[v4]: .default breaks optional fields in json schema

[dariozachow]

When using the z.toJSONSchema optional fields will be marked as required when using the .default on this field.

Example:

z.toJSONSchema(
    z.object({
      test: z.string().optional().default('abc')
    })
  )

will result in the following json schema:

  '$schema': 'https://json-schema.org/draft/2020-12/schema',
  type: 'object',
  properties: { test: { default: 'abc', type: 'string' } },
  required: [ 'test' ],
  additionalProperties: false
}

Even though test is not required, because its marked as optional.

When using optional without default works perfectly fine:

  z.toJSONSchema(
    z.object({
      test: z.string().optional()
    })
  )

will become

{
  '$schema': 'https://json-schema.org/draft/2020-12/schema',
  type: 'object',
  properties: { test: { type: 'string' } },
  additionalProperties: false
}

Using zod 3.25.64

[dosubot]

Hi @dariozachow! I'm Dosu and I’m helping the zod team.

This is expected behavior in Zod v4: when you use .default() on a field, it will be marked as required in the generated JSON schema, even if you also use .optional(). This is intentional and aligns with one interpretation of the JSON Schema spec—fields with defaults are considered required because the schema assumes a value will always be present, either from input or by applying the default [source].

If you want the field to remain optional in the JSON schema but still have a default at runtime, you can use .prefault() instead of .default(). This method allows you to specify a default value without marking the field as required in the schema [source].

You can find more details and migration guidance in the Zod v4 changelog. If this answers your question, feel free to close the issue!

To reply, just mention @dosu.

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[samchungy]

By default the io is set to output. You'll need to pass in input as the second argument to toJSONSchema.

eg.

z.toJSONSchema(mySchema, { io: 'input' });

[dariozachow]

thanks, that works :)