Return value in a Bash function

Question

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?

Answer 1

As explained by @andreas-spindler above you have two channels of communication:

1. echo in function (become STDOUT+STDERR of function). You can get it in invoker code through "res=$(func1)"
2. exit code of function. You can get it in invoker code through "res=$?"

Example of code:

function fun1() {
        echo 35
        return 34
    }
function fun2() {
        res1=$(fun1) # <- this will get all "fun1" STDOUT and STDERR
        res2=$?  # <- this will get exit code of "fun1" executed in previous line
        echo "res1=$res1"
        echo "res2=$res2"
    }

if you then run in shell "fun2" you will get such output:

res1=35
res1=34

Answer 2

Faced with the same issue, but in my case I have to use -set e So my workaround is to use combination of && and ||

#!/bin/bash -e

function func1() {
   return ${1:-0}
}

function func2() {
   local RVAL=""

   RVAL=$(func1 ${1:-0} && echo 0 || echo $?)
   if [[ $RVAL != "0" ]]; then
      echo "Some logic on exit code $RVAL"
      exit $RVAL
   fi

   echo "Some logic on exit code 0"
}

func2 ${1:-0}

Some basic tests

$ ./test.sh
Some logic on exit code 0

$ echo $?
0

$ ./test.sh 128
Some logic on exit code 128

$ echo $?
128

Answer 3

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!

Answer 4

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:

1. Print to stdout/stderr within the function scope (without returning it)
2. Return (set) multiple variables.

Answer 5

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.

Answer 6

Another way to achieve this is name references (requires Bash 4.3+).

function example {
  local -n VAR=$1
  VAR=foo
}

example RESULT
echo $RESULT

Answer 7

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:

Answer 8

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"
}

Answer 9

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.

Answer 10

$(...) 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
}

Answer 11

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.

Answer 12

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

Answer 13

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
}