Use nonstrict with unknown instead of any

Currently when .nonstrict() is used, [k: string]: any appears on the inferred type. Is it possible to use [k: string]: unknown instead? The problem of using any is that TypeScript allows you to do anything on any property. If you have a typo on the property name, an unexpected undefined will appear at runtime. If unknown is used, at least TypeScript can report an error when you try to use it or assign it to a different variable.

This feature request is similar to https://github.com/vriad/zod/issues/104. I am not sure if this issue will be considered a duplicate of it, but I think .nonstrict() will be much safer to use if it uses unknown instead of any by default.

[tup1tsa]

I think not having index signatures at all is a better idea. I don’t want to use unknown keys that weren’t validated by zod schema. Having [index: string]: unknown can still cause some problems. You can make a typo and unknown type will be used. If the function not that strict, you may not even notice it. Good example is a string constructor (via backquotes)

type SchemaResult = {
  propA: string;
  propB: number;
  [index: string]: unknown;
};
const result: SchemaResult = {
  propA: "",
  propB: 23,
};
const stringWithUnknown = `${result.prop3}`;

This will not cause typescript to stall here.

[colinhacks]

Zod 2 is now in beta and sidesteps this issue by getting rid of index signatures altogether. Objects are “nonstrict” by default (in that they allow unknown keys) but the inferred type contains no index signature unless you use the .catchall() method to provide one: https://github.com/vriad/zod/tree/v2#catchall