optional().default() getter doesn't allow returning undefined

Given the following snippet:

import { z } from "zod";

const checkIfOnPath = (executableName: string) => false;

const parsed = z.string()
  .nonempty()
  .optional()
  .default(() => ["a", "b", "c"].find(checkIfOnPath))
  .parse(undefined);

I get this error:

No overload matches this call.
  Overload 1 of 2, '(def: string): ZodDefault<ZodOptional<ZodString>>', gave the following error.
    Argument of type '() => string | undefined' is not assignable to parameter of type 'string'.
  Overload 2 of 2, '(def: () => string): ZodDefault<ZodOptional<ZodString>>', gave the following error.
    Type 'string | undefined' is not assignable to type 'string'.
      Type 'undefined' is not assignable to type 'string'.(2769)

(TS Playground link.)

My expectation is that .default() with a function on ZodOptional<ZodString> should be able to return string | undefined, but it only accepts a function that returns string.

Further, at runtime this works just fine – if the default getter returns undefined, .parse() doesn’t throw, and its return value is undefined as I expect.

[scotttrinh]

@lostfictions

But just to make sure I understand, this means that .optional().default() and .default() are effectively the same thing? Given this constraint, the former is effectively redundant and there’s never a reason one would reach for it, right?

Correct!

[scotttrinh]

The spirit and intention of default is to return a value of the base type (in this case string), as you’ve noticed. Since you’d like to define an optional default here, you can use transform instead:

CodeSandbox

const schema = z
  .string()
  .nonempty()
  .optional()
  .transform(
    (maybeString) => maybeString ?? ["a", "b", "c"].find(checkIfOnPath)
  );