Extending Non Generic Interface with Generic Interface

Question

I'm probably over thinking this. Basically, I would like to have an non-generic interface that implements a few properties, then create a second type that implements that interface with it's own generic property.

interface Car {
  model: string;
}

interface Car<T> extends Car {
  transmission: T;
}

TypeScript errors:

"All declarations of 'Car' must have identical type parameters."

The only way I can find is to name the non-generic interface something different, like CarBase. Suggestions?

What are you trying to do? Why would you separate it into two different interfaces with the same name? It's equivalent to interface Car<T> { model: string; transmission: T } — Nitzan Tomer
– Nitzan Tomer, Commented Feb 21, 2017 at 14:04
I want other code to use the non-generic reference. Currenly Im doing Car<any> to bypass the type checking, when I just want to use Car. — cgatian
– cgatian, Commented Feb 21, 2017 at 14:08

Nitzan Tomer · Accepted Answer · 2017-02-21 14:09:55Z

1

You can use a type alias to save you the trouble of writing Car<any>:

interface Car<T> {
    model: string;
    transmission: T;
}

type AnyCar = Car<any>;

answered Feb 21, 2017 at 14:09

Nitzan Tomer

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2 Comments

cgatian Over a year ago

Is this just the same as writing: interface CarBase implements Car<any> ?

Nitzan Tomer Over a year ago

It's pretty much the same, but there are differences between interfaces and type aliases. It is explained in Interfaces vs. Type Aliases

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