microsoft / microsoft/TypeScript

array filter/find has inconsistent results and incorrect constraint specifications

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#56,572 2 comentarios 0 reacciones 0 asignados Ver en GitHub

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Domain: check: Type Inference Possible Improvement
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Descripción

🔎 Search Terms

type predicate error 2677
union of arrays type predicate
array filter result
infer return type from constraint

One error noted in this bug report (2345 on the rhs of a1 in the code) is a known problem with design-limitation status, I believe. That is incidental and not the main focus of this bug report.

🕗 Version & Regression Information
  • 5.2.2 / 5.3.2
  • I was unable to compare the result on prior versions because there were other worse problems in prior versions
⏯ Playground Link

playground

💻 Code
interface Fizz {
    id: number;
    fizz: string;
}

interface Buzz {
    id: number;
    buzz: string;
}

declare function isFizz(x: unknown): x is Fizz;
const x1 = ([] as Fizz[]|Buzz[]).find(isFizz);
//    ?^  x1: Fizz | undefined
const x2 = ([] as Buzz[]).find(isFizz);
//    ?^  x2: Buzz | undefined

const y1 = ([] as Fizz[]|Buzz[]).filter(isFizz);
//   ?^  y1: Fizz[]
const y2 = ([] as Buzz[]).filter(isFizz);
//   ?^  y2: Buzz[]


// Different results obtained for this non library version
declare function filter2<T,S extends T>(predicate:(t:T)=>t is S, arr:T[]): S[] ;
declare function filter2<T>(predicate:(t:T)=>unknown, arr:T[]): T[] ;

const z1 = filter2(isFizz, ([] as Fizz[]|Buzz[]));
//   ?^  z1: Fizz[]
const z2 = filter2(isFizz, ([] as Buzz[]));
//   ?^  z2: Fizz[]

// Related: If the constraint "S extends T" is ommitted, an error occurs.
declare function filter3<T,S>(predicate:(t:T)=>t is S, arr:T[]): (T&S)[];
//                                                  ~ Type 'S' is not assignable to type 'T'. 2677


// Assuming the object type T is open (i.e., could have other keys, which is the TS default assumption)
// then the filter result should be (T&S)[], so try defining that result explicitly (without an overload, to avoid confusion).
// This should work:
declare function typePredicateFilter4<D, T extends D,S extends D>(predicate:(t:D)=>t is S, arr:T[]): (T&S)[];

const a1 = typePredicateFilter4(isFizz, ([] as Fizz[]|Buzz[])); // error
//   ?^  z1: Fizz[] (invalidated by error)
const a2 = typePredicateFilter4(isFizz, ([] as Buzz[]));
//   ?^  z2: (Buzz & Fizz)[]
🙁 Actual behavior
  1. The types of y2 (Buzz[]) and z2 (Fizz) differ.

  2. As shown in function3, omitting the constraint S extends T results in error 2677.

  3. The result a2 is (Fizz & Buzz)[] as desired (good), but the result of a1 invalidated because of the error occurring on Fizz[]|Buzz[]. So typePredicateFilter4 is not currently usable in the general case.

🙂 Expected behavior
  1. The types of y2 and z2 match.

  2. Want to be able to omit the that constraint S extends T which triggers the error.
    2.1. One reason we want to omit it is because S extends T triggers the inference logic we don't need if (T&S) is explicitly provided as the return type.
    2.2. The other reason is that S extends T unnecessarily constrains the domain of the predicate type function. See typePredicateFilter4 for the correct constraints - D is the domain of the predicate function, and both T and S independently extends D.

  3. typePredicateFilter4 should not produce an error when Fizz[]|Buzz[] is the array argument type. This is an already known issue. 1 and 2 can be solved without solving this problem #3.

Additional information about the issue

Some special type inference logic to calculate the return type is triggered by the presence of the type predicate function arg is S as an argument, and that logic uses the constraint S extends T to infer the return type. If the requirement of having a constraint is dropped, and return type was specified explicitly (as in function filter3 and typePredicateFilter4, then that logic would not be required.

Would such a change be good? If (T & S) is what is actually expected to pass the type predicate function, then I think yes.

Moreover, the entire result signature output of typePredicateFilter4is "linear" with respect to it type inputs. This allows faster resolution.

Guía de contribución

Abrir la guía de contribución

Primeros pasos

  1. Lee el issue completo y luego la guía de contribución del proyecto.
  2. Comenta en el issue que vas a ocuparte — evita que dos personas hagan lo mismo.
  3. Haz un fork del repositorio y trabaja en una rama.
  4. Abre un pull request que haga referencia al número del issue.

Línea de trabajo

Comienza con la reproducción enlazada de TypeScript Playground y compara los ejemplos de filter/find, especialmente y2 frente a z2 y las llamadas a typePredicateFilter4. Rastrea la inferencia de predicados de tipo y la comprobación de restricciones para las declaraciones mostradas. Se considera terminado cuando los tipos de resultado indicados coincidan y la llamada Fizz[]|Buzz[] ya no produzca un error, sin introducir regresiones en el comportamiento indicado.

Escrito por el modelo de indexación a partir del texto del issue.

Evaluación

Stack tecnológico
typescript
Área
compilers
Tipo de issue
Error
Dificultad
5/5
Tiempo estimado
Más de una semana
Estado de actividad
Estancado
Claridad
Bastante claro
Aptitud para principiantes
25/100

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