Shell Script: Is there a difference between "local foo" and "local foo="? - linux

I found following code in /etc/init.d/functions on CentOS.
status() {
local base pid lock_file= pid_file=
...
4 variables are declared.
Two of them are not initialized, base and pid.
But rest of them are initialized with empty value, lock_file and pid_file.
I tested following code and found no differences.
local a b=
echo "a is $a, length is ${#a}"
echo "b is $b, length is ${#b}"
Is there any differences between them?

Yes, there is a difference. Consider the following function:
x() {
local a b=
echo ${a-X}
echo ${b-X}
}
Calling this function in bash-4.x results in this output:
$ x
X
$
The ${parameter−word} parameter expansion expands to the expansion of word (in this case X) if the parameter is unset, or to the parameter value if it is set.
From the example output, it is obvious that local a leaves the variable a unset, while local b= explicitly sets it to the empty (null) string.
EDIT:
On the other hand, on bash-3.x you get this:
$ x
$
A call to set within the function verifies that local a in bash-3.x initializes that variable to the empty string. This, however, seems to have been a bug. From the bash changelog:
This document details the changes between this version, bash-4.0-beta,
and the previous version, bash-4.0-alpha.
...
e. Fixed a bug that caused local variables to be created with the empty
string for a value rather than no value.

Related

Assigning one variable to another in Bash?

I have a doubt. When i declare a value and assign to some variable, I don't know how to reassign the same value to another variable. See the code snippet below.
#/bin/sh
#declare ARG1 to a
a=ARG1
#declaring $a to ARG2
ARG2=$`$a`
echo "ARG 2 = $ARG2"
It should display my output as
ARG 2 = ARG1
...but instead the actual output is:
line 5: ARG1: command not found
ARG 2 = $
To assign the value associated with the variable dest to the variable source, you need simply run dest=$source.
For example, to assign the value associated with the variable arg2 to the variable a:
a=ARG1
arg2=$a
echo "ARG 2 = $arg2"
The use of lower-case variable names for local shell variables is by convention, not necessity -- but this has the advantage of avoiding conflicts with environment variables and builtins, both of which use all-uppercase names by convention.
You may also want to alias rather than copy the variable. For example, if you need mutation. Or if you want to run a function multiple times on different variables. Here's how it works
Example:
C=cat
declare -n VAR=C
VAR+=" says Hi"
echo "$C" # prints "cat says Hi"
Example with arrays/dictionaries:
A=(a a a)
declare -n VAR=A # "-n" stands for "name", e.g. a new name for the same variable
VAR+=(b)
echo "${A[#]}" # prints "a a a b"
That is, VAR becomes effectively the same as the original variable. Instead of copying, you're adding an alias. Here's an example with functions:
function myFunc() {
local -n VAR="$1"
VAR="Hello from $2"
echo "I've set variable '$1' to value '$VAR'"
}
myFunc Inbox Bob # I've set variable 'Inbox' to value 'Hello from Bob'
myFunc Luke Leia # I've set variable 'Luke' to value 'Hello from Leia'
echo "$Luke" # Hello from Leia
Whether you should use these approaches is a question. Generally, immutable code is easier to read and to reason about (in almost any programming language). However, sometimes you really need to get stuff done in a certain way. Hope this answer helps you then.

Bash, referring to array by value?

Is there some way to access a variable by referring to it by a value?
BAR=("hello", "world")
function foo() {
DO SOME MAGIC WITH $1
// Output the value of the array $BAR
}
foo "BAR"
Perhaps what you're looking for is indirect expansion. From man bash:
If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the
value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value
is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The
exceptions to this are the expansions of ${!prefix*} and ${!name[#]} described below. The exclamation point must immediately fol‐
low the left brace in order to introduce indirection.
Related docs: Shell parameter expansion (Bash Manual) and Evaluating indirect/reference variables (BashFAQ).
Here's an example.
$ MYVAR="hello world"
$ VARNAME="MYVAR"
$ echo ${!VARNAME}
hello world
Note that indirect expansion for arrays is slightly cumbersome (because ${!name[#]} means something else. See linked docs above):
$ BAR=("hello" "world")
$ v="BAR[#]"
$ echo ${!v}
hello world
$ v="BAR[0]"
$ echo ${!v}
hello
$ v="BAR[1]"
$ echo ${!v}
world
To put this in context of your question:
BAR=("hello" "world")
function foo() {
ARR="${1}[#]"
echo ${!ARR}
}
foo "BAR" # prints out "hello world"
Caveats:
Indirect expansion of the array syntax will not work in older versions of bash (pre v3). See BashFAQ article.
It appears you cannot use it to retrieve the array size. ARR="#${1}[#]" will not work. You can however work around this issue by making a copy of the array if it is not prohibitively large. For example:
function foo() {
ORI_ARRNAME="${1}[#]"
local -a ARR=(${!ORI_ARRNAME}) # make a local copy of the array
# you can now use $ARR as the array
echo ${#ARR[#]} # get size
echo ${ARR[1]} # print 2nd element
}
BAR=("hello", "world")
function foo() {
eval echo "\${$1[#]}"
}
foo "BAR"
You can put your arrays into a dictionary matched with their names. Then you can look up this dictionary to find your array and display its contents.

shell script function return a string

I am new to shell scripts, I am trying to create a simple function which will return the concatenated two strings that are passed as parameters. I tried with below code
function getConcatenatedString() {
echo "String1 $1"
echo "String2 $2"
str=$1/$2
echo "Concatenated String ${str}"
echo "${str}"
}
//I am calling the above function
constr=$(getConcatenatedString "hello" "world")
echo "printing result"
echo "${constr}"
echo "exit"
I see the below output when running the script with above code,
printing result
String1 hello
String2 world
Concatenated String hello/world
hello/world
exit
If you look at the code I am first calling the function and then I am echoing "printing result" statement, but the result is first comes the "printing result" and echos the statement inside the function. Is the below statement calling the function
constr=$(getConcatenatedString "hello" "world")
or
echo ${constr}
is calling the function ?
Because if I comment out #echo ${constr} then nothing is getting echoed !!! Please clarify me.
The first is calling the function and storing all of the output (four echo statements) into $constr.
Then, after return, you echo the preamble printing result, $constr (consisting of four lines) and the exit message.
That's how $() works, it captures the entire standard output from the enclosed command.
It sounds like you want to see some of the echo statements on the console rather than capturing them with the $(). I think you should just be able to send them to standard error for that:
echo "String1 $1" >&2
paxdiablo's solution is correct. You cannot return a string from a function, but you can capture the output of the function or return an integer value that can be retrieved by the caller from $?. However, since all shell variables are global, you can simply do:
getConcatenatedString() { str="$1/$2"; }
getConcatenatedString hello world
echo "Concatenated String ${str}"
Note that the function keyword is redundant with (), but function is less portable.
A more flexible, but slightly harder to understand approach is to pass a variable name, and use eval so that the variable becomes set in the caller's context (either a global or a function local). In bash:
function mylist()
{
local _varname=$1 _p _t
shift
for _p in "$#"; do
_t=$_t[$_p]
done
eval "$_varname=\$_t"
}
mylist tmpvar a b c
echo "result: $tmpvar"
On my Linux desktop (bash-3.2) it's approx 3-5x faster (10,000 iterations) than using ``, since the latter has process creation overheads.
If you have bash-4.2, its declare -g allows a function to set a global variable, so you can replace the unpretty eval with:
declare -g $_varname="$_t"
The eval method is similar to TCL's upvar 1, and declare -g is similar to upvar #0.
Some shell builtins support something similar, like bash's printf with "-v", again saving process creation by assigning directly to a variable instead of capturing output (~20-25x faster for me).

How to return a string value from a Bash function

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=

ksh: assigning function output to an array

Why doesn't this work???
#!/bin/ksh
# array testfunc()
function testfunc {
typeset -A env
env=( one="motherload" )
print -r $env
return 0
}
testfunc # returns: ( one=motherload )
typeset -A testvar # segfaults on linux, memfaults on solaris
testvar=$(testfunc) # segfaults on linux, memfaults on solaris
print ${testvar.one}
note: I updated the above script to print ${testvar.one} from print $testvar to show more precisely what I am trying to accomplish.
I am sure this has been asked before, but I am not sure what to search on and everything I have been trying to use for keywords is not bringing me any answers that relate to my problem.
ksh version:
linux: version sh (AT&T Research) 1993-12-28 s+
solaris: version sh (AT&T Research) 93s+ 2008-01-31
Update:
So another question is, this will run in ksh 93t+ without giving an error, but, it doesn't assign the array properly. I would I go about assigning an array from a function? I tried assigning the array like this also:
typeset -A testvar=$(testfunc)
print ${testvar.one}
But that also didn't work properly.
EDIT
So what is happening here?
typeset -A env=( one="motherload" two="vain" )
print ${env.one}
print ${env.two}
I thought this was how you defined associative arrays, maybe what I was looking at was old but who knows.... seems odd behaviour since this prints out "motherload" and "vain"
Your script works fine for me on Linux with ksh 93t+.
Since it's the same script and you're getting similar errors in two different environments, I would suspect stray characters in the file. Try one of these to show any stray characters that might be present:
hd filename
cat -v filename
hexdump -C filename
If it's simply a matter of DOS line endings, then this will fix that:
dos2unix filename
Edit:
Here's one way to create and populate an associative array in ksh:
$ typeset -A testvar
$ testvar=([one]="motherlode" [two]="vein" [waste]="tailings")
$ echo ${testvar[two]}
vein
$ testvar[ore]="gold"
$ echo ${!testvar[#]} # print the indices of the array
one two waste ore
$ typeset -p testvar # show the current definition of the array
typeset -A testvar=([one]="motherlode" [two]="vein" [waste]="tailings" [ore]="gold")
As you can see, ksh uses bracketed subscripts for arrays. Dotted notation is used for accessing members of a compound variable.
I don't believe ksh functions can return arrays. However, you can use the print technique you have in your function (but add square brackets around the index name) and use eval to do the assignment.
$ typeset -A testvar
$ eval "testvar=($(testfunc))"
or to append to an existing array:
$ eval "testvar+=($(testfunc))"
Unless your function is using associative arrays internally, you don't necessarily need to use them to build your output.
However, if you do, you can parse from the result of typeset -p:
$ result=$(typeset -p env)
$ result=${result#*\(}
$ result=${result%\)*}
$ print result
or iterate through the array:
$ for index in ${!env[#]}; do print -n "[$index]=${env[$index]} "; done; print
You may want to consult the documentation concerning discipline functions and type variables
Here is an alternative to getting any return value from a function using name reference. The value returned will be stored in a variable defined as the first positional argument of the function (not declaring the variable beforehand will work but the variable will be global):
#################################
# Example using compound variable
#################################
function returnCompound {
typeset -n returnVal="$1"
returnVal=( one="motherloadCompound" )
return 0
}
# Declaring the variable to keep it in this scope
# Useful for calling nested functions whitout messing
# with the global scope
typeset myNewCompoundVar
returnCompound myNewCompoundVar
echo "Compound: ${myNewCompoundVar.one}"
##################################
# Example using asssociative array
##################################
function returnMap {
typeset -n myNewMapVar="$1"
myNewMapVar=( [one]="motherloadMap" )
typeset nestedCompoundVar
returnCompound nestedCompoundVar
echo "Compound (Nested) from inside: ${nestedCompoundVar.one}"
return 0
}
# Declaring the variable to keep it in this scope
# Useful for calling nested functions whitout messing
# with the global scope
typeset myNewMapVar
returnMap myNewMapVar
echo "Associative array: ${myNewMapVar[one]}"
echo "Compound (Nested) from outside: ${nestedCompoundVar.one}"
Output:
Compound: motherloadCompound
Compound (Nested) from inside: motherloadCompound
Associative array: motherloadMap
Compound (Nested) from outside:
Important side notes:
Function declarations must be done using the function keyword or else the concept of local scope variable won't be taken under account. In which case the name of your reference variable and global variable might clash if they happen to be the same, resulting in a typeset: invalid self reference error. This can be tested by changing the declaration of the 'returnMap' function.
If you do not declare the return variable before the function call, the variable to which is assigned the return value will be created globally and not limited to the calling scope.

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