Zod - Using optional() with default() infers the wrong type

Question

I'm using Zod with a field that is optional, but has a default value set, however, the inferred type says it could be undefined, even though it has a default value

const Schema = z.object({
  page: z.number().positive().optional().default(1)
})

type SchemaType = z.infer<typeof Schema>
// page?: number | undefined;

I tried using z.input and z.output, but they don't work either.

Is this intentional or is there another infer helper, which infers the parsed result?

Answer 1

Have you enabled strict mode in your tsconfig.json? It's not enabled by default (the default value for --strict is false) but Zod's installation requirements say that it must be enabled:

• TypeScript 4.5+!

• You must enable strict mode in your tsconfig.json. This is a best practice for all TypeScript projects.

  // tsconfig.json
  {
    // ...
    "compilerOptions": {
      // ...
      "strict": true
    }
  }
  

On my machine, using your code and the latest version of Zod (v3.21.4), SchemaType is inferred as:

• { page: number } when strict mode is enabled.
• { page?: number } when strict mode is disabled.