Set shell variable then invoke subprocess fails

Question

I was playing with getopts, which has a shell variable OPTERR that influences its behavior. I wanted to change the value of OPTERR and invoke getopts on the same line, to affect the behavior of getopts but restore OPTERR to its default value thereafter. So, given that I'm a little fuzzy on the process of mixing a variable change a command invocation, I thought I'd experiment with just a humble echo. Here is a series of shell commands with their outputs, some of which don't make sense to me.

$ echo OPTERR
1
$ OPTERR=0 echo $OPTERR
1

Now, let's stop there. Shouldn't the output for the second command be '0'? So before asking the question, I wracked my brains a bit and thought it must be because following command should be in a sub shell. So then I tried...

$ OPTERR=0 (echo $OPTERR)
-bash: syntax error near unexpected token `('

This was an even bigger surprise. How does one change the value of shell variable for the run of a sub shell, if the above construct is a syntax error? (I also tried putting a terminal ";" after the echo command with the same result). I also tried...

$OPTERR=0 bash -c 'echo $OPTERR'
1

So, no joy there either. I took a different tack...

$function sf() { echo $OPTERR }
$sf
1
$OPTERR=0 sf
0
$sf
1

Success!! But why? Functions are specifically not sub shells (I don't think) and this isn't really what I want (though, I could probably make it work for my application). Here's another formula that worked (sort of):

$function sf() { (OPTERR=0; echo $OPTERR;) }
$sf
0
$echo $OPTERR
1

This is PRETTY close to what I want, but of course the intended target for the actual script (not my trivial echo) is getopts, and getopts has to change variables that the rest of the script can see. So, I changed my trivial echo script to a simple variable set, as in ...

$function sf() { foo='bar'; (foo='check'; echo $foo;); echo $foo; }
$sf
check
bar

D'oh!! So I can call getopts with the environment I want, but I can't get any values back out of the sub shell. I tried exporting foo and all kinds of things too embarrassing to list here, but here's my question: if you put all my experiments together, they must reveal a fundamentally wrong model of how all this should be working. What is the missing piece to my understanding, and what's the right thing to do?

Just to re-iterate, all I'm trying to do is build a function, that uses getopts with OPTERR set to 0, ideally without setting OPTERR on the line where the function is invoked, and in a way that OPTERR is restored outside the function, and most importantly in a way that allows me to actually process the output of getopts. It sounds like a lot to ask, but it's pretty sensible and the existence of this OPTERR variable suggests it should be possible.

Answer 1

$ echo $VAR

This gives the old value of $VAR because the shell applies variable assignments only after the expansions on the command line.

The IFS= read -r foo construct you often see here on the site works, however, since read doesn't take IFS from the command line, but uses it internally. Similarly, OPTERR=0 getopts ... should also work.

$ cat arg.sh
#!/bin/bash
OPTERR=0 getopts a:b var
echo "var: $var OPTERR: $OPTERR"

$ bash arg.sh -z
var: ? OPTERR: 1

No error message even though -z is an invalid option.

---

$ OPTERR=0 bash -c 'echo $OPTERR'

This is a bit of a special case, because OPTERR is special. Bash's man page says:

OPTERR If set to the value 1, Bash displays error messages generated by the getopts builtin command. OPTERR is initialized to 1 each time the shell is invoked or a shell script is executed.

The idea is right, though, it works with regular variables:

$ FOO=123 bash -c 'echo $FOO'
123

---

all I'm trying to do is build a function, that uses getopts

If you're parsing the arguments to the function, and calling it multiple times from the main program (e.g. dothis -a; dothis -b -c), and not using the function to parse arguments to the shell (parse_args "$@"), then you'll also need to remember that getopts modifies the OPTIND variable, too. It will also need to be made local to the function.

E.g. here, the second function call doesn't see any of its options:

foo() { while getopts "abcd" opt; do echo "$opt"; done; };
foo -a -a
foo -b -b

Combined with the previous, this function would work, silently:

foo() {
    local OPTIND=1
    while OPTERR=0 getopts "abcd" opt; do
        echo "$opt"
    done
}

Answer 2

If you need OPTERR to be 0 during a function call, set it as a local variable, as Gordon Davisson indicated:

#!/bin/bash

myfunc() {
  local OPTERR=0
  printf "Inside myfunc, OPTERR=%d\n" "$OPTERR"
  while getopts ":a:" opt; do
    echo $opt is $OPTARG
  done
}

printf "Before calling myfunc, OPTERR=%d\n" "$OPTERR"
myfunc
printf "After calling myfunc, OPTERR=%d\n" "$OPTERR"

The result is:

Before calling myfunc, OPTERR=1
Inside myfunc, OPTERR=0
After calling myfunc, OPTERR=1