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Support custom typeof functions

See original GitHub issue

Search Terms

  • custom typeof
  • custom typeof function
  • custom type guard


When using a custom typeof‑like function, I’d like TypeScript compiler to be able to infer the correct type in the scope guarded by the custom typeof‑like function.

Use Cases

I need this to provide proper type information for the type function from Blissfuljs and similar projects.

The current approach requires defining the type(…) function as type(obj: any): string and then doing a type cast every time something is accessed within the if block:

/// <reference types="blissfuljs"/>
declare var something: any;
if ($.type(something) === "array") {
	(something as any[]).forEach(v => {/* stuff */})


 * @param obj The variable to check the type of.
 * @return The result of `typeof obj` or the class name in lowercase for objects.
 *         In the case of numbers, if the value is `NaN`, then the result is `nan`.
declare function type(obj: null): "null";
declare function type(obj: undefined): "undefined";

// This is to ensure that the type system short-circuits when it encounters primitive types.
declare function type(obj: number): "number" | "nan";
declare function type(obj: string): "string";
declare function type(obj: symbol): "symbol";
declare function type(obj: boolean): "boolean";

// Needed to ensure proper return values when wrapper objects are used.
/* tslint:disable:ban-types */
declare function type(obj: number | Number): "number" | "nan";
declare function type(obj: string | String): "string";
declare function type(obj: symbol | Symbol): "symbol";
declare function type(obj: boolean | Boolean): "boolean";
declare function type(obj: Function): "function";
/* tslint:enable:ban-types */

declare function type(obj: any[]): "array";
declare function type(obj: RegExp): "regexp";
declare function type(obj: any): string;
export = type;


import type = require("./type-func");
declare var something: any;

if (type(something) === "array") {
	// $ExpectType any[]
} else if (type(somthing) === "number") {
	// $ExpectType number


My suggestion meets these guidelines:

  • This wouldn’t be a breaking change in existing TypeScript/JavaScript code
  • This wouldn’t change the runtime behavior of existing JavaScript code
  • This could be implemented without emitting different JS based on the types of the expressions
  • This isn’t a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
  • This feature would agree with the rest of TypeScript’s Design Goals.

Issue Analytics

  • State:open
  • Created 4 years ago
  • Reactions:1
  • Comments:13 (8 by maintainers)

github_iconTop GitHub Comments

dragomirtitiancommented, Apr 2, 2019

@ExE-Boss There isn’t enough in the type system to know that 'regex' is tied to the Regex type, and overloads don’t seem like a good choice to encode this behavior . I would propose an extension to the type guard syntax. Add the capability to specify the return value when a parameter is of certain type

declare function type(obj: any): 
    obj is number? "number" | "nan" :
    obj is string ? "string":
    obj is symbol ? "symbol"

After ? you can have a literal type or a union of literal types. Omitting ? defaults to obj is T ? true: obj is Exclude<typeof obj, T>? false which is the current behavior.

</wacky-syntax-proposal >

ExE-Bosscommented, Apr 1, 2019

Because I’m using a JavaScript library that doesn’t work that way:

Read more comments on GitHub >

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