What are the reasons behind TypeScript prohibiting the use of concat in the string | string[] type?

Question

What are the reasons behind TypeScript prohibiting the use of concat in the string | string[] type?

const s: string = 'foo';

const pass1 = (origin: string) => origin.concat(s);
const pass2 = (origin: string[]) => origin.concat(s);

type S = string | string[];

const error = (origin: S) => origin.concat(s);

In the code snippet above, I am able to call concat on both a string and a string[]. However, TypeScript prohibits calling concat on a string | string[] type.

The error message states:

Cannot invoke an expression whose type lacks a call signature.
Type '((...strings: string[]) => string) | { (...items: ConcatArray<string>[]): string[]; (...items: (s...'
has no compatible call signatures.

This could be due to the different return types. Even though I believe TypeScript should be able to infer that the type of error is S, it seems to be an intentional design choice. But why?

javascript typescript type-systems duck-typing

Answer 1

Answer №1

In the case of the `concat` method, even though it is shared between two types, their signatures differ significantly. This prevents TypeScript from merging the method declarations seamlessly. One way to handle this situation, although not ideal, is to use a type guard to differentiate between the two types:

const s: string = 'foo';
type S = string | string[];

const error = (origin: S) => typeof origin === 'string' ?
    origin.concat(s) :
    origin.concat(s);

Alternatively, you can simply assert as `any` :

const s: string = 'foo';
type S = string | string[];

const error = (origin: S) =>  (origin as any).concat(s) as S

Another approach is transforming the union of signatures into an intersection of signatures. While this may be effective in some scenarios, it may not work well in others:

const s: string = 'foo';
type S = string | string[];

type UnionToIntersection<U> = 
(U extends any ? (k: U)=>void : never) extends ((k: infer I)=>void) ? I : never

function mergeSignature<T, K extends keyof T> (value: T, method: K) : UnionToIntersection<T[K]>{
    return ((...args: any[]) => (value[method] as any as Function).apply(value, args)) as any;
}

const error = (origin: S) => mergeSignature(origin, 'concat')(s);

Answer 2

In the case of the `concat` method, even though it is shared between two types, their signatures differ significantly. This prevents TypeScript from merging the method declarations seamlessly. One way to handle this situation, although not ideal, is to use a type guard to differentiate between the two types:

const s: string = 'foo';
type S = string | string[];

const error = (origin: S) => typeof origin === 'string' ?
    origin.concat(s) :
    origin.concat(s);

Alternatively, you can simply assert as `any` :

const s: string = 'foo';
type S = string | string[];

const error = (origin: S) =>  (origin as any).concat(s) as S

Another approach is transforming the union of signatures into an intersection of signatures. While this may be effective in some scenarios, it may not work well in others:

const s: string = 'foo';
type S = string | string[];

type UnionToIntersection<U> = 
(U extends any ? (k: U)=>void : never) extends ((k: infer I)=>void) ? I : never

function mergeSignature<T, K extends keyof T> (value: T, method: K) : UnionToIntersection<T[K]>{
    return ((...args: any[]) => (value[method] as any as Function).apply(value, args)) as any;
}

const error = (origin: S) => mergeSignature(origin, 'concat')(s);

What are the reasons behind TypeScript prohibiting the use of concat in the string | string[] type?

Answer №1

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