ZodType is not contravariant to the input

[mstniy]

Background

Contravariance is a property associated of generic types on their type parameters where the generic type G<A> extends G<B> if B extends A. It is usually associated with type parameters used as inputs.

Each ZodType is composed of an input and an output type. Consider, for instance, the following Zod schema:

const s = z.object({
    a: z.string().default("foo")
});

Its input type has a as an optional field, but it is required on the output type:

type A = z.input<typeof s>; // { a?: string | undefined; }
type B = z.output<typeof s>; // {a: string}

Description

Given that the input and output types are type parameters to ZodType, one might expect ZodType to be covariant on the output type and contravariant on the input type. While the former is true, the latter, unfortunately, is not.

Example

function f<S extends z.ZodType<string, any, any>>(_: S) {}
f(z.literal("a")); // Works fine

function g<S extends z.ZodType<any, any, "a">>(_: S) {}
g(z.string()); // Errors out :(

Expected behaviour

ZodType should be contravariant with respect to the input type

Actual behaviour

ZodType is not contravariant with respect to the input type