Zod default without making input type optional

[BrendanC23]

I'm looking for a way to have a default value while keeping the input type unchanged. I want the default value to take effect when parse is called, but not when explicitly defining an object using the type.

const schemaWithDefault = z.object({
    a: z.string(),
    b: z.string().default(""),
});

// I want { a: string; b: string; }, not { a: string; b?: string | undefined; }
type SchemaWithDefaultType = z.input<typeof schemaWithDefault>;

const data: unknown = { a: "a" };
const parsedData = schemaWithDefault.parse(data); // .default() works, b is set to "" as expected

const ExplicitlyDefinedType: SchemaWithDefaultType = {
    a: z.string(),
    // I want a TypeScript error because b is not defined
};

I've also looked at catch(), but that makes the input type unknown.

const schemaWithDefault = z.object({
    a: z.string(),
    b: z.string().catch(""),
});

// I want { a: string; b: string; }, not { a: string; b: unknown; }
type SchemaWithDefaultType = z.input<typeof schemaWithDefault>; 

Is there a way to accomplish this?

[ru-van-urk]

@BrendanC23 Did you find a solution?

[BrendanC23]

No. This would be simple if there was a way to provide default values when calling parse instead of when defining the schema, but for now there's no built-in way that I've found.

[BrendanC23]

I ended up using this (Zod V3):

function defaultButRequired<T extends z.ZodTypeAny>(schema: T, defaultValue: z.output<T>) {
    return schema.optional().default(defaultValue) as unknown as T;
}

const schema = z.object({
    name: defaultButRequired(z.string(), "default"),
});

type Input = z.input<typeof schema>; // { name: string }
type Output = z.output<typeof schema>; // { name: string }

const data = schema.parse({}); // { name: "default" }