I run my own script to dump databases into files on a nightly basis.
I wanted to count time (in seconds) it takes to dump each database, so I was trying to write some functions to help me achieve it, but I'm running into problems.
I am no expert in scripting in bash, so if I'm doing it plain wrong, just say so and ideally suggest alternative, please.
Here's the script:
#!/bin/bash
declare -i time_start
function get_timestamp {
declare -i time_curr=`date -j -f "%a %b %d %T %Z %Y" "\`date\`" "+%s"`
echo "get_timestamp:" $time_curr
return $time_curr
}
function timer_start {
get_timestamp
time_start=$?
echo "timer_start:" $time_start
}
function timer_stop {
get_timestamp
declare -i time_curr=$?
echo "timer_stop:" $time_curr
declare -i time_diff=$time_curr-$time_start
return $time_diff
}
timer_start
sleep 3
timer_stop
echo $?
The code should really be quite self-explanatory. echo commands are only for debugging.
I expect the output to be something like this:
$ bash timer.sh
get_timestamp: 1285945972
timer_start: 1285945972
get_timestamp: 1285945975
timer_stop: 1285945975
3
Now this is not the case unfortunately. What I get is:
$ bash timer.sh
get_timestamp: 1285945972
timer_start: 116
get_timestamp: 1285945975
timer_stop: 119
3
As you can see, the value that local var time_curr gets from the command is a valid timestamp, but returning this value causes it to be changed to an integer between 0 and 255.
Can someone please explain to me why this is happening?
PS. This obviously is just my timer test script without any other logic.
UPDATE
Just to be perfectly clear, I want this to be part of a bash script very similar to this one, where I want to measure each loop cycle.
Unless of course I can do it with time, then please suggest a solution.
You don't need to do all this. Just run time <yourscript> in the shell.
$? is used to hold the exit status of a command and can only hold a value between 0 and 255. If you pass an exit code outside this range (say, in a C program calling exit(-1)), the shell will still receive a value in that range and set $? accordingly.
As a workaround, you could just set a different value in your bash function:
function get_timestamp {
declare -i time_curr=`date -j -f "%a %b %d %T %Z %Y" "\`date\`" "+%s"`
echo "get_timestamp:" $time_curr
get_timestamp_return_value=$time_curr
}
function timer_start {
get_timestamp
#time_start=$?
time_start=$get_timestamp_return_value
echo "timer_start:" $time_start
}
...
I believe you should be able to use the existing "time" function.
After Update to the question:
This was the bit of script from your link which was doing a for loop.
# dump each database in turn
for db in $databases; do
echo $db
$MYSQLDUMP --force --opt --user=$USER --password=$PASSWORD
--databases $db > "$OUTPUTDIR/$db.bak"
done
You could extract the inner portion of the loop into a new script (call it dump_one_db.sh)
and do this inside the loop:
# dump each database in turn
for db in $databases; do
time dump_one_db.sh $db
done
Make sure to write the output of the time against the db name into some file.
This is happening because return codes need to be between 0-255. You can't return an arbitrary number. If you continue to refuse to use the builtin time function and roll your own, change your functions to echo their stamp and use a process expansion ($()) to grab the value.
Related
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]: $f "
echo
##----------------------------------------------##
read -p "Press Enter To Exit:"
Expected output:
FN_MULTIPLE_RETURN_VALUES EXAMPLES:
-------------------------------------------
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[0]: Hello
RET_ARR[1]: World
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[2]: Hello
RET_ARR[3]: World
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[4]: Hello
RET_ARR[5]: World
Press Enter To Exit:
I am working with a bash script and I want to execute a function to print a return value:
function fun1(){
return 34
}
function fun2(){
local res=$(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]: $f "
echo
##----------------------------------------------##
read -p "Press Enter To Exit:"
Expected output:
FN_MULTIPLE_RETURN_VALUES EXAMPLES:
-------------------------------------------
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[0]: Hello
RET_ARR[1]: World
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[2]: Hello
RET_ARR[3]: World
Running: FN_MULTIPLE_RETURN_VALUES
RET_ARR[4]: Hello
RET_ARR[5]: World
Press Enter To Exit:
Hellow fellow Stack users,
I am using a function "setup_date" to change date command for my custom execution. This is to mock it and test some bash scripts, their execution has to use always the same date in order to compare results.
So this approach worked very good with ssh or sftp command mocking. But this time, just after "date" command substitution, the script's execution becomes very slow! What is the reason for that? Is "date" command called frequently by linux system for internal uses?
Regards,
#
#replace a command with previousy defined mock one
#
mock_cmd() {
local command="${1:-}"
local override="${2:-}"
# Remove target function if one is already set
unset ${command}
# Create a wrapper function called "${command}"
eval "${command}() { ${override} \${#}; }"
}
#mock the date command
#1- date formatting
#There has to be a variable: dateFile!
date_mock_SP() {
date "${1}" -r ${dateFile}
}
#
#1- date ex: 201203101513
#2- dateFile path
setup_date() {
touch -t "${1}" ${2}/dateFile
export dateFile=${2}/dateFile
}
EXECUTION :
mock_cmd "date" "date_mock_SP"
setup_date "201203101513" ${pwd}/in
Date=$(date +"%y%j")
echo $Date
Date=$(date +"%y%j")
echo $Date
exit 1
mock_cmd is brittle and more complicated than you need. You are already defining the function date_mock_SP; just name it date, and the function will override the command. Inside the function, use command date to avoid infinite recursion.
date () { command date "$1" -r "$dateFile"; }
setup_date "201203101513" "$pwd/in" # uses the function, not the executable, date
unset -f date
What I try to achieve, is to define global variables in my script. These variables can be reused in a loop (preferably a while loop..) and with every iteration, the loop should get a new set a variables.
My script (so far):
PACKAGE_ASSET_ID=AUTO`date +%s`000001
TITLE_ASSET_ID=AUTO`date +%s`000002
MOVIE_ASSET_ID=AUTO`date +%s`000003
PREVIEW_ASSET_ID=AUTO`date +%s`000004
POSTER_ASSET_ID=AUTO`date +%s`000005
while read name; do
#DATE=`date +%s`
#PACKAGE_ASSET_ID="AUTO${DATE}000001"
#TITLE_ASSET_ID="AUTO${DATE}000002"
#MOVIE_ASSET_ID="AUTO${DATE}000003"
#PREVIEW_ASSET_ID="AUTO${DATE}000004"
#POSTER_ASSET_ID="AUTO${DATE}000005"
echo $PACKAGE_ASSET_ID
echo $TITLE_ASSET_ID
echo $MOVIE_ASSET_ID
echo $PREVIEW_ASSET_ID
echo $POSTER_ASSET_ID
done <names.txt
Within the file names.txt, there are 15 entries. For every entry, the while loop needs to process these sets of variables. Giving me something like
AUTO1521884581000001
AUTO1521884581000002
AUTO1521884581000003
AUTO1521884581000004
AUTO1521884581000005
AUTO1521884592000001
AUTO1521884592000002
AUTO1521884592000003
AUTO1521884592000004
AUTO1521884592000005
As you can see in the script, I tried putting it into the while loop and with different syntax but regretfully, without success. The results I get are always the same set of variables, for all the 15 entries.
Did you really expect that bash (even bash!) would need more than a second to read one line?? Try adding the nanoseconds.
while read name; do
DATE=$(date +%s%N)
PACKAGE_ASSET_ID="AUTO${DATE}000001"
TITLE_ASSET_ID="AUTO${DATE}000002"
MOVIE_ASSET_ID="AUTO${DATE}000003"
PREVIEW_ASSET_ID="AUTO${DATE}000004"
POSTER_ASSET_ID="AUTO${DATE}000005"
echo $PACKAGE_ASSET_ID
echo $TITLE_ASSET_ID
echo $MOVIE_ASSET_ID
echo $PREVIEW_ASSET_ID
echo $POSTER_ASSET_ID
done <names.txt
I am working with a bash script and I want to execute a function to print a return value:
function fun1(){
return 34
}
function fun2(){
local res=$(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: