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44185520cf
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.
r? ``````@cjgillot``````
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