Function behaviour on shell(ksh) script - shell

Here are 2 different versions of a program:
this
Program:
#!/usr/bin/ksh
printmsg() {
i=1
print "hello function :)";
}
i=0;
echo I printed `printmsg`;
printmsg
echo $i
Output:
# ksh e
I printed hello function :)
hello function :)
1
and
Program:
#!/usr/bin/ksh
printmsg() {
i=1
print "hello function :)";
}
i=0;
echo I printed `printmsg`;
echo $i
Output:
# ksh e
I printed hello function :)
0
The only difference between the above 2 programs is that printmsg is 2times in the above program while printmsg is called once in the below program.
My Doubt arises here: To quote
Be warned: Functions act almost just like external scripts... except
that by default, all variables are SHARED between the same ksh
process! If you change a variable name inside a function.... that
variable's value will still be changed after you have left the
function!!
But we can clearly see in the 2nd program's output that the value of i remains unchanged. But we are sure that the function is called as the print statement gets the the output of the function and prints it. So why is the output different in both?

When you use backticks (or $(...)), you execute it in a subshell.
That is, a new shell is started (which inherits from the current one) and then exists.
Edit: I checked your link, if you read the bottom of it, the very last section, you'll see this explained.

Related

Why can't I store the return value of a shell function by variable assignment? [duplicate]

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:

How to use the numeric value returned by a function? [duplicate]

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:

Return value in a Bash function

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:

Bash Function is not getting called, unless I echo the return value

In my program I am trying to return a value from a function, the return value is string. Everything works fine(atleast some part), if I echo the value once it is returned, but it is not even calling the function, if I dont return.... Consider the code below....
#!/bin/bash
function get_last_name() {
echo "Get Last Name"
ipath=$1
IFS='/'
set $ipath
for item
do
last=$item
done
echo $last
}
main() {
path='/var/lib/iscsi/ifaces/iface0'
current=$(get_last_name "$path")
echo -n "Current="
echo $current
}
main
It gives me an output like this
OUTPUT
Current=Get Last Name iface0
If I comment the echo $current, then the I am not even seeing the "Get Last Name", which makes to come to conclusion, that it is not even calling the function. Please let me know what mistake I am making. But one thing I am sure, bash is the ugliest language I have ever seen.......
Functions do not have return values in bash. When you write
current=$(get_last_name "$path")
you are not assigning a return value to current. You are capturing the standard output of get_last_name (written using the echo command) and assigning it to current. That's why you don't see "Get last name"; that text does not make it to the terminal, but is stored in current.
Detailed explanation
Let's walk through get_last_name first (with some slight modifications to simplify the explanation):
function get_last_name () {
ipath=$1
local IFS='/'
set $ipath
for item
do
last=$item
done
echo "Get Last Name"
echo $last
}
I added the local command before IFS so that the change is confined to the body of get_last_name, and I moved the first echo to the end to emphasize the similarity between the two echo statements. When get_last_name is called, it processes its single argument (a string containing a file path), then echoes two strings: "Get Last Name" and the final component of the file path. If you were to run execute this function from the command line, it would appear something like this:
$ get_last_name /foo/bar/baz
Get Last Name
baz
The exit code of the function would be the exit code of the last command executed, in this case echo $last. This will be 0 as long as the write succeeds (which it almost certainly will).
Now, we look at the function main, which calls get_last_name:
main() {
path='/var/lib/iscsi/ifaces/iface0'
current=$(get_last_name "$path")
echo -n "Current="
echo $current
}
Just like with get_last_name, main will not have a return value; it will produce an exit code which is the exit code of echo $current. The function begins by calling get_last_name inside a command substitution ($(...)), which will capture all the standard output from get_last_name and treat it as a string.
DIGRESSION
Note the difference between the following:
current=$(get_last_name "$path")
sets the value of current to the accumulated standard output of get_last_name. (Among other things, newlines in the output are replaced with spaces, but the full explanation of how whitespace is handled is a topic for another day). This has nothing to do with return values; remember, the exit code (the closet thing bash has to "return values") is a single integer.
current=get_last_name "$path"
would not even call get_last_name. It would interpret "$path" as the name of a command and try to execute it. That command would have a variable current with the string value "get_last_name" in its environment.
The point being, get_last_name doesn't "return" anything that you can assign to a variable. It has an exit code, and it can write to standard output. The $(...) construct lets you capture that output as a string, which you can then (among other things) assign to a variable.
Back to main
Once the value of current is set to the output generated by get_last_name, we execute
two last echo statements to write to standard output again. The first writes "Current=" without a newline, so that the next echo statement produces text on the same line as the first. The second just echoes the value of current.
When you commented out the last echo of main, you didn't stop get_last_name from being executed (it had already been executed). Rather, you just didn't print the contents of the current variable, where the output of get_last_name was placed rather than on the terminal.

Bash: pass a function as parameter

I need to pass a function as a parameter in Bash. For example, the following code:
function x() {
echo "Hello world"
}
function around() {
echo "before"
eval $1
echo "after"
}
around x
Should output:
before
Hello world
after
I know eval is not correct in that context but that's just an example :)
Any idea?
If you don't need anything fancy like delaying the evaluation of the function name or its arguments, you don't need eval:
function x() { echo "Hello world"; }
function around() { echo before; $1; echo after; }
around x
does what you want. You can even pass the function and its arguments this way:
function x() { echo "x(): Passed $1 and $2"; }
function around() { echo before; "$#"; echo after; }
around x 1st 2nd
prints
before
x(): Passed 1st and 2nd
after
I don't think anyone quite answered the question. He didn't ask if he could echo strings in order. Rather the author of the question wants to know if he can simulate function pointer behavior.
There are a couple of answers that are much like what I'd do, and I want to expand it with another example.
From the author:
function x() {
echo "Hello world"
}
function around() {
echo "before"
($1) <------ Only change
echo "after"
}
around x
To expand this, we will have function x echo "Hello world:$1" to show when the function execution really occurs. We will pass a string that is the name of the function "x":
function x() {
echo "Hello world:$1"
}
function around() {
echo "before"
($1 HERE) <------ Only change
echo "after"
}
around x
To describe this, the string "x" is passed to the function around() which echos "before", calls the function x (via the variable $1, the first parameter passed to around) passing the argument "HERE", finally echos after.
As another aside, this is the methodology to use variables as function names. The variables actually hold the string that is the name of the function and ($variable arg1 arg2 ...) calls the function passing the arguments. See below:
function x(){
echo $3 $1 $2 <== just rearrange the order of passed params
}
Z="x" # or just Z=x
($Z 10 20 30)
gives: 30 10 20, where we executed the function named "x" stored in variable Z and passed parameters 10 20 and 30.
Above where we reference functions by assigning variable names to the functions so we can use the variable in place of actually knowing the function name (which is similar to what you might do in a very classic function pointer situation in c for generalizing program flow but pre-selecting the function calls you will be making based on command line arguments).
In bash these are not function pointers, but variables that refer to names of functions that you later use.
there's no need to use eval
function x() {
echo "Hello world"
}
function around() {
echo "before"
var=$($1)
echo "after $var"
}
around x
You can't pass anything to a function other than strings. Process substitutions can sort of fake it. Bash tends to hold open the FIFO until a command its expanded to completes.
Here's a quick silly one
foldl() {
echo $(($(</dev/stdin)$2))
} < <(tr '\n' "$1" <$3)
# Sum 20 random ints from 0-999
foldl + 0 <(while ((n=RANDOM%999,x++<20)); do echo $n; done)
Functions can be exported, but this isn't as interesting as it first appears. I find it's mainly useful for making debugging functions accessible to scripts or other programs that run scripts.
(
id() {
"$#"
}
export -f id
exec bash -c 'echowrap() { echo "$1"; }; id echowrap hi'
)
id still only gets a string that happens to be the name of a function (automatically imported from a serialization in the environment) and its args.
Pumbaa80's comment to another answer is also good (eval $(declare -F "$1")), but its mainly useful for arrays, not functions, since they're always global. If you were to run this within a function all it would do is redefine it, so there's no effect. It can't be used to create closures or partial functions or "function instances" dependent on whatever happens to be bound in the current scope. At best this can be used to store a function definition in a string which gets redefined elsewhere - but those functions also can only be hardcoded unless of course eval is used
Basically Bash can't be used like this.
A better approach is to use local variables in your functions. The problem then becomes how do you get the result to the caller. One mechanism is to use command substitution:
function myfunc()
{
local myresult='some value'
echo "$myresult"
}
result=$(myfunc) # or result=`myfunc`
echo $result
Here the result is output to the stdout and the caller uses command substitution to capture the value in a variable. The variable can then be used as needed.
You should have something along the lines of:
function around()
{
echo 'before';
echo `$1`;
echo 'after';
}
You can then call around x
eval is likely the only way to accomplish it. The only real downside is the security aspect of it, as you need to make sure that nothing malicious gets passed in and only functions you want to get called will be called (along with checking that it doesn't have nasty characters like ';' in it as well).
So if you're the one calling the code, then eval is likely the only way to do it. Note that there are other forms of eval that would likely work too involving subcommands ($() and ``), but they're not safer and are more expensive.

Resources