Why does TypeScript complaint about missing property in generic inheritance?

Question

Consider the following test case :

interface BaseFoo {}

interface FooAdapter {
  method<F extends BaseFoo>(foo:F):string;
}


interface ConcreteFoo extends BaseFoo {
  value:string;
}

class ConcreteFooAdapter implements FooAdapter {
  method(foo: ConcreteFoo): string {
    return foo.value;
  }
}

The only error is with the method signature, where TypeScript complaints that :

Property 'value' is missing in type 'BaseFoo' but required in type 'ConcreteFoo'.

Why would value be present in BaseFoo since the generic F is supposed to extend it?

But, more importantly, how to fix this so there is no error?

Edit

Here is an alternative solution I was trying to investigate, but with similar failure:

interface BarAdapter {
  method<F>(bar:F):string;
}

type Bar = {
  value:string;
}

class ConcreteBarAdapter implements BarAdapter {
  method(bar:Bar):string {
    return bar.value;
  }
}

It complaints that F is not assignable to type Bar, and I don't understand why.

I'd say you've misplaced the generic type parameter scope. You probably mean to have FooAdapter be a generic interface instead of having FooAdapter have a generic method. Like this playground link shows. With your version, callers choose the type argument F. Does that address your question fully? If so I could write up an answer explaining; if not, what am I missing? — jcalz
– jcalz, Commented Dec 4, 2022 at 4:53
Your updated edit has the same problem; a BarAdapter's method must work for any F the caller chooses, so a caller should be able to write barAdapter.method(123). But ConcreteBarAdapter's method only accepts Bars, so it's not a valid BarAdapter. Again, I think you want to make BarAdapter a generic interface with a specific method, instead of a specific interface with a generic method, as shown in this playground link. Does that work for you or not? If it works I will write up an answer (probably tomorrow because it's my bedtime now 😴) — jcalz
– jcalz, Commented Dec 4, 2022 at 5:15
I was hoping to have the generics at the method level, but it does work, yes. — Yanick Rochon
– Yanick Rochon, Commented Dec 4, 2022 at 5:21
Can you explain why you wanted the generics there? Is there a use case for that? — jcalz
– jcalz, Commented Dec 4, 2022 at 5:22
Oh well, I see that @GuerricP has posted essentially this answer, so I won't bother with mine. — jcalz
– jcalz, Commented Dec 4, 2022 at 14:11

Guerric P · Accepted Answer · 2022-12-04 09:40:07Z

1

If your only criteria is that the parameter should extend BaseFoo, and the return value should be a string, you may not need generics at all, this would be enough:

interface BaseFoo { }

interface FooAdapter {
  method(foo: BaseFoo): string;
}

interface ConcreteFoo extends BaseFoo {
  value: string;
}

class ConcreteFooAdapter implements FooAdapter {
  method(foo: ConcreteFoo): string {
    return foo.value;
  }
}

This provide as strong typing as in your attempt with generics. TypeScript constraints method in the implementor to extend method(foo: BaseFoo): string.

However, if you need to be able to use the adapters as implementors of a specific method signature, then you'd have to add the generic parameter on the interface, and then provide the type explicitly when implementing it:

interface BaseFoo { }

interface FooAdapter<F extends BaseFoo>  {
  method(foo: F): string;
}


interface ConcreteFoo extends BaseFoo {
  value: string;
}

class ConcreteFooAdapter implements FooAdapter<ConcreteFoo> {
  method(foo: ConcreteFoo): string {
    return foo.value;
  }
}

answered Dec 4, 2022 at 9:40

Guerric P

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1 Comment

Yanick Rochon Over a year ago

Ah! Yes, even if generics are fancy and really useful, sometimes the simple solution is best. Thanks!

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