Narrowing does not work for variables typed with type parameters

interface Base { b }
interface Derived extends Base { d }

declare function isDerived(x: Base): x is Derived;

function f<T extends Base>(x: T, y: T) {
    if (isDerived(x) && isDerived(y)) {
        return x.d === y.d;
    }

    return false;
}

Currently, this errors because both x and y still have type T, so they are missing the d property.

Wesley Wigham (@weswigham)

getNarrowedType doesn't check whether the type it's narrowing is a type parameter with a constraint. It just calls isTypeAssignableTo(Derived, T), which is false.

getNarrowedType needs to instead check isTypeAssignableTo(Derived, Base), but I'm not sure whether it's appropriate to narrow to Derived at that point, or to T extends Derived. Probably the latter.

Wesley Wigham (@weswigham), do you have an idea of the right thing to do here?

this is handled by #8010

Fixed in #8010.

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Narrowing does not work for variables typed with type parameters #6445

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Narrowing does not work for variables typed with type parameters #6445

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sandersn commented on Mar 7, 2016

mhegazy commented on Apr 21, 2016

ahejlsberg commented on Apr 22, 2016

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