How to enforce generic types typescript function

Issue

I have the following function signature:

const renderMessage = <T extends A | B > 
(
  a: T,
  b: Map<T, Array<string>>, 
  c: T,
) => {

My problem is that I can call this function as follows:

type A = "A" | "B"
type B = "Z" | "Y"
let myA: A = "A"
let myB: B = "Z"
let myAMap: Map<A,Array<string>> = new Map(...)
let myBMap: Map<B,Array<string>> = new Map(...)
renderMessage(myA, myBMap, myA) // I want error here
renderMessage(myB, myAMap, my B) // I want error here

The reason for being able to do this is explained here. If I understood correctly, the structure of my Map object does not change based on T being A or B, so typescript won't complain. However, I am struggling to figure out how to enforce this.

EDIT

They are types and not classes. See Playground Link

Solution

The T type is too ambiguous for the TypeScript inference. It resolves to the most narrow type that satisfies the whole signature: A | B.

I guess you just have to provide it yourself between angle brackets:

TypeScript playground

Answered By - Guerric P