How to allow Typescript Generic parameter to have default

Issue

I am trying to set a default for a generic function in TS. Here is a simplistic example:

type Size = "small" | "medium" | "large";
type Detail<S extends Size> = S extends "small" ? "noDetail" : S extends "medium" ? "partialDetail" : "fullDetail"

function getDetail<S extends Size>(size:S = "small"):Detail<S>{
    if(size === "small")
        return "noDetail" as Detail<S>;
    if(size === "medium")
        return "partialDetail" as Detail<S>
    return "fullDetail" as Detail<S>;
}

It causes the error:

Type '"small"' is not assignable to type 'S'.
  '"small"' is assignable to the constraint of type 'S', but 'S' could be instantiated with a different subtype of constraint 'Size'.ts(2322)

I understand the problem (ex: someone could try getDetail<"large">(); ) and I've read several posts on SO to try and resolve.

However,

1. I don't want to force people to pass a parameter.
2. I want to return the conditional type (not unconstrained string, not union)

How can I do something like this?

Solution

I would add "small" as type for size and I would also add "small" as default for the generic type.

function getDetail<S extends Size = "small">(size: S | "small" = "small"):Detail<S>{
    if(size === "small")
        return "noDetail" as Detail<S>;
    if(size === "medium")
        return "partialDetail" as Detail<S>
    return "fullDetail" as Detail<S>;
}

Playground

Answered By - Tobias S.