Implicit return in bash functions?

Question

Say I have a bash function like so:

gmx(){
  echo "foo";
}

will this function implicitly return the exit value of the echo command, or is using return necessary?

gmx(){
  echo "foo";
  return $?
}

I assume that the way bash works, the exit status of the final command of the bash function is the one that gets "returned", but not 100% certain.

Answer 1

return does an explicit return from a shell function or "dot script" (a sourced script). If return is not executed, an implicit return is made at the end of the shell function or dot script.

If return is executed without a parameter, it is equivalent of returning the exit status of the most recently executed command.

That is how return works in all POSIX shells.

For example,

gmx () {
  echo 'foo'
  return "$?"
}

is therefore equivalent to

gmx () {
  echo 'foo'
  return
}

which is the same as

gmx () {
  echo 'foo'
}

In general, it is very seldom that you need to use $? at all. It is really only needed if you need to save it for future use, for example if you need to investigate its value multiple times (in which case you would assign its value to a variable and perform a series of tests on that variable).

Answer 2

From the bash(1) man page:

When executed, the exit status of a function is the exit status of the last command executed in the body.

Answer 3

I'll just add a few notes of caution to the answers already provided:

• Even though return has a very special meaning to the shell, from a syntax point of view, it is a shell builtin command and a return statement is parsed like any other simple command. So, that means that like in the argument of any other command, $? when unquoted, would be subject to split+glob

  So you need to quote that $? to avoid it:

  return "$?"
  

• return usually doesn't accept any option (ksh93's accepts the usual --help, --man, --author... though). The only argument it expects (optional) is the return code. The range of accepted return codes varies from shell to shell, and whether any value outside 0..255 is properly reflected in $? also varies from shell to shell. See Default exit code when process is terminated? for details on that.

  Most shells accept negative numbers (after all, the argument passed to the _exit()/exitgroup() system call is an int, so with values encompassing at least -2³¹ to 2³¹-1, so it only makes sense that shells accept the same range for its functions).

  Most shells use the waitpid() and co. API to retrieve that exit status however in which case, it's truncated to a number between 0 and 255 when stored in $?. Even though invoking a function does not involve spawning a process and used waitpid() to retrieve its exit status as all is done in the same process, many shells also mimic that waitpid() behaviour when invoking functions. Which means that even if you call return with a negative value, $? will contain a positive number.

  Now, among those shells whose return accepts negative numbers (ksh88, ksh93, bash, zsh, pdksh and derivatives other than mksh, yash), there are a few (pdksh and yash) which need it written as return -- -123 as otherwise that -123 is taken as three -1, -2, -3 invalid options.

  As pdksh and its derivatives (like OpenBSD sh or posh) preserve the negative number in $?, that means that doing return "$?" would fail when $? contains a negative number (which would happen when the last run command was a function that returned a negative number).

  So return -- "$?" would be better in those shells. However note that while supported by most shells, that syntax is not POSIX and in practice not supported by mksh and ash derivatives.

  So, to sum up, with pdksh-based shells, you may use negative numbers in arguments to functions, but if you do, return "$@" won't work. In other shells, return "$@" will work and you should avoid using negative numbers (or numbers outside of 0..255) as arguments to return.

• In all shells that I know, calling return from inside a subshell running inside a function will cause the subshell to exit (with the provided exit status if any or that of the last command run), but will not otherwise cause a return from the function (to me, it's unclear whether POSIX gives you that warranty, some argue that exit should be used instead of exit subshells inside functions). For instance

  f() {
    (return 3)
    echo "still inside f. Exit status: $?"
  }
  f
  echo "f exit status: $?"
  

  will output:

  still inside f. Exit status: 3
  f exit status: 0
  

Answer 4

Yes, the implicit return value of a function is the exit status of the last executed command. That is also true at any point of any shell script. At any point in the script execution sequence, the present exit status is the exit status of the last command executed. Even command executed as part of a variable assignment: var=$(exit 34). The difference with functions is that a function could change the exit status at the end of execution of the function.

The alternative way to change the "present exit status" is to start a sub shell and exit it with any needed exit status:

$ $(exit 34)
$ echo "$?"
34

And yes, the exit status expansion do need to be quoted:

$ IFS='123'
$ $(exit 34)
$ echo $?
4

A (exit 34) also work.
Some may argue that a more robust construct should be $(return 34), and that an exit should "exit" the script being executed. But $(return 34) does not work with any version of bash. So, it is not portable.

The safest way to set an exit status is to use it as it was designed to work, define and return from a function:

exitstatus(){ return "${1:-"$?"}"; }

So, in the end of a function. it is exactly equivalent to have either nothing or return or return "$?". The end of a function does not need to mean the "last code line of a function".

#!/bin/sh
exitstatus(){ a="${1:-"$?"}"; return "$a"; }
gmx(){
    if     [ "$1" = "one" ]; then
           printf 'foo ';
           exitstatus 78
           return "$?"
    elif   [ "$1" = "two" ]; then
           printf 'baz ';
           exitstatus 89
           return
    else
           printf 'baz ';
           exitstatus 90
    fi
}  

Will print:

$ ./script
foo 78
baz 89
baz 90

The only practical use for "$?" is to either print its value: echo "$?" or to store it in a variable (as it is an ephemeral value and change with every command executed): exitstatus=$? (remember to quote the variable in commands like export EXITSTATUS="$?".

In the return command, the valid range of values is generally 0 to 255, but understand that values of 126 + n are used by some shells to signal special exit status, so, the general recommendation is to use 0-125.