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Resolve deferred conditional types to their true branch when instantiated with a type parameter constrained to the tested typeΒ #52144

@kaczor6418

Description

@kaczor6418

Bug Report

πŸ”Ž Search Terms

conditional-type, distributive-type, generics

πŸ•— Version & Regression Information

Version 4.9.4 and nightly

⏯ Playground Link

TS resolves type to different type even if I add constraint that should resolve to given type

πŸ’» Code

type PrimitiveDataType = string | number | bigint | boolean | symbol | undefined | null;

type ConditionalType<T> = [T] extends [PrimitiveDataType]
  ? (v: T) => void
  : T extends Array<unknown>
  ? (v: T, t: number) => void
  : never;

abstract class AbstractClass<T> {
  abstract value: T;
  protected conditionalFunctions: Map<ConditionalType<T>, number | undefined> = new Map();
}

class SomeClass<T extends PrimitiveDataType> extends AbstractClass<T> {
  value: T;

  constructor(value: T) {
    super();
    this.value = value;
  }

  someMethod() {
    for (const someFn of this.conditionalFunctions.keys()) {
      someFn(this.value);
    }
  }
}

In the above code I have created a PrimitiveDataType which is a union of all primitive data types in JavaScript. Then I have created a ConditionalType<T> that will resolve to some callback only if T is one of PrimitiveDataType. Then I have created an abstract generic class that have a field which type(ConditionalType<T>) depends on generic value of this class. In the end, I have crated a SomeClass that extends AbstractClass<T> and add constraints that generic parameter T have to extend PrimitiveDataType.

πŸ™ Actual behavior

For someMethod of this SomeClass conditionalFunctions.keys() is resolved to (v: T, t: number) => void what would implicate that my generic T extends an Array type.

πŸ™‚ Expected behavior

For someMethod of this SomeClass I have expected that TS should resolve conditionalFunctions.keys() to (v: T) => void to my surprise it is resolved to (v: T, t: number) => void what would implicate that my generic T extends an Array type. This doesn't make sens to me because T has constraint that has to extends PrimitiveDataType.

Example of what I would expect:

Works for regular variables

type PrimitiveDataType = string | number | bigint | boolean | symbol | undefined | null;

type ConditionalType<T> = [T] extends [PrimitiveDataType]
  ? (v: T) => void
  : T extends Array<unknown>
  ? (v: T, t: number) => void
  : never;

const x: PrimitiveDataType = 12;
const y: ConditionalType<typeof x> = (param: number) => undefined;

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