I wrote this code
interface Foo { abcdef: number; } let x: Foo | string; if (x instanceof Foo) { // ... } But TypeScript gave me this error:
'Foo' only refers to a type, but is being used as a value here. Why is this happening? I thought that instanceof could check whether my value has a given type, but TypeScript seems not to like this.
instanceof works with classes, not interfaces nor type aliases.
The issue is that instanceof is a construct from JavaScript, and in JavaScript, instanceof expects a value for the right-side operand. Specifically, in x instanceof Foo JavaScript will perform a runtime check to see whether Foo.prototype exists anywhere in the prototype chain of x.
However, in TypeScript, interfaces have no emit. The same is true of type aliases. That means that neither Foo nor Foo.prototype exist at runtime, so this code will definitely fail.
TypeScript is trying to tell you this could never work. Foo is just a type, it's not a value at all!
If you're coming from another language, you might have meant to use a class here. Classes do create values at runtime, but there are some notes about that that you may want to read about below.
instanceof if I still want a type or interface?"You can look into type guards and user-defined type guards.
interface to a class?"You might be tempted to switch from an interface to a class, but you should realize that in TypeScript's structural type system (where things are primarily shape based), you can produce any an object that has the same shape as a given class:
class C { a: number = 10; b: boolean = true; c: string = "hello"; } let x = new C() let y: C = { a: 10, b: true, c: "hello", } // Works! x = y; y = x; In this case, you have x and y that have the same type, but if you try using instanceof on either one, you'll get the opposite result on the other. So instanceof won't really tell you much about the type if you're taking advantage of structural types in TypeScript.
instanceof works with classes, not interfaces. Thought that needed to be emphasized.To do type checking at runtime with an interface is using type guards, if interfaces you wish to check have different properties/functions.
Example
let pet = getSmallPet(); if ((pet as Fish).swim) { (pet as Fish).swim(); } else if ((pet as Bird).fly) { (pet as Bird).fly(); } 
Duck, you type guard it becomes Fish, but still no runtime exception when you invoke swim(). Suggest you create 1 level of common interface (e.g. Swimmable) and move you swim() functions there, then type guard still looks good with ((pet as Swimmable).swim.
'swim' in pet condition. It will narrow it down to a subset that has to have swim defined (ex: Fish | Mammal)When it is about checking if an object conforms to an interface signature, then I think the appropriate approach is considering using "type predicates": https://www.typescriptlang.org/docs/handbook/2/narrowing.html#using-type-predicates
Daniel Rosenwasser might be right and dandy but i feel like doing an amendment to his answer. It is fully possible to check instance of x, see the code snippet.
But it's equally easy to assign x = y. Now x would not be an instance of C as y only had the shape of C.
class C { a: number = 10; b: boolean = true; c: string = "hello"; } let x = new C() let y = { a: 10, b: true, c: "hello", } console.log('x is C? ' + (x instanceof C)) // return true console.log('y is C? ' + (y instanceof C)) // return false You can use the in operator narrowing for checking if the element you need is in the object.
With this method, you can verify if x is a string or Foo
if ('abcdef' in x) { // x is instance of Foo } instanceof is usually the correct answer.
In my case however, I was working with a primitive type and both instanceof Number and instanceof number errored in the ts compiler:
TS2322: Type 'Number' is not assignable to type 'number'. TS2693: 'number' only refers to a type, but is being used as a value here. typeof is the correct answer when primitive types are in use:
if (typeof value === 'number') {this.propertyValue = value} 
Foo | string.