Comparing constrained generic types/substitution types to conditional types

When comparing a generic type to a conditional type whose checkType is the same type, we have additional information that we are not utilizing.. for instance:

function f<T extends number>(x: T) {
    var y: T extends number ? number : string;

    // `T` is not assignable to `T extends number ? number : string`
    y = x;
 }

Ignoring intersections, we should be able to take the true branch all the time based on the constraint.

The intuition here is that T in the example above is really T extends number ? T : never which is assignable to T extends number ? number : string.

Similarly, with substitution types, we have additional information that we can leverage, e.g.:

declare function isNumber(a: any): a is number;

function map<T extends number | string>(
    o: T,
    map: (value: T extends number ? number : string) => any
): any {
    if (isNumber(o)) {
        // `T & number` is not assignable to `T extends number ? number : string`
        return map(o);
    }
}

L’issue non indica file né test. Inizia riproducendo entrambi gli esempi TypeScript e segui la gestione dell’assignability da parte del compilatore per i tipi generici vincolati e i tipi di sostituzione rispetto ai tipi condizionali; il lavoro è completo quando le assegnazioni mostrate e la chiamata passano il type-check come descritto.