Can a TypeScript function be structured to return never (or throw) if a generic type extends a subtype without requiring casting?

(This code snippet is purely for demonstration purposes, as no real use-case exists here)

I am attempting to create a function that throws an error if the input string is equal to "fish". I have achieved this using the as keyword, but I am curious if there is another way to accomplish this without relying on as.

function throwIfFish<S extends string>(string: S): S extends "fish" ? never : S {
    if (string === "fish") {
        throw new Error("fish");
    }
    return string as S extends "fish" ? never : S;
    //            ^ Is there an alternative approach?
}

Typescript Playground

Answer №1

The conditional type presented here is quite straightforward, and the resulting type seems to be satisfactory enough for an index signature type to function in this case:


type UniqueAnimals<T extends string> = {
    animal: unique;
    [u: string]: T;
}

function checkIfUniqueAnimal<T extends string>(str: T): UniqueAnimals<T>[T] {
    if (str === "unique") {
        throw new Error("unique");
    } else { 
        return str; 
    }
}


let result1 = checkIfUniqueAnimal("dog");    // "dog"
let result2 = checkIfUniqueAnimal("unique"); // unique


Answer №2

Transfer the signature to a different overload:

function validateNoFish<S extends string>(string: S): S extends "fish" ? never : S;
function validateNoFish(string: string) {
    if (string === "fish") {
        throw new Error("fish");
    }

    return string;
}

However, this will cause TypeScript to no longer enforce conformity with the overload's signature. Another option is to use a cast with reduced redundancy:

return string as ReturnType<typeof validateNoFish<S>>;

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