I am working with a bash script and I want to execute a function to print a return value:
function fun1(){
return 34
}
function fun2(){
local res=$(fun1)
echo $res
}
When I execute fun2, it does not print "34". Why is this the case?
Although Bash has a return statement, the only thing you can specify with it is the function's own exit status (a value between 0 and 255, 0 meaning "success"). So return is not what you want.
You might want to convert your return statement to an echo statement - that way your function output could be captured using $() braces, which seems to be exactly what you want.
Here is an example:
function fun1(){
echo 34
}
function fun2(){
local res=$(fun1)
echo $res
}
Another way to get the return value (if you just want to return an integer 0-255) is $?.
function fun1(){
return 34
}
function fun2(){
fun1
local res=$?
echo $res
}
Also, note that you can use the return value to use Boolean logic - like fun1 || fun2 will only run fun2 if fun1 returns a non-0 value. The default return value is the exit value of the last statement executed within the function.
Functions in Bash are not functions like in other languages; they're actually commands. So functions are used as if they were binaries or scripts fetched from your path. From the perspective of your program logic, there shouldn't really be any difference.
Shell commands are connected by pipes (aka streams), and not fundamental or user-defined data types, as in "real" programming languages. There is no such thing like a return value for a command, maybe mostly because there's no real way to declare it. It could occur on the man-page, or the --help output of the command, but both are only human-readable and hence are written to the wind.
When a command wants to get input it reads it from its input stream, or the argument list. In both cases text strings have to be parsed.
When a command wants to return something, it has to echo it to its output stream. Another often practiced way is to store the return value in dedicated, global variables. Writing to the output stream is clearer and more flexible, because it can take also binary data. For example, you can return a BLOB easily:
encrypt() {
gpg -c -o- $1 # Encrypt data in filename to standard output (asks for a passphrase)
}
encrypt public.dat > private.dat # Write the function result to a file
As others have written in this thread, the caller can also use command substitution $() to capture the output.
Parallely, the function would "return" the exit code of gpg (GnuPG). Think of the exit code as a bonus that other languages don't have, or, depending on your temperament, as a "Schmutzeffekt" of shell functions. This status is, by convention, 0 on success or an integer in the range 1-255 for something else. To make this clear: return (like exit) can only take a value from 0-255, and values other than 0 are not necessarily errors, as is often asserted.
When you don't provide an explicit value with return, the status is taken from the last command in a Bash statement/function/command and so forth. So there is always a status, and return is just an easy way to provide it.
$(...) captures the text sent to standard output by the command contained within. return does not output to standard output. $? contains the result code of the last command.
fun1 (){
return 34
}
fun2 (){
fun1
local res=$?
echo $res
}
The problem with other answers is they either use a global, which can be overwritten when several functions are in a call chain, or echo which means your function cannot output diagnostic information (you will forget your function does this and the "result", i.e. return value, will contain more information than your caller expects, leading to weird bugs), or eval which is way too heavy and hacky.
The proper way to do this is to put the top level stuff in a function and use a local with Bash's dynamic scoping rule. Example:
func1()
{
ret_val=hi
}
func2()
{
ret_val=bye
}
func3()
{
local ret_val=nothing
echo $ret_val
func1
echo $ret_val
func2
echo $ret_val
}
func3
This outputs
nothing
hi
bye
Dynamic scoping means that ret_val points to a different object, depending on the caller! This is different from lexical scoping, which is what most programming languages use. This is actually a documented feature, just easy to miss, and not very well explained. Here is the documentation for it (emphasis is mine):
Variables local to the function may be declared with the local
builtin. These variables are visible only to the function and the
commands it invokes.
For someone with a C, C++, Python, Java,C#, or JavaScript background, this is probably the biggest hurdle: functions in bash are not functions, they are commands, and behave as such: they can output to stdout/stderr, they can pipe in/out, and they can return an exit code. Basically, there isn't any difference between defining a command in a script and creating an executable that can be called from the command line.
So instead of writing your script like this:
Top-level code
Bunch of functions
More top-level code
write it like this:
# Define your main, containing all top-level code
main()
Bunch of functions
# Call main
main
where main() declares ret_val as local and all other functions return values via ret_val.
See also the Unix & Linux question Scope of Local Variables in Shell Functions.
Another, perhaps even better solution depending on situation, is the one posted by ya.teck which uses local -n.
Another way to achieve this is name references (requires Bash 4.3+).
function example {
local -n VAR=$1
VAR=foo
}
example RESULT
echo $RESULT
The return statement sets the exit code of the function, much the same as exit will do for the entire script.
The exit code for the last command is always available in the $? variable.
function fun1(){
return 34
}
function fun2(){
local res=$(fun1)
echo $? # <-- Always echos 0 since the 'local' command passes.
res=$(fun1)
echo $? #<-- Outputs 34
}
As an add-on to others' excellent posts, here's an article summarizing these techniques:
set a global variable
set a global variable, whose name you passed to the function
set the return code (and pick it up with $?)
'echo' some data (and pick it up with MYVAR=$(myfunction) )
Returning Values from Bash Functions
I like to do the following if running in a script where the function is defined:
POINTER= # Used for function return values
my_function() {
# Do stuff
POINTER="my_function_return"
}
my_other_function() {
# Do stuff
POINTER="my_other_function_return"
}
my_function
RESULT="$POINTER"
my_other_function
RESULT="$POINTER"
I like this, because I can then include echo statements in my functions if I want
my_function() {
echo "-> my_function()"
# Do stuff
POINTER="my_function_return"
echo "<- my_function. $POINTER"
}
The simplest way I can think of is to use echo in the method body like so
get_greeting() {
echo "Hello there, $1!"
}
STRING_VAR=$(get_greeting "General Kenobi")
echo $STRING_VAR
# Outputs: Hello there, General Kenobi!
Instead of calling var=$(func) with the whole function output, you can create a function that modifies the input arguments with eval,
var1="is there"
var2="anybody"
function modify_args() {
echo "Modifying first argument"
eval $1="out"
echo "Modifying second argument"
eval $2="there?"
}
modify_args var1 var2
# Prints "Modifying first argument" and "Modifying second argument"
# Sets var1 = out
# Sets var2 = there?
This might be useful in case you need to:
Print to stdout/stderr within the function scope (without returning it)
Return (set) multiple variables.
Git Bash on Windows is using arrays for multiple return values
Bash code:
#!/bin/bash
## A 6-element array used for returning
## values from functions:
declare -a RET_ARR
RET_ARR[0]="A"
RET_ARR[1]="B"
RET_ARR[2]="C"
RET_ARR[3]="D"
RET_ARR[4]="E"
RET_ARR[5]="F"
function FN_MULTIPLE_RETURN_VALUES(){
## Give the positional arguments/inputs
## $1 and $2 some sensible names:
local out_dex_1="$1" ## Output index
local out_dex_2="$2" ## Output index
## Echo for debugging:
echo "Running: FN_MULTIPLE_RETURN_VALUES"
## Here: Calculate output values:
local op_var_1="Hello"
local op_var_2="World"
## Set the return values:
RET_ARR[ $out_dex_1 ]=$op_var_1
RET_ARR[ $out_dex_2 ]=$op_var_2
}
echo "FN_MULTIPLE_RETURN_VALUES EXAMPLES:"
echo "-------------------------------------------"
fn="FN_MULTIPLE_RETURN_VALUES"
out_dex_a=0
out_dex_b=1
eval $fn $out_dex_a $out_dex_b ## <-- Call function
a=${RET_ARR[0]} && echo "RET_ARR[0]: $a "
b=${RET_ARR[1]} && echo "RET_ARR[1]: $b "
echo
## ---------------------------------------------- ##
c="2"
d="3"
FN_MULTIPLE_RETURN_VALUES $c $d ## <--Call function
c_res=${RET_ARR[2]} && echo "RET_ARR[2]: $c_res "
d_res=${RET_ARR[3]} && echo "RET_ARR[3]: $d_res "
echo
## ---------------------------------------------- ##
FN_MULTIPLE_RETURN_VALUES 4 5 ## <--- Call function
e=${RET_ARR[4]} && echo "RET_ARR[4]: $e "
f=${RET_ARR[5]} && echo "RET_ARR[5]: $f "
echo
##----------------------------------------------##
read -p "Press Enter To Exit:"
Expected output:
FN_MULTIPLE_RETURN_VALUES EXAMPLES:
-------------------------------------------
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[0]: Hello
RET_ARR[1]: World
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[2]: Hello
RET_ARR[3]: World
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[4]: Hello
RET_ARR[5]: World
Press Enter To Exit:
Can someone explain to me why this simple example below does not work.
In this example, the "helper" function contains a different function as a parameter ("setV" and "getV"). Within the "setV" function, the value of the variable is updated. Still, the value within the "getV" function remains the old value. What is the reason for this?
vari="Oh no... I'm old."
function init() {
helper setV
helper getV
}
function helper() {
($1)
}
function setV() {
vari="Hey! I'm new!"
}
function getV() {
echo $vari
}
init
The parens in ($1) cause $1 to be executed in a subshell. This means that any environment/variable changes are lost when the subshell exits.
Observe:
$ x=; setx() { x=Y; }; echo "1 x=$x"; (setx); echo "2 x=$x"; setx; echo "3 x=$x"
1 x=
2 x=
3 x=Y
If you want the variable changes to survive, don't put the command in a subshell.
Documentation
From man bash:
(list) list is executed in a subshell environment (see COMMAND
EXECUTION ENVIRONMENT below). Variable assignments and builtin
commands that affect the shell's environment do not remain in effect
after the command completes. The return status is the exit status of
list. [Emphasis added]
I got two variables in a bash script. One contains the name of a function within the script while the other one is an array containing KEY=VALUE or KEY='VALUE WITH SPACES' pairs. They are the result of parsing a specific file, and I can't change this.
What I want to do is to invoke the function whose name I got. This is quite simple:
# get the value for the function
myfunc="some_function"
# invoke the function whose name is stored in $myfunc
$myfunc
Consider the function foo be defined as
function foo
{
echo "MYVAR: $MYVAR"
echo "MYVAR2: $MYVAR2"
}
If I get the variables
funcname="foo"
declare -a funcenv=(MYVAR=test "MYVAR2='test2 test3'")
How would I use them to call foo with the pairs of funcenv being added to the environment? A (non-variable) invocation would look like
MYVAR=test MYVAR2='tes2 test3' foo
I tried to script it like
"${funcenv[#]}" "$funcname"
But this leads to an error (MYVAR=test: command not found).
How do I properly call the function with the arguments of the array put in its environment (I do not want to export them, they should just be available for the invoked function)?
You can do like this:
declare -a funcenv=(MYVAR=test "MYVAR2='test2 test3'")
for pairs in "${funcenv[#]}"; do
eval "$pairs"
done
"$funcname"
Note however that the variables will be visible outside the function too.
If you want to avoid that, then you can wrap all the above in a (...) subshell.
why don't you pass them as arguments to your function?
function f() { echo "first: $1"; echo "second: $2"; }
fn=f; $fn oneword "two words"
This is the code I am dealing with:
function execute {
task="$1"
servername="$2"
"$task" "${servername[#]}"
}
function someOtherThing {
val=$1
echo "$val"
}
function makeNecessaryDirectory {
arr=("$#")
echo "${arr[#]}"
}
dem=(1 2 3 4 5)
execute someOtherThing 1
execute makeNecessaryDirectory "${dem[#]}"
Output:
1
1
Expected output:
1
1 2 3 4 5
How to achieve this? I found no error logically.
Side question:
Is it safe to always receive 2nd parameter as an array inside execute so that it can deal with both of the dependent functions, or i should have an explicit check inside execute?
As explained in my comment
You are passing the array as individual args to execute and then only passing the first one the makeNecessaryDirectory, so $# is just the single argument passed which is 1.
I would do it this way, I have added comments to the parts i have changed.
It is only minor changes but should hopefully work for you.
#!/bin/bash
function execute {
task="$1"
servername="$2"
"$task" "$servername"
#No longer pass array here just pass servername as the name of array
}
function someOtherThing {
val=$1
echo "$val"
}
function makeNecessaryDirectory {
local arr=("${!1}")
#Indirect reference to the first arg passed to function which is now the
#name of the array
for i in "${arr[#]}";
do
echo "$i"
done
}
dem=(1 2 3 4 5)
execute someOtherThing 1
execute makeNecessaryDirectory 'dem[#]' #Pass the array name instead of it's contents
Is this the correct syntax for parameterized functions?
#!/bin/bash
twoPow()
{
prod=1
for((i=0;i<$1;i++));
do
prod=$prod*2
done
return prod
}
echo "Enter a number"
read num
echo `twoPow $num`
Output:
bash sample.sh
Enter a number
3
sample.sh: line 10: return: prod: numeric argument required
Part 2:
I removed the return, but what should I do if I want to run multiple times and store results like below? How can I make this work?
#!/bin/bash
tp1=1
tp2=1
twoPow()
{
for((i=0;i<$1;i++));
do
$2=$(($prod*2))
done
}
twoPow 3 tp1
twoPow 2 tp2
echo $tp1+$tp2
In Bash scripts you can't return values to the calling code.
The simplest way to emulate "returning" a value as you would in other languages is to set a global variable to the intended result.
Fortunately in bash all variables are global by default. Just try outputting the value of prod after calling that function.
A sample Bash function definition and call with a few parameters and return values. It may be useful and it works.
#!/bin/sh
## Define function
function sum()
{
val1=$1
val2=$2
val3=`expr $val1 + $val2`
echo $val3
}
# Call function with two parameters and it returns one parameter.
ret_val=$(sum 10 20)
echo $ret_val