The const assertion seems that the type can't be assigned to the type any[] for the generic parameter

Issue

export const CHRONIC_CHANNEL_LEVELS = [
  { label: 'one', value: '1' },
  { label: 'two', value: '2' },
] as const;

export type ElementType<T extends any[]> = T extends (infer U)[] ? U : never;

type ChronicChannelLevel = ElementType<typeof CHRONIC_CHANNEL_LEVELS>['value']; // throw error

TSC throws an error:

Type 'readonly [{ readonly label: "one"; readonly value: "1"; }, { readonly label: "two"; readonly value: "2"; }]' does not satisfy the constraint 'any[]'.
  The type 'readonly [{ readonly label: "one"; readonly value: "1"; }, { readonly label: "two"; readonly value: "2"; }]' is 'readonly' and cannot be assigned to the mutable type 'any[]'.(2344)

The const assertion seems that the type can't be assigned to the type any[] for the generic parameter.

I want the type ChronicChannelLevel to be a union type '1' | '2'.

I know I can use typeof CHRONIC_CHANNEL_LEVELS[number]['value'], but I want to create a utility type like ElementType to get the element type from an array.

TypeScript Playground

Solution

By accepting that tuple/array and readonly tuple/array are fundamentally different types, we can make progress:

export type ElementType<T extends (any[] | readonly any[])> =
  T extends (infer U)[] ?
  U :
  T extends readonly (infer U)[] ?
  U :
  never

Playground Link

...or by introducing a new type:

export type ArrayType<T> = T[] | readonly T[]

export type ElementType<T extends ArrayType<any>> =
  T extends ArrayType<infer U> ?
  U :
  never

we can simplify the amount of inference and branching required for the conditional type

Playground Link

Answered By - spender