TypeScript complains about non-existent properties in generic function

Question

I am using TypeScript Generics to specify the parameter types a function accepts. As seeing below, Foo and Bar, both have a function called isDirectory but seems to be invalid in the generic function. Can anyone explain me what I am missing here?

class Foo {

    isDirectory(): boolean {
        return true;
    }
}

class Bar {

    isDirectory(): boolean {
        return true;
    }
}

type XYZ = Foo | Bar;

function bas<XYZ>(x: XYZ) {
    !!!!!!!! Property 'isDirectory' does not exist on type 'XYZ'.(2339)
    return x.isDirectory();
}

I know that I can extend the generic type with extends, but why would that be needed in this case? Is there a solution without manually typing an interface?

https://www.typescriptlang.org/play?#code/MYGwhgzhAEBiD29oG8BQrqegSwgEWwCcBTYAF3kIE8AKASgC5oAjREYsAOxQyz5LIBXQtzKFBxANy9MAX1TzUoSDABCYQjxk58RUhWr0mreOy48+-YkJHQxE6X3mKyVAA7FoADQCaALWgAXjhEaAAfaHVCaVQAM0FOcmx4bmZIAB5fPwA+GgAPJiy6CywBYW48gDpcAhJySlo6aVkgA

Nicholas Tower · Accepted Answer · 2021-04-06 16:33:07Z

2

You're using the same name for multiple things. So while you think the XYZ's are all related, they're not. Let me give them unique names so it's clear what is what:

type XYZ = Foo | Bar;

function bas<T>(x: T) {
  return x.isDirectory();
}

The function bas is a generic which accepts any type T. It has no relationship to the XYZ you defined a couple lines earlier. Since T can be anything (say, a number), there's no guarantee it would have an isDirectory function on it.

What you probably meant to do was to restrict T so it can only be an XYZ, or something extending from it:

type XYZ = Foo | Bar;

function bas<T extends XYZ>(x: T) {
  return x.isDirectory();
}

Playground link

answered Apr 6, 2021 at 16:33

Nicholas Tower

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1 Comment

HelloWorld Over a year ago

Oh stupid me, that makes so much sense!! Faceplam! Thanks for the explanation! Accepted once I can

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