Type guard should infer the type of parent object when applied on a property

Suggestion

🔍 Search Terms

Type guard, parent object, infer, inference

✅ Viability Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

⭐ Suggestion

Typescript should be able to infer the type of an object, if a type guard was checked on a property of the said object. Currently, Typescript does correctly infer the type of the property, but not its parent object.

📃 Motivating Example

The following example is very common in data oriented design.

interface Type1 { list: string[] }

interface Type2 { list: { [key: string]: string } }

declare const obj: Type1 | Type2;

if(Array.isArray(obj.list)) {
    const list: string[] = obj.list; // WORKS
    const map: { [key: string]: string } = obj.list; // ERROR, as expected
    const objCasted: Type1 = obj; // ERROR, unexpectedly
} else {
    const map: { [key: string]: string } = obj.list; // WORKS
    const list: string[] = obj.list; // ERROR, as expected
    const objCasted: Type2 = obj; // ERROR, unexpectedly
}

The following example works and that is good because it is an equally common case in this type of design.

interface Type3 { type: 'type3', data: boolean }

interface Type4 { type: 'type4', data: string }

declare const obj2: Type3 | Type4;

if(obj2.type === 'type3') {
    const objCasted: Type3 = obj2; // WORKS
} else {
    const objCasted: Type4 = obj2; // WORKS
}

So I believe the type guards should work the same way. As far as I see, this does not cause any inconsistency in the language or the type system. It is an improvement without any downsides.

💻 Use Cases

See the full example.

Empieza reproduciendo los dos ejemplos motivadores en el playground de TypeScript enlazado y compara el narrowing actual de la propiedad con el objeto padre. No se nombran archivos fuente ni tests; se considera terminado cuando las comprobaciones de propiedades estrechan la unión padre correspondiente en ambas ramas sin cambiar la salida en tiempo de ejecución.