Question

看下面的代码：

type Shape = null | string | string[] | { [key: string]: string | string[] }

interface ICfg<InShape extends Shape, OutShape extends Shape> {
  shapes?: readonly [InShape, OutShape]
  handle?(x: NonNullable<this["shapes"]>[0]): NonNullable<this["shapes"]>[1]
}

function apply<InShape extends Shape, OutShape extends Shape>(cfg: () => ICfg<InShape, OutShape>) {
  return cfg()
}

const shape = {
  type: "abc",
  date: "qwe",
}

var x = apply(() => ({
  shapes: [shape, shape],
  handle: x => null as any,
}))

变量x的类型为ICfg<{ type: string; date: string; }, { type: string; date: string; }>，这很完美。

但是让我们稍微更改一下函数语法：从() => ({ ... })更改为() => { return { ... } }：

var y = apply(() => {
  return {
    shapes: [shape, shape],
    handle: x => null as any,
  }
})

现在y的类型是ICfg<Shape, Shape>，这很糟糕。

实际上，我遇到了更多困难，因为我确实需要将变量shape放入函数内部（实际上，函数apply具有参数，而shape则基于参数）： / p>

var z = apply(() => {
  const shape = {
    type: "abc",
    date: "qwe",
  }

  return {
    shapes: [shape, shape],
    handle: x => null as any,
  }
})

此代码会为变量ICfg<Shape, Shape>产生相同的z。

我需要变量z像ICfg<{ type: string; date: string; }, { type: string; date: string; }>一样具有类型x。没有显式指定apply的通用参数，有什么方法可以实现？

Full code in the playground.

函数具有显式返回时的窄推断类型

0 个答案: