How do I specify a type for a function parameter that optionally includes a given method? - typescript-generics

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Updated Question
I want to define a function named bsearch() to do binary searches against arrays of arbitrary object types. When I invoke the function, I want it to check whether or not the Type of the array contains a compare() method and use it, if it does. If it does not, I want it to fall back to using < and === (so it will work with strings and numbers).
What should the function declaration look like? (I don't need an actual implementation, just the syntax for a type-safe solution.)
Or maybe I'm going about this all wrong? How can I create a function that uses a method built into a parameter type if it exists, or use some other function when it doesn't?
Original Question
This is the original question, but I've replaced it with the above as it seems this wasn't getting my point across.
I want to define a function named bsearch() to do binary searches against arrays of arbitrary object types. So I'd like to do something like this:
type Comparator = <Type>(a: Type, b: Type) => -1 | 0 | 1;
static bsearch<Type extends { compare?: Comparator }>(
ary: Type[],
value: Type
): number { ... }
My goal is to specify that Type must extend a type that may or may not include the compare method. In my function, I will check whether the compare method exists on the value parameter and call if it does, or use a generic function (that uses < and ===) if it does not.
The definition of bsearch() does not produce any warnings or errors, but attempts to invoke it from my unit test does:
class Person {
name: string;
length: number;
compare: Comparator<Person>; // What goes here?
}
describe('Utils tests', () => {
const arrayOfInt = [10, 20, 30, 40];
const arrayOfStr = ['Alfred', 'Bob', 'Chuck'];
const arrayOfPersons: Person = [
{name:'Barney',length:2},
{name:'Fred',length:6}
{name:'Wilma',length:12},
];
it('can find integer in an array of integers', () => {
let search_for = 30;
let result = Utils.bsearch(arrayOfInt, search_for)
expect(result).to.be.equal(2);
});
it('can find string in an array of strings', () => {
let search_for = 'Bob';
let result = Utils.bsearch(arrayOfStr, search_for)
expect(result).to.be.equal(1);
});
it('can find Person in an array of Persons', () => {
// This one uses Person.compare() to do the search.
// The previous two tests used the fallback technique.
let search_for = {name:'Fred',length:6};
let result = Utils.bsearch(arrayOfPersons, search_for)
expect(result).to.be.equal(1);
});
});
The error message is:
TS2345: Argument of type 'number[]' is not assignable to parameter of type '{ compare?: Comparator | undefined; }[]'.   Type 'number' has no properties in common with type '{ compare?: Comparator | undefined; }'.
I would appreciate pointers to other techniques if there is a better way to accomplish this (I'm still a TypeScript newbie).

Answer 1

Your generic is:
Type extends { compare?: Comparator }
Which means that Type must fulfill { compare?: Comparator } type. While passing object value, for example { name: 'Barney', length: 2, comparator: /* snip */}, is obviously correct, it's not the case for primitives like 10 and Bob. You need to include information about primitive types in the generic, for example:
Type extends ({ compare?: Comparator }) | number | string
Also, you'd probably want to enrich a bit the object typing:
{[key: string]: unknown, compare?: () => void } | number | string
Because, based on your description, you'd also want to accept also objects that do not have compare function in their type signature at all. If it does sound strange, I recommend reading about excess property checking.