Doubts about type inference when using picks

Hello everyone, when calling the pick method of a zodObject in a function by passing in a reference, how to set the type of T to ensure the correct type of return value？

function zodPickDemo<T extends ZodObject<ZodRawShape>>(src: T) {
  return src.pick({ id: true });
}

const dto = z.object({ id: z.string(), name: z.string() });
const picked = zodPickDemo(dto);
// In this case the type of picked is not the expected type

In the above code, the type expected to be obtained is a zodObject object type containing only the id, but what is obtained is a type such that:

z.ZodObject<Pick<z.ZodRawShape, never>, "strip", z.ZodTypeAny, {}, {}>

Playground: link

[AustinShelby]

I managed to get the correct functionality with this code:

import type { ZodObject, ZodTypeAny } from 'zod';
import { z } from 'zod';

function zodPickDemo<T extends ZodTypeAny>(src: ZodObject<{id: T}>): ZodObject<{id: T}> {
  return src.pick({ id: true });
}

const dto = z.object({ id: z.string(), name: z.string() });
const picked = zodPickDemo(dto);
picked.shape.id; // z.ZodString

Does this solve your problem?

[scotttrinh]

Thanks @AustinShelby for providing the working code, that does seem to be the correct way to address the original question.