Optional parameter to function type

Question

Say I have the following (trivial) flow.js example:

type Example = (d: number, b?: string) => void

const f:Example = (d: number, boo: string) => {}

This fails to compile with:

const f:Example = (d: number, boo: string) => {}
                                      ^ Cannot assign function to `f` because string [1] is incompatible with undefined [2] in the second argument.
References:
3: const f:Example = (d: number, boo: string) => {}
                                      ^ [1]
1: type Example = (d: number, b?: string) => void
                                  ^ [2]

I'm trying to understand why? I thought the second parameter (b) was annotated as optional, eg if it does not exist that's fine, but if it exists it has to be a string?

Clearly this is not accurate. What's the actual explanation of what's going on here? And how can I have the behavior I'm after (eg, a function which takes either one or two arguments, and when the second argument exists it must be a string)

Try flow link: https://flow.org/try/#0C4TwDgpgBAogHgQwLZgDbQLxQBQBMBcUAdgK5IBGEATgDRTkD8hAzsFQJZEDmAlFBgD4oANwD27XAChJAY1FFWUAGb54yNJhwFiZSrXqjRLNp178hAbwC+QA

Answer 1

An optional parameter ? means the caller may or may not provide the argument. It does not mean that a particular callee can choose whether or not to ignore the parameter.

The error is telling you that the function f always expects a string argument even though it is not always provided, according to the Example type.

What you probably meant to do was:

type Example = (d: number, b: string) => void

const f: Example = (d: number, boo: string) => {}
const g: Example = (d: number) => {}

Try it out

This is valid because the string argument is always provided, but the callee can choose whether or not to use the argument.