javascript - 将 Array.prototype.concat.apply([], [x]) 重构为 [].concat(x) 会引发错误

Question

将 Array.prototype.concat.apply([], [x]) 重构为 [].concat(x) 会引发此错误:

No overload matches this call.
  Overload 1 of 2, '(...items: ConcatArray<never>[]): never[]', gave the following error.
    Argument of type 'Moment | [Moment, Moment]' is not assignable to parameter of type 'ConcatArray<never>'.
      Type 'Moment' is missing the following properties from type 'ConcatArray<never>': length, join, slice
  Overload 2 of 2, '(...items: ConcatArray<never>[]): never[]', gave the following error.
    Argument of type 'Moment | [Moment, Moment]' is not assignable to parameter of type 'ConcatArray<never>'.
      Type 'Moment' is not assignable to type 'ConcatArray<never>'.ts(2769) 

我遗漏了什么问题以及如何有效地重构它？

Answer 1

这是因为 [] 的类型为 never[]。但由于您在 Array.prototype 的第一个示例中调用 concat (即 any[])，因此串联发生在 any[] 之间 和 (typeof x)[]

Argument of type 'Moment | [Moment, Moment]'

所以你想将其标准化为Moment[]？ 你可以这样做:

let y:Moment[];

y = ([] as any[]).concat(x);

// or something like

y = "length" in x ? x: [x];

// or something completely different, as you are already refactoring.

const [
  from = x,
  to = x
] = x as any;

或者将其放入函数中

function foo<T>(arg: T|T[]): T[] {
  return Array.isArray(arg) ? arg : [arg];
}