Allow using guard with Refinement

[JalilArfaoui]

IMHO this should work

interface A {
  a: string;
}
interface B extends A {
  b: string;
}
declare const aIsB: Refinement<A, B>;
declare const a: A;
declare const doSomethingWithB : (b: B) => string;

guard<A, string>([[aIsB, doSomethingWithB]])(constant(''))(a);

Right now I have to :

guard<A, string>([[aIsB, (a) => doSomethingWithB(a as B)]])(constant(''))(a);

to make TypeScript happy …

[samhh]

I suspect we'd need to rewrite the function to be overloaded so that each branch can infer its own generics, but there might be a smarter way with newer type system features.

Taking a step back, have you considered using Options instead? Here's an alternative approach with altAllBy and function composition:

const guardA: (x: A) => Option<string> = altAllBy([
  flow(getB, O.map(doSomethingWithB)),
  flow(getB, O.map(doSomethingElseWithB)),
  constant(O.of('foo')),
])

[JalilArfaoui]

Thanks @samhh

I still prefer my workaround as I don't need an Option as result …

But thanks for altAllBy that I didn't knew …