CODEWITHSUNDEEP
typescript
Vinz243
Vinz243

Reputation: 9938

Cannot invoke an expression whose type lacks a call signature. Type has no compatible call signatures

i'm trying to make a definition for docuri

export type URI<K extends RouteParams> = string;

export interface RouteParams {
  [key: string]: (string | number | boolean)
}

export interface Document {
  [key: string]: (string | number | boolean)
}

/**
 * Create a URI from a document properties
 * @param the props to build the URI from
 * @return the URI
 */
export type RouteCreator<K extends RouteParams> = (props: K) => string;

/**
 * Parses a URI and returns the props
 * @param uri the URI to parse
 * @return the params parsed from URI
 */
export type RouteParser<K extends RouteParams> = (uri: string) => K;

export type Route<T extends RouteParams> = RouteParser<T> | RouteCreator<T>;

/**
 * Creates a Route which is a function that either parse or stringify object/string
 * @param route the route uri
 * @return the Route
 */
export type RouteFactory<K extends RouteParams> = (route: string) => Route<K>;

export interface DocURI<K extends RouteParams> {
  route: RouteFactory<K>;
}

And then to use it:

import {DocURI, Document, RouteParams, URI, RouteFactory} from './Definitions';

const docuri = require('docuri');

function getRoute <T extends Document> (): DocURI<T> {
  return (docuri as DocURI<T>);
}
...
const artistURI = getRoute<ArtistParams>().route('artist/name');

const parsed = artistURI(album.artist); // Cannot invoke an expression whose type lacks a call signature. Type 'Route<ArtistParams>' has no compatible call signatures.

And I get on the last line:

 Cannot invoke an expression whose type lacks a call signature. Type 'Route<ArtistParams>' has no compatible call signatures.

What am I doing wrong?

Upvotes: 1

Views: 7989

Answers (1)

Nitzan Tomer
Nitzan Tomer

Reputation: 164137

Check out this issue: Call signatures of union types:

This is currently by design because we don't synthesize an intersectional call signature when getting the members of a union type -- only call signatures which are identical appear on the unioned type

So for now you'll need to either make both of the function signatures the same, or cast the result:

const artistURI = getRoute<ArtistParams>().route('artist/name') as (str: string) => string;

const parsed = artistURI(album.artist); // should be ok

Upvotes: 3

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