
two)
two
exit 0
;;
esac
one
two
This is all assuming you are not passing any other arguments to the script and that $1 would be used for the desired function. The case statement would become more complex otherwise.
Given the following two executable scripts:
----- file1.sh
#!/bin/sh
. file2.sh
some_routine data
----- file2.sh
#!/bin/sh
some_routine()
{
#get the data passed in
localVar=$1
}
I can pass 'data' to a subroutine in another script, but I would also like to return data.
Is it possible to return information from some_routine?
e.g: var = some_routine data
Have the subroutine output something, and then use $() to capture the output:
some_routine() {
echo "foo $1"
}
some_var=$(some_routine bar)
It's not allowed, just set the value of a global variable (..all variables are global in bash)
if
some_routine() {
echo "first"
echo "foo $1"
}
some_var=$(some_routine "second")
echo "result: $some_var"
they are ok.But the result seems to be decided by the first "echo".Another way is use "eval".
some_var return "first"
some_routine()
{
echo "cmj"
eval $2=$1
}
some_routine "second" some_var
echo "result: $some_var"
in this way, some_var return "second".The bash don't return a string directly.So we need some tricks.
I'd like to return a string from a Bash function.
I'll write the example in java to show what I'd like to do:
public String getSomeString() {
return "tadaa";
}
String variable = getSomeString();
The example below works in bash, but is there a better way to do this?
function getSomeString {
echo "tadaa"
}
VARIABLE=$(getSomeString)
There is no better way I know of. Bash knows only status codes (integers) and strings written to the stdout.
You could have the function take a variable as the first arg and modify the variable with the string you want to return.
#!/bin/bash
set -x
function pass_back_a_string() {
eval "$1='foo bar rab oof'"
}
return_var=''
pass_back_a_string return_var
echo $return_var
Prints "foo bar rab oof".
Edit: added quoting in the appropriate place to allow whitespace in string to address #Luca Borrione's comment.
Edit: As a demonstration, see the following program. This is a general-purpose solution: it even allows you to receive a string into a local variable.
#!/bin/bash
set -x
function pass_back_a_string() {
eval "$1='foo bar rab oof'"
}
return_var=''
pass_back_a_string return_var
echo $return_var
function call_a_string_func() {
local lvar=''
pass_back_a_string lvar
echo "lvar='$lvar' locally"
}
call_a_string_func
echo "lvar='$lvar' globally"
This prints:
+ return_var=
+ pass_back_a_string return_var
+ eval 'return_var='\''foo bar rab oof'\'''
++ return_var='foo bar rab oof'
+ echo foo bar rab oof
foo bar rab oof
+ call_a_string_func
+ local lvar=
+ pass_back_a_string lvar
+ eval 'lvar='\''foo bar rab oof'\'''
++ lvar='foo bar rab oof'
+ echo 'lvar='\''foo bar rab oof'\'' locally'
lvar='foo bar rab oof' locally
+ echo 'lvar='\'''\'' globally'
lvar='' globally
Edit: demonstrating that the original variable's value is available in the function, as was incorrectly criticized by #Xichen Li in a comment.
#!/bin/bash
set -x
function pass_back_a_string() {
eval "echo in pass_back_a_string, original $1 is \$$1"
eval "$1='foo bar rab oof'"
}
return_var='original return_var'
pass_back_a_string return_var
echo $return_var
function call_a_string_func() {
local lvar='original lvar'
pass_back_a_string lvar
echo "lvar='$lvar' locally"
}
call_a_string_func
echo "lvar='$lvar' globally"
This gives output:
+ return_var='original return_var'
+ pass_back_a_string return_var
+ eval 'echo in pass_back_a_string, original return_var is $return_var'
++ echo in pass_back_a_string, original return_var is original return_var
in pass_back_a_string, original return_var is original return_var
+ eval 'return_var='\''foo bar rab oof'\'''
++ return_var='foo bar rab oof'
+ echo foo bar rab oof
foo bar rab oof
+ call_a_string_func
+ local 'lvar=original lvar'
+ pass_back_a_string lvar
+ eval 'echo in pass_back_a_string, original lvar is $lvar'
++ echo in pass_back_a_string, original lvar is original lvar
in pass_back_a_string, original lvar is original lvar
+ eval 'lvar='\''foo bar rab oof'\'''
++ lvar='foo bar rab oof'
+ echo 'lvar='\''foo bar rab oof'\'' locally'
lvar='foo bar rab oof' locally
+ echo 'lvar='\'''\'' globally'
lvar='' globally
All answers above ignore what has been stated in the man page of bash.
All variables declared inside a function will be shared with the calling environment.
All variables declared local will not be shared.
Example code
#!/bin/bash
f()
{
echo function starts
local WillNotExists="It still does!"
DoesNotExists="It still does!"
echo function ends
}
echo $DoesNotExists #Should print empty line
echo $WillNotExists #Should print empty line
f #Call the function
echo $DoesNotExists #Should print It still does!
echo $WillNotExists #Should print empty line
And output
$ sh -x ./x.sh
+ echo
+ echo
+ f
+ echo function starts
function starts
+ local 'WillNotExists=It still does!'
+ DoesNotExists='It still does!'
+ echo function ends
function ends
+ echo It still 'does!'
It still does!
+ echo
Also under pdksh and ksh this script does the same!
Bash, since version 4.3, feb 2014(?), has explicit support for reference variables or name references (namerefs), beyond "eval", with the same beneficial performance and indirection effect, and which may be clearer in your scripts and also harder to "forget to 'eval' and have to fix this error":
declare [-aAfFgilnrtux] [-p] [name[=value] ...]
typeset [-aAfFgilnrtux] [-p] [name[=value] ...]
Declare variables and/or give them attributes
...
-n Give each name the nameref attribute, making it a name reference
to another variable. That other variable is defined by the value
of name. All references and assignments to name, except for⋅
changing the -n attribute itself, are performed on the variable
referenced by name's value. The -n attribute cannot be applied to
array variables.
...
When used in a function, declare and typeset make each name local,
as with the local command, unless the -g option is supplied...
and also:
PARAMETERS
A variable can be assigned the nameref attribute using the -n option to the
declare or local builtin commands (see the descriptions of declare and local
below) to create a nameref, or a reference to another variable. This allows
variables to be manipulated indirectly. Whenever the nameref variable is⋅
referenced or assigned to, the operation is actually performed on the variable
specified by the nameref variable's value. A nameref is commonly used within
shell functions to refer to a variable whose name is passed as an argument to⋅
the function. For instance, if a variable name is passed to a shell function
as its first argument, running
declare -n ref=$1
inside the function creates a nameref variable ref whose value is the variable
name passed as the first argument. References and assignments to ref are
treated as references and assignments to the variable whose name was passed as⋅
$1. If the control variable in a for loop has the nameref attribute, the list
of words can be a list of shell variables, and a name reference will be⋅
established for each word in the list, in turn, when the loop is executed.
Array variables cannot be given the -n attribute. However, nameref variables
can reference array variables and subscripted array variables. Namerefs can be⋅
unset using the -n option to the unset builtin. Otherwise, if unset is executed
with the name of a nameref variable as an argument, the variable referenced by⋅
the nameref variable will be unset.
For example (EDIT 2: (thank you Ron) namespaced (prefixed) the function-internal variable name, to minimize external variable clashes, which should finally answer properly, the issue raised in the comments by Karsten):
# $1 : string; your variable to contain the return value
function return_a_string () {
declare -n ret=$1
local MYLIB_return_a_string_message="The date is "
MYLIB_return_a_string_message+=$(date)
ret=$MYLIB_return_a_string_message
}
and testing this example:
$ return_a_string result; echo $result
The date is 20160817
Note that the bash "declare" builtin, when used in a function, makes the declared variable "local" by default, and "-n" can also be used with "local".
I prefer to distinguish "important declare" variables from "boring local" variables, so using "declare" and "local" in this way acts as documentation.
EDIT 1 - (Response to comment below by Karsten) - I cannot add comments below any more, but Karsten's comment got me thinking, so I did the following test which WORKS FINE, AFAICT - Karsten if you read this, please provide an exact set of test steps from the command line, showing the problem you assume exists, because these following steps work just fine:
$ return_a_string ret; echo $ret
The date is 20170104
(I ran this just now, after pasting the above function into a bash term - as you can see, the result works just fine.)
Like bstpierre above, I use and recommend the use of explicitly naming output variables:
function some_func() # OUTVAR ARG1
{
local _outvar=$1
local _result # Use some naming convention to avoid OUTVARs to clash
... some processing ....
eval $_outvar=\$_result # Instead of just =$_result
}
Note the use of quoting the $. This will avoid interpreting content in $result as shell special characters. I have found that this is an order of magnitude faster than the result=$(some_func "arg1") idiom of capturing an echo. The speed difference seems even more notable using bash on MSYS where stdout capturing from function calls is almost catastrophic.
It's ok to send in a local variables since locals are dynamically scoped in bash:
function another_func() # ARG
{
local result
some_func result "$1"
echo result is $result
}
You could also capture the function output:
#!/bin/bash
function getSomeString() {
echo "tadaa!"
}
return_var=$(getSomeString)
echo $return_var
# Alternative syntax:
return_var=`getSomeString`
echo $return_var
Looks weird, but is better than using global variables IMHO. Passing parameters works as usual, just put them inside the braces or backticks.
The most straightforward and robust solution is to use command substitution, as other people wrote:
assign()
{
local x
x="Test"
echo "$x"
}
x=$(assign) # This assigns string "Test" to x
The downside is performance as this requires a separate process.
The other technique suggested in this topic, namely passing the name of a variable to assign to as an argument, has side effects, and I wouldn't recommend it in its basic form. The problem is that you will probably need some variables in the function to calculate the return value, and it may happen that the name of the variable intended to store the return value will interfere with one of them:
assign()
{
local x
x="Test"
eval "$1=\$x"
}
assign y # This assigns string "Test" to y, as expected
assign x # This will NOT assign anything to x in this scope
# because the name "x" is declared as local inside the function
You might, of course, not declare internal variables of the function as local, but you really should always do it as otherwise you may, on the other hand, accidentally overwrite an unrelated variable from the parent scope if there is one with the same name.
One possible workaround is an explicit declaration of the passed variable as global:
assign()
{
local x
eval declare -g $1
x="Test"
eval "$1=\$x"
}
If name "x" is passed as an argument, the second row of the function body will overwrite the previous local declaration. But the names themselves might still interfere, so if you intend to use the value previously stored in the passed variable prior to write the return value there, be aware that you must copy it into another local variable at the very beginning; otherwise the result will be unpredictable!
Besides, this will only work in the most recent version of BASH, namely 4.2. More portable code might utilize explicit conditional constructs with the same effect:
assign()
{
if [[ $1 != x ]]; then
local x
fi
x="Test"
eval "$1=\$x"
}
Perhaps the most elegant solution is just to reserve one global name for function return values and
use it consistently in every function you write.
As previously mentioned, the "correct" way to return a string from a function is with command substitution. In the event that the function also needs to output to console (as #Mani mentions above), create a temporary fd in the beginning of the function and redirect to console. Close the temporary fd before returning your string.
#!/bin/bash
# file: func_return_test.sh
returnString() {
exec 3>&1 >/dev/tty
local s=$1
s=${s:="some default string"}
echo "writing directly to console"
exec 3>&-
echo "$s"
}
my_string=$(returnString "$*")
echo "my_string: [$my_string]"
executing script with no params produces...
# ./func_return_test.sh
writing directly to console
my_string: [some default string]
hope this helps people
-Andy
You could use a global variable:
declare globalvar='some string'
string ()
{
eval "$1='some other string'"
} # ---------- end of function string ----------
string globalvar
echo "'${globalvar}'"
This gives
'some other string'
To illustrate my comment on Andy's answer, with additional file descriptor manipulation to avoid use of /dev/tty:
#!/bin/bash
exec 3>&1
returnString() {
exec 4>&1 >&3
local s=$1
s=${s:="some default string"}
echo "writing to stdout"
echo "writing to stderr" >&2
exec >&4-
echo "$s"
}
my_string=$(returnString "$*")
echo "my_string: [$my_string]"
Still nasty, though.
The way you have it is the only way to do this without breaking scope. Bash doesn't have a concept of return types, just exit codes and file descriptors (stdin/out/err, etc)
Addressing Vicky Ronnen's head up, considering the following code:
function use_global
{
eval "$1='changed using a global var'"
}
function capture_output
{
echo "always changed"
}
function test_inside_a_func
{
local _myvar='local starting value'
echo "3. $_myvar"
use_global '_myvar'
echo "4. $_myvar"
_myvar=$( capture_output )
echo "5. $_myvar"
}
function only_difference
{
local _myvar='local starting value'
echo "7. $_myvar"
local use_global '_myvar'
echo "8. $_myvar"
local _myvar=$( capture_output )
echo "9. $_myvar"
}
declare myvar='global starting value'
echo "0. $myvar"
use_global 'myvar'
echo "1. $myvar"
myvar=$( capture_output )
echo "2. $myvar"
test_inside_a_func
echo "6. $_myvar" # this was local inside the above function
only_difference
will give
0. global starting value
1. changed using a global var
2. always changed
3. local starting value
4. changed using a global var
5. always changed
6.
7. local starting value
8. local starting value
9. always changed
Maybe the normal scenario is to use the syntax used in the test_inside_a_func function, thus you can use both methods in the majority of cases, although capturing the output is the safer method always working in any situation, mimicking the returning value from a function that you can find in other languages, as Vicky Ronnen correctly pointed out.
The options have been all enumerated, I think. Choosing one may come down to a matter of the best style for your particular application, and in that vein, I want to offer one particular style I've found useful. In bash, variables and functions are not in the same namespace. So, treating the variable of the same name as the value of the function is a convention that I find minimizes name clashes and enhances readability, if I apply it rigorously. An example from real life:
UnGetChar=
function GetChar() {
# assume failure
GetChar=
# if someone previously "ungot" a char
if ! [ -z "$UnGetChar" ]; then
GetChar="$UnGetChar"
UnGetChar=
return 0 # success
# else, if not at EOF
elif IFS= read -N1 GetChar ; then
return 0 # success
else
return 1 # EOF
fi
}
function UnGetChar(){
UnGetChar="$1"
}
And, an example of using such functions:
function GetToken() {
# assume failure
GetToken=
# if at end of file
if ! GetChar; then
return 1 # EOF
# if start of comment
elif [[ "$GetChar" == "#" ]]; then
while [[ "$GetChar" != $'\n' ]]; do
GetToken+="$GetChar"
GetChar
done
UnGetChar "$GetChar"
# if start of quoted string
elif [ "$GetChar" == '"' ]; then
# ... et cetera
As you can see, the return status is there for you to use when you need it, or ignore if you don't. The "returned" variable can likewise be used or ignored, but of course only after the function is invoked.
Of course, this is only a convention. You are free to fail to set the associated value before returning (hence my convention of always nulling it at the start of the function) or to trample its value by calling the function again (possibly indirectly). Still, it's a convention I find very useful if I find myself making heavy use of bash functions.
As opposed to the sentiment that this is a sign one should e.g. "move to perl", my philosophy is that conventions are always important for managing the complexity of any language whatsoever.
In my programs, by convention, this is what the pre-existing $REPLY variable is for, which read uses for that exact purpose.
function getSomeString {
REPLY="tadaa"
}
getSomeString
echo $REPLY
This echoes
tadaa
But to avoid conflicts, any other global variable will do.
declare result
function getSomeString {
result="tadaa"
}
getSomeString
echo $result
If that isn’t enough, I recommend Markarian451’s solution.
They key problem of any 'named output variable' scheme where the caller can pass in the variable name (whether using eval or declare -n) is inadvertent aliasing, i.e. name clashes: From an encapsulation point of view, it's awful to not be able to add or rename a local variable in a function without checking ALL the function's callers first to make sure they're not wanting to pass that same name as the output parameter. (Or in the other direction, I don't want to have to read the source of the function I'm calling just to make sure the output parameter I intend to use is not a local in that function.)
The only way around that is to use a single dedicated output variable like REPLY (as suggested by Evi1M4chine) or a convention like the one suggested by Ron Burk.
However, it's possible to have functions use a fixed output variable internally, and then add some sugar over the top to hide this fact from the caller, as I've done with the call function in the following example. Consider this a proof of concept, but the key points are
The function always assigns the return value to REPLY, and can also return an exit code as usual
From the perspective of the caller, the return value can be assigned to any variable (local or global) including REPLY (see the wrapper example). The exit code of the function is passed through, so using them in e.g. an if or while or similar constructs works as expected.
Syntactically the function call is still a single simple statement.
The reason this works is because the call function itself has no locals and uses no variables other than REPLY, avoiding any potential for name clashes. At the point where the caller-defined output variable name is assigned, we're effectively in the caller's scope (technically in the identical scope of the call function), rather than in the scope of the function being called.
#!/bin/bash
function call() { # var=func [args ...]
REPLY=; "${1#*=}" "${#:2}"; eval "${1%%=*}=\$REPLY; return $?"
}
function greet() {
case "$1" in
us) REPLY="hello";;
nz) REPLY="kia ora";;
*) return 123;;
esac
}
function wrapper() {
call REPLY=greet "$#"
}
function main() {
local a b c d
call a=greet us
echo "a='$a' ($?)"
call b=greet nz
echo "b='$b' ($?)"
call c=greet de
echo "c='$c' ($?)"
call d=wrapper us
echo "d='$d' ($?)"
}
main
Output:
a='hello' (0)
b='kia ora' (0)
c='' (123)
d='hello' (0)
You can echo a string, but catch it by piping (|) the function to something else.
You can do it with expr, though ShellCheck reports this usage as deprecated.
bash pattern to return both scalar and array value objects:
definition
url_parse() { # parse 'url' into: 'url_host', 'url_port', ...
local "$#" # inject caller 'url' argument in local scope
local url_host="..." url_path="..." # calculate 'url_*' components
declare -p ${!url_*} # return only 'url_*' object fields to the caller
}
invocation
main() { # invoke url parser and inject 'url_*' results in local scope
eval "$(url_parse url=http://host/path)" # parse 'url'
echo "host=$url_host path=$url_path" # use 'url_*' components
}
Although there were a lot of good answers, they all did not work the way I wanted them to. So here is my solution with these key points:
Helping the forgetful programmer
Atleast I would struggle to always remember error checking after something like this: var=$(myFunction)
Allows assigning values with newline chars \n
Some solutions do not allow for that as some forgot about the single quotes around the value to assign. Right way: eval "${returnVariable}='${value}'" or even better: see the next point below.
Using printf instead of eval
Just try using something like this myFunction "date && var2" to some of the supposed solutions here. eval will execute whatever is given to it. I only want to assign values so I use printf -v "${returnVariable}" "%s" "${value}" instead.
Encapsulation and protection against variable name collision
If a different user or at least someone with less knowledge about the function (this is likely me in some months time) is using myFunction I do not want them to know that he must use a global return value name or some variable names are forbidden to use. That is why I added a name check at the top of myFunction:
if [[ "${1}" = "returnVariable" ]]; then
echo "Cannot give the ouput to \"returnVariable\" as a variable with the same name is used in myFunction()!"
echo "If that is still what you want to do please do that outside of myFunction()!"
return 1
fi
Note this could also be put into a function itself if you have to check a lot of variables.
If I still want to use the same name (here: returnVariable) I just create a buffer variable, give that to myFunction and then copy the value returnVariable.
So here it is:
myFunction():
myFunction() {
if [[ "${1}" = "returnVariable" ]]; then
echo "Cannot give the ouput to \"returnVariable\" as a variable with the same name is used in myFunction()!"
echo "If that is still what you want to do please do that outside of myFunction()!"
return 1
fi
if [[ "${1}" = "value" ]]; then
echo "Cannot give the ouput to \"value\" as a variable with the same name is used in myFunction()!"
echo "If that is still what you want to do please do that outside of myFunction()!"
return 1
fi
local returnVariable="${1}"
local value=$'===========\nHello World\n==========='
echo "setting the returnVariable now..."
printf -v "${returnVariable}" "%s" "${value}"
}
Test cases:
var1="I'm not greeting!"
myFunction var1
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var1:\n%s\n" "${var1}"
# Output:
# setting the returnVariable now...
# myFunction(): SUCCESS
# var1:
# ===========
# Hello World
# ===========
returnVariable="I'm not greeting!"
myFunction returnVariable
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "returnVariable:\n%s\n" "${returnVariable}"
# Output
# Cannot give the ouput to "returnVariable" as a variable with the same name is used in myFunction()!
# If that is still what you want to do please do that outside of myFunction()!
# myFunction(): FAILURE
# returnVariable:
# I'm not greeting!
var2="I'm not greeting!"
myFunction "date && var2"
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var2:\n%s\n" "${var2}"
# Output
# setting the returnVariable now...
# ...myFunction: line ..: printf: `date && var2': not a valid identifier
# myFunction(): FAILURE
# var2:
# I'm not greeting!
myFunction var3
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var3:\n%s\n" "${var3}"
# Output
# setting the returnVariable now...
# myFunction(): SUCCESS
# var3:
# ===========
# Hello World
# ===========
#Implement a generic return stack for functions:
STACK=()
push() {
STACK+=( "${1}" )
}
pop() {
export $1="${STACK[${#STACK[#]}-1]}"
unset 'STACK[${#STACK[#]}-1]';
}
#Usage:
my_func() {
push "Hello world!"
push "Hello world2!"
}
my_func ; pop MESSAGE2 ; pop MESSAGE1
echo ${MESSAGE1} ${MESSAGE2}
agt#agtsoft:~/temp$ cat ./fc
#!/bin/sh
fcall='function fcall { local res p=$1; shift; fname $*; eval "$p=$res"; }; fcall'
function f1 {
res=$[($1+$2)*2];
}
function f2 {
local a;
eval ${fcall//fname/f1} a 2 3;
echo f2:$a;
}
a=3;
f2;
echo after:a=$a, res=$res
agt#agtsoft:~/temp$ ./fc
f2:10
after:a=3, res=
This question already has answers here:
How do I list the functions defined in my shell? [duplicate]
(8 answers)
Closed 4 years ago.
I have a Bourne Shell script that has several functions in it, and allows to be called in the following way:
my.sh <func_name> <param1> <param2>
Inside, func_name() will be called with param1 and param2.
I want to create a help function that would just list all available functions, even without parameters.
The question: how do I get a list of all function names in a script from inside the script?
I'd like to avoid having to parse it and look for function patterns. Too easy to get wrong.
Update: the code. Wanted my help() function be like main() - a function added to the code is added to the help automatically.
#!/bin/sh
# must work with "set -e"
foo ()
{
echo foo: -$1-$2-$3-
return 0
}
# only runs if there are parameters
# exits
main ()
{
local cmd="$1"
shift
local rc=0
$cmd "$#" || rc=$?
exit $rc
}
if [[ "$*" ]]
then
main "$#"
die "how did we get here?"
fi
You can get a list of functions in your script by using the grep command on your own script. In order for this approach to work, you will need to structure your functions a certain way so grep can find them. Here is a sample:
$ cat my.sh
#!/bin/sh
function func1() # Short description
{
echo func1 parameters: $1 $2
}
function func2() # Short description
{
echo func2 parameters: $1 $2
}
function help() # Show a list of functions
{
grep "^function" $0
}
if [ "_$1" = "_" ]; then
help
else
"$#"
fi
Here is an interactive demo:
$ my.sh
function func1() # Short description
function func2() # Short description
function help() # Show a list of functions
$ my.sh help
function func1() # Short description
function func2() # Short description
function help() # Show a list of functions
$ my.sh func1 a b
func1 parameters: a b
$ my.sh func2 x y
func2 parameters: x y
If you have "private" function that you don't want to show up in the help, then omit the "function" part:
my_private_function()
{
# Do something
}
typeset -f returns the functions with their bodies, so a simple awk script is used to pluck out the function names
f1 () { :; }
f2 () { :; }
f3 () { :; }
f4 () { :; }
help () {
echo "functions available:"
typeset -f | awk '/ \(\) $/ && !/^main / {print $1}'
}
main () { help; }
main
This script outputs:
functions available:
f1
f2
f3
f4
help
You call this function with no
arguments and it spits out a
"whitespace" separated list of
function names only.
function script.functions () {
local fncs=`declare -F -p | cut -d " " -f 3`; # Get function list
echo $fncs; # not quoted here to create shell "argument list" of funcs.
}
To load the functions into an array:
declare MyVar=($(script.functions));
Of course, common sense dictates that
any functions that haven't been
sourced into the current file before
this is called will not show up in the
list.
To Make the list read-only and
available for export to other scripts
called by this script:
declare -rx MyVar=($(script.functions));
To print the entire list as newline
separated:
printf "%s\n" "${MyVar[#]}";
The best thing to do is make an array (you are using bash) that contains functions that you want to advertise and have your help function iterate over and print them.
Calling set alone will produce the functions, but in their entirety. You'd still have to parse that looking for things ending in () to get the proverbial symbols.
Its also probably saner to use something like getopt to turn --function-name into function_name with arguments. But, well, sane is relative and you have not posted code :)
Your other option is to create a loadable for bash (a fork of set) that accomplishes this. Honestly, I'd prefer going with parsing before writing a loadable for this task.