Remove `Option<!>` return types.

Several compiler functions have `Option<!>` for their return type.
That's odd. The only valid return value is `None`, so why is this type
used?

Because it lets you write certain patterns slightly more concisely. E.g.
if you have these common patterns:
```
    let Some(a) = f() else { return };
    let Ok(b) = g() else { return };
```
you can shorten them to these:
```
    let a = f()?;
    let b = g().ok()?;
```
Huh.

An `Option` return type typically designates success/failure. How should
I interpret the type signature of a function that always returns (i.e.
doesn't panic), does useful work (modifying `&mut` arguments), and yet
only ever fails? This idiom subverts the type system for a cute
syntactic trick.

Furthermore, returning `Option<!>` from a function F makes things
syntactically more convenient within F, but makes things worse at F's
callsites. The callsites can themselves use `?` with F but should not,
because they will get an unconditional early return, which is almost
certainly not desirable. Instead the return value should be ignored.
(Note that some of callsites of `process_operand`, `process_immedate`,
`process_assign` actually do use `?`, though the early return doesn't
matter in these cases because nothing of significance comes after those
calls. Ugh.)

When I first saw this pattern I had no idea how to interpret it, and it
took me several minutes of close reading to understand everything I've
written above. I even started a Zulip thread about it to make sure I
understood it properly. "Save a few characters by introducing types so
weird that compiler devs have to discuss it on Zulip" feels like a bad
trade-off to me. This commit replaces all the `Option<!>` return values
and uses `else`/`return` (or something similar) to replace the relevant
`?` uses. The result is slightly more verbose but much easier to
understand.

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