`optional` is not inferred as optional field in TypeScript types

When using object with an optional field, that field is not optional in the inferred TypeScript type:

const TestStruct = object({
  foo: optional(string())
});

type Type = Infer<typeof TestStruct>; // { to: string }

I would expect the type to be inferred as { to?: string } instead.

[thesunny]

Hi @ianstormtaylor

My comment was ambiguous so to clarify…

I mean to suggest to add to the documentation that SuperStruct won’t work specifically with the default tsconfig. In other words, one must manually enable the option as a step on a new project. Because a lot of presets enable strictNullChecks or strict mode (which also enables strictNullChecks) it can be easy to forget that strict mode is not on by default.

Any chance to give me contributor access to this project? I know I can do a PR but I find I contribute more often when I can just pull and push.

[thesunny]

Hi @ianstormtaylor

I like that wording. I added a few details. Here’s what I ended up with which I updated and also added to the optional section of reference/types.md:

🤖 Warning: You must enable TypeScript’s strictNullChecks option in your tsconfig.json for Superstruct to work properly with “optional” types. Note that strictNullChecks is disabled by default. If you enable strict, strictNullChecks will automatically be enabled.

I think that’s clear enough. I’m closing this issue.