Ensuring that a function in Typescript returns the same data type it was

Is there a way to ensure that a function returns a value with the same type as the argument given to it, while still maintaining broader argument type restrictions than the original value?

For instance: I need 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 a non-optional field of type string | null. If I try to trim this field and assign the trimmed value back to the original variable, I receive a compiler error because the trimOrOriginal function can potentially return undefined, even though the argument provided will never be undefined.

I could utilize generics:

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

However, this approach isn't very clean. I would prefer to specify that the argument type must be specifically "string".

Answer №1

If you want to determine the return type based on the argument type, you can use generic types and type constraints.

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

To achieve this, you will need to implement a function like this:

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

Using conditional return types helps 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 practice, it's simpler to just use s?.trim() :-)

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