let x? allows initial undefined value + type inference; let x? = y for simple y

[edemaine]

Previously, let x? compiled to the invalid TypeScript let x?. Now it compiles to let x = undefined. As this playground shows, this seems to allow for the initial undefined value (no error about access before initialized) as well as type inference.

This PR also makes let x? = 5 compile to let x = 5 (instead of leaving the invalid ? in there). I wanted to compile it to let x: undefined | typeof 5 = 5, but sadly this isn't valid TypeScript:

TypeScript intentionally limits the sorts of expressions you can use typeof on. Specifically, it’s only legal to use typeof on identifiers (i.e. variable names) or their properties.

[STRd6]

A good improvement. We could check for common literal types: string, number, bigint, and generate the corresponding type | undefined in those cases.

[edemaine]

Ah, good idea. I had actually written code to do the rewriting into typeof before realizing that it didn't always work, so I could use that, modified to work in just the good cases (member expression and identifier). And I added a new literalType helper for literal types, and error in the other cases.