使用' as const '只读T[]不可赋值给T[]时的不便-如何从数组中删除只读修饰符

Question

我一直遇到这个问题(playground link):

const arr = [1,2,3] as const
const doSomethingWithNumbers = <T extends number[]>(numbers: T) => {}
doSomethingWithNumbers(arr)
//                     ^
// Argument of type 'readonly [1, 2, 3]' is not assignable to parameter of type 'number[]'.
//  The type 'readonly [1, 2, 3]' is 'readonly' and cannot be assigned to the mutable type 'number[]'.

我知道我可以修改函数也采取readonly数组:

export type AllowReadonly<T extends any[]> = T | readonly T[number][]
const doSomethingWithNumbers = <T extends <number[]>>(numbers: AllowReadonly<T>) => {}

但随后我在doSomethingWithNumbers中遇到了完全相同的问题，因为numbers参数现在也可以是readonly。

解决这个问题的好方法是什么?

Answer 1

EDIT:

我在一个问题中发现了这个更好的答案，该问题旨在不仅从数组中删除readonly修饰符，而且还从对象中删除。

这个方法对数组和对象都有效:

---

原始答:

我发现的一种方法是使用此函数在传递到函数时删除readonly修饰符:

const removeReadonly = <T extends any>(arr: T[] | readonly T[]): T[] => arr as T[]

使用(运动场链接):

const arr = [1,2,3] as const
const doSomethingWithNumbers = <T extends number[]>(numbers: T) => {}
const removeReadonly = <T extends any>(arr: T[] | (readonly T[])): T[] => arr as T[];
const nonReadonlyArr = removeReadonly(arr)  // (1 | 2 | 3)[];
doSomethingWithNumbers(nonReadonlyArr)      // No error;