Does declaring an object property as unknown make it optional?

[aryzing]

Moved to https://discord.com/channels/1252985447273992222/1255508856608522341/1255508856608522341

[harshagarwalnyu]

No — unknown() and optional() are independent concepts in Valibot:

• v.unknown() — the property is required but accepts any value (no type constraint). It must be present in the input.
• v.optional(schema) — the property may be absent from the input (or explicitly undefined). The wrapped schema only runs when the value is present.

So { key: v.unknown() } requires key to exist but accepts any value for it. If key is missing, parsing fails.

To make a property accept any value AND be optional:

import * as v from 'valibot';

const Schema = v.object({
  required_any: v.unknown(),           // required, any value
  optional_typed: v.optional(v.string()), // optional, must be string if present
  optional_any: v.optional(v.unknown()),  // optional, any value if present
});

// All three parse successfully:
v.parse(Schema, { required_any: 42 });
v.parse(Schema, { required_any: 42, optional_typed: 'hello' });
v.parse(Schema, { required_any: 42, optional_any: { nested: true } });

// This fails — required_any is missing:
v.parse(Schema, {});

Summary:

Schema	Required?	Type constraint
v.unknown()	Yes	None (any value)
v.optional(v.string())	No	string if present
v.optional(v.unknown())	No	None (any value if present)
v.nullable(v.string())	Yes	string or null