z.infer vs z.input; the secret is in the types

Zod is great. Concise and simple and TypeScript oriented.

I recently ran into a scenario with an object with an optional property with a default. Thinking this schema would just work, I wrote up the schema and the tests and then suddenly it all fell a part.

For example, suppose you have a User with a name and a role. Then, name is required while role will default to guest if nothing else is supplied.

Here’s an zod schema:

const UserSchema = z.object({
  name: z.string(),
  role: z.string().default("guest"),

Here’s a small test, that fails:

const testUser: z.infer<typeof UserSchema> = {
  name: "name",
// Error
// Property 'role' is missing in type '{ name: string; }' but required in type '{ name: string; role: string; }'.

In theory, with .default having been set on role, not providing role in the object should be OK. And yet we have an error here.

I thought this error was because .optional was not defined in the schema:

const UserSchema = z.object({
  name: z.string(),
- role: z.string().default("guest"),
+ role: z.string().default("guest").optional(),

This seemed to make progress, getting rid of the type error, however the output was wrong:


Actually, I used the wrong input type. In fact, I did not use the input type, I used the output type. There’s a few hints in the docs, but it wasn’t very obvious to me either. The original schema was fine, but the test was wrong. Here’s how it should look:

- const testUser: z.infer<typeof UserSchema> = {
+ const testUser: z.input<typeof UserSchema> = {
  name: "name",

With z.input, you may safely omit the role property from the object.

Finally, the resulting shape is what we wanted all along:


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