Why does 'instanceof' in TypeScript give me the error “'Foo' only refers to a type, but is being used as a value here.”?

114
votes

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.

4
javascripttypescriptinstanceof
See the answer below @4castle. Otherwise, you're right, I'll make it Foo | string. – Daniel Rosenwasser
Possible duplicate of Interface type check with Typescript – Cerbrus
And possible duplicate of Check if variable is a specific interface type in a typescript union (I don't really want to single-handedly hammer this) – Cerbrus
@Jenny O'Reilly, now that's definitely a duplicate of a Possible duplicate! – marckassay

4 Answers

123
votes

instanceof works with classes, not interfaces.

What's going on

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. 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!

"What can I do instead of instanceof?"

You can look into type guards and user-defined type guards.

"But what if I just switched from an 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 = {
    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.

10
votes

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();
}
2
votes

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
0
votes

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