Fill

挑战

Fill, a common JavaScript function, now let us implement it with types. Fill<T, N, Start?, End?>, as you can see,Fill accepts four types of parameters, of which T and N are required parameters, and Start and End are optional parameters. The requirements for these parameters are: T must be a tuple, N can be any type of value, Start and End must be integers greater than or equal to 0.

type exp = Fill<[1, 2, 3], 0> // expected to be [0, 0, 0]

In order to simulate the real function, the test may contain some boundary conditions, I hope you can enjoy it :)

解答

解答跟着我的思路一步步来，最后解出最终答案

初始状态：

type Fill<
  T extends unknown[],
  N,
  Start extends number = 0,
  End extends number = T['length'],
> = any

我们需要用T extends去遍历整个数组

type Fill<
  T extends unknown[],
  N,
  Start extends number = 0,
  End extends number = T['length'],
> = 
  T extends [infer L,...infer R]
  ? never
  : never

一开始的思路是我们需要有一个 A 元组来传递我们每一步的结果，然后靠A['length']来判断 Start 和 End 的边缘状态，大概像下面这样

type Fill<
  T extends unknown[],
  N,
  Start extends number = 0,
  End extends number = T['length'],
  A extends unknown[] = [] 
> = 
  T extends [infer L,...infer R]
   ? A['length'] extends Start
     ? never
     : A['length'] extends End
       ? never
       : never
  : A

但是这样显然不够，因为我们只知道A['length']刚好等于 Start 和 End的情况，所以我们需要有一个 boolean 的 In 来判断我们是否在Start和End内，如果是在内，那么我们只需要判断End，除此之外，我们只需要判断Start

type Fill<
  T extends unknown[],
  N,
  Start extends number = 0,
  End extends number = T['length'],
  In extends boolean = false,
  A extends unknown[] = [] 
> = 
  T extends [infer L,...infer R]
  ? 
    In extends true 
    ? A['length'] extends End
      ? never
      : never
    : A['length'] extends Start
      ? never
      : never
  : A

然后考虑
在内的时候我们要压入的是N，在外我们要压入L
以及过边缘In的状态转换

就写出了能通过大部分样例的版本

type Fill<
  T extends unknown[],
  N,
  Start extends number = 0,
  End extends number = T['length'],
  In extends boolean = false,
  A extends unknown[] = [] 
> = 
  T extends [infer L,...infer R]
  ? 
    In extends true 
    ? A['length'] extends End
      ? Fill<R,N,Start,End,false,[...A,L]>
      : Fill<R,N,Start,End,true,[...A,N]>
    : A['length'] extends Start
      ? Fill<R,N,Start,End,true,[...A,N]>
      : Fill<R,N,Start,End,false,[...A,L]>
  : A

但是我们发现样例里当Start和End相等时，是不做替换的
所以加入Start extends End的判断就做出了我们的最终版本：

type Fill<
  T extends unknown[],
  N,
  Start extends number = 0,
  End extends number = T['length'],
  In extends boolean = false,
  A extends unknown[] = [] 
> = 
Start extends End
? T
: T extends [infer L,...infer R]
  ? 
    In extends true 
    ? A['length'] extends End
      ? Fill<R,N,Start,End,false,[...A,L]>
      : Fill<R,N,Start,End,true,[...A,N]>
    : A['length'] extends Start
      ? Fill<R,N,Start,End,true,[...A,N]>
      : Fill<R,N,Start,End,false,[...A,L]>
  : A