Why do type aliases fulfill a constraint while interfaces do not in Typescript?

I encountered a peculiar situation. I defined a conditional type where a type alias satisfies the extends constraint, but an interface with identical structure does not.

I'm perplexed by this discrepancy. To see it in action, visit the playground.

interface Constraint {
  [key: string]: string | number | boolean
}

type ATypeAlias = {
  str: string
  num: number
  bool: boolean
}

interface SameInterface {
  str: string
  num: number
  bool: boolean
}

type expectToBeTrue = ATypeAlias extends Constraint ? true : false

// Strange!
type butWhyAmIFalse = SameInterface extends Constraint ? true : false

Answer №1

You've encountered a well-known issue, referenced as microsoft/TypeScript#15300, where implicit index signatures are automatically derived for type aliases but not for interfaces. This is an interesting case where type aliases and interfaces differ in their type analysis. According to @RyanCavanaugh (Development lead for the TypeScript team at Microsoft), this behavior is intentionally designed, as stated in this explanation:

In simple terms, the current behavior is intentional because interfaces have the ability to be extended by additional declarations while type aliases cannot. Therefore, it is considered "safer" (with emphasis) to automatically infer an implicit index signature for type aliases instead of interfaces.

Although there were suggestions to change this behavior, the proposal was ultimately rejected with the following response:

Making such a change would be highly disruptive without significant benefits. To clarify, interfaces should explicitly declare an index signature if they are meant to be accessed using indices.

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