How to check the object type on runtime in TypeScript?
Edit: I want to point out to people coming here from searches that this question is specifically dealing with non-class types, ie object shapes as defined by
interface
ortype
alias. For class types you can use JavaScript'sinstanceof
to determine the class an instance comes from, and TypeScript will narrow the type in the type-checker automatically.
Types are stripped away at compile-time and do not exist at runtime, so you can't check the type at runtime.
What you can do is check that the shape of an object is what you expect, and TypeScript can assert the type at compile time using a user-defined type guard that returns true (annotated return type is a "type predicate" of the form arg is T
) if the shape matches your expectation:
interface A {
foo: string;
}
interface B {
bar: number;
}
function isA(obj: any): obj is A {
return obj.foo !== undefined
}
function isB(obj: any): obj is B {
return obj.bar !== undefined
}
function func(obj: any) {
if (isA(obj)) {
// In this block 'obj' is narrowed to type 'A'
obj.foo;
}
else if (isB(obj)) {
// In this block 'obj' is narrowed to type 'B'
obj.bar;
}
}
Example in Playground
How deep you take the type-guard implementation is really up to you, it only needs to return true or false. For example, as Carl points out in his answer, the above example only checks that expected properties are defined (following the example in the docs), not that they are assigned the expected type. This can get tricky with nullable types and nested objects, it's up to you to determine how detailed to make the shape check.
"I'm trying to find a way to pass an object to function in and check it type in a runtime".
Since a class instance is just an object
, the "native" answer is to use a class instance and instanceof
when runtime type checking is needed, use an interface when not in order to keep a contract and decouple your application, make reduce signature size on methods/ctors, while not add any additional size. In my humble opinion this is one of a few main considerations for me in TypeScript when deciding to use a class vs type/interface. One other main driving factor is whether the object will ever need to be instantiated vs if it for instance defines a POJO.
In my codebase, I will typically have a class which implements an interface and the interface is used during compilation for pre-compile time type safety, while classes are used to organize my code and allow for ease in passing data between functions, classes and method as well as do runtime type checks in typescript.
Works because routerEvent is an instance of NavigationStart class
if (routerEvent instanceof NavigationStart) {
this.loading = true;
}
if (routerEvent instanceof NavigationEnd ||
routerEvent instanceof NavigationCancel ||
routerEvent instanceof NavigationError) {
this.loading = false;
}
Will not work
// Must use a class not an interface
export interface IRouterEvent { ... }
// Fails
expect(IRouterEvent instanceof NavigationCancel).toBe(true);
Will not work
// Must use a class not a type
export type RouterEvent { ... }
// Fails
expect(IRouterEvent instanceof NavigationCancel).toBe(true);
As you can see by the code above, classes are used to compare the instance to the types NavigationStart|Cancel|Error
within the Angular library and if you have used the router before you a project I am willing to be that you have done similar if not identical checks within your own codebase in order to determine application state during runtime.
Using instanceof
on a Type or Interface is not possible, since the ts compiler strips away these attributes during its compilation process and prior to being interpreted by JIT or AOT. Classes are a great way to create a type which can be used pre-compilation as well as during the JS runtime.
Update 2022
In addition to my original response to this, you can leverage The TypeScript Reflect Metadata API or roll your own solution using the TypeScript compiler to do static analysis of your code and parse the AST, querying like so:
switch (node.kind) {
case ts.SyntaxKind.InterfaceDeclaration:
// ...
break;
case ts.SyntaxKind.TypeDeclaration:
// ...
break;
}
See this solution for additonal details
Expanding on Aaron's answer, I've made a transformer that generates the type guard functions at compile time. This way you don't have to manually write them.
For example:
import { is } from 'typescript-is';
interface A {
foo: string;
}
interface B {
bar: number;
}
if (is<A>(obj)) {
// obj is narrowed to type A
}
if (is<B>(obj)) {
// obj is narrowed to type B
}
You can find the project here, with instructions to use it:
https://github.com/woutervh-/typescript-is