function isObject(v: any): v is Object{
return v instanceof Object && !Array.isArray(v);
}
let v: string | Object;
if (isObject(v))
v.hasOwnProperty("x");
else
if (typeof v === 'string')
v.trim(); //Error: Property 'trim' does not exist on type 'never'
I am confused with that error in the last line, apparently, v gets narrowed to type 'never" but I can't see the logic of it. (Typescript version 4.2.3)
This is how Object
interface defined in typescript:
interface Object {
constructor: Function;
toString(): string;
toLocaleString(): string;
valueOf(): Object;
hasOwnProperty(v: PropertyKey): boolean;
isPrototypeOf(v: Object): boolean;
propertyIsEnumerable(v: PropertyKey): boolean;
}
Almost any type including primitives like string
is assignable to Object
(yes string have all of the above properties).
object (lowercase) is a type that represents the non-primitive type, i.e. anything that is not number
, string
, boolean
, symbol
, null
, or undefined
function isObject(v: any): v is object {
return v instanceof Object && !Array.isArray(v);
}
declare let v: string | object;
if (isObject(v))
v.hasOwnProperty("x");
else
v.trim(); // v narrowed to string
Playground
Do's and Don'ts
let v: string | Object;
While executing the instanceOf(type checker) is considering v variable type as string
because of you mentioned string as first type and Object is a second type of that variable. So in this case without having any value of that variable how the instanceOf know the type?. So it will check null instanceof Object
. Which means null is not an object type. This is the reason to why isObject()
return false. See
console.log(null instanceof Object)
If you assign any value, then only the instanceOf will be considered the type based on value because of that variable have two data types.
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