What could be causing TypeScript to forego type inference and default to 'any' in this scenario?

While refactoring parts of our React app in TypeScript, I encountered a challenge that required me to use what I consider to be a less than ideal double type argument. I'm unsure if this is a bug in TypeScript or if there is some type ambiguity causing the callback parameter to default to type 'any'.

To address this issue, I had to conceal the original type within another type argument to correct the inference of the wrong type for the callback method due to the "extends string" condition.

Here is the original code snippet:

type Optional<TItem> = TItem extends string
  ? { get?: (item: TItem) => string }
  : { get: (item: TItem) => string };

type TItem = { hello: string } | string;

const obj: Optional<TItem> = {
  get: item => typeof item === 'string' ? item : item.hello
};

The error output from npx tsc with the original code using TypeScript 5.4.2 was:

error TS7006: Parameter 'item' implicitly has an 'any' type.

18   get: item => typeof item === 'string' ? item : item.hello

Below is the workaround applied:

type TItem = { hello: string } | string;

type Optional<TItem, TItemType> = TItemType extends string
  ? { get?: (item: TItem) => string }
  : { get: (item: TItem) => string };

const obj: Optional<TItem, TItem> = {
  get: item => typeof item === 'string' ? item : item.hello
};

Answer №1

Your initial Optional<T> represents a distributive conditional type, which means that if T is a union, then Optional<T> also becomes a union of objects. This results in the get method being a union of methods, making it unclear what the item method parameter should be. Currently, unions of methods outside of discriminated unions lack contextual types for their parameters. Although an issue was raised for improved contextual typing in TypeScript at microsoft/TypeScript#48460, it was closed as working correctly. Some discussions have debated whether the contextual type should be a union, intersection, or something else, but the TypeScript team does not find it clear and has decided not to invest resources into resolving this.

If you are set on having a union of methods, you will need to manually annotate the parameters to explicitly specify the expected type.


However, if your intention is not to have a union in the scenario where

type TItem = { hello: string } | string;

You actually want Optional<TItem> to result in a single object type

{get: (item: TItem) => string}
, correct? In that case, you can disable the distributive object type by utilizing the wrap-as-a-single-element-tuple-type technique described in the documentation for distributive conditional types:

type Optional<T> = [T] extends [string]
  ? { get?: (item: T) => string }
  : { get: (item: T) => string };

With this modification, Optional<T> will no longer break down T into pieces before performing operations on it. As a result, you will achieve the desired contextual type:

const obj: Optional<TItem> = {
  get: item => typeof item === 'string' ? item : item.hello // okay
};

Link to code on TypeScript Playground

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