> ; doesn't do this, afaict

It's chaining together transitions in the state machine that is your program.

> when you're writing imperative code it's important that control flow (return) is explicitly visible

Control flow is visible with `?` or other do-notation variants. If I want to error out in a `Result` context, I explicitly return `Err(bad stuff)`. And if I don't, I explicitly return `Ok(return value)` instead. If I want to introduce a new asynchronous value in js, I explicitly call `new Promise`. And so on.

What's not visible in a do-block is the implementation of control flow. Which is fine, because this isn't the code that controls it - `Ok(Err(x))` is reduced in exactly the same way no matter what `x` is or where it came from. Traditional imperative code is the same way: the runtime system always works the same way, no matter which statements you ask it to execute.

If you do choose to expose the underlying mechanisms of your control flow everywhere, you get continuation-passing style, which is useful in small doses but more or less impossible to reason about at scale.