Assert function returns original type in Typescript

Issue

Is there a way to assert that a function returns a value that has type of the value given as an argument to the function, yet keep the wider argument type restrictions than of the original value?

Example: I want to trim a string value, but if the value is null or undefined, I want it to return the original null or undefined.

function trimOrOriginal(value?: string | null) {
    return value ? value.trim() : value;
};

Let's say I have non-optional field of type string | null. If I want to trim this field and assign the trimmed value to the original variable, I get a compiler error because the trimOrOriginal can return undefined even though I know it can never be undefined as the value I gave for the parameter is never undefined.

I could use generics:

const trimOrOriginal = <T>(value: T) => {
    if (typeof value === "string") {
        return value.trim();
    }
    return value;
};

But this doesn't look nice. I would like to specify that the argument type has to be "string".

Solution

You want to infer the return type based on the argument type.

const example1 = trimOrOriginal(null) // inferred type null
const example2 = trimOrOriginal("bla") // inferred type string
let input: string | undefined = "blub"
const example3 = trimOrOriginal(input) // inferred type string | undefined

For this you need to use generic types and type constraints. There is no way to do this without generics. That said, an implementation could look like this:

const trimOrOriginal = <T extends string | null | undefined >(s: T) =>
  (typeof s === 'string' ? s.trim() : s) as T extends string ? string : T

The conditional return type is necessary to widen string literals to strings:

type SL = "a " | "b " // constructed example
const a: SL = "a "
const example4 = trimOrOriginal(a) // inferred type is now string, not SL!
const sl: SL = trimOrOriginal(a) // type error, string is not assignable to SL

In real life, just use s?.trim() :-)

Answered By - Christian Rodemeyer