javascript - 测试变量类型或抛出 TypeScript 的函数？

Question

我想要一个函数来测试类变量是否不是 null 并在后续函数调用中使用它。但是得到了 TS 的投诉。封装这个验证函数，因为我需要在我的许多方法中调用它。

class A {
  point: Point | null
  validatePoint() {
    if (!this.point) throw new Error()
  }

  update(p: Point | null) {
    this.validatePoint()
    // ts complains this.point can be null
    doSomething(this.point)
  }
}

Answer 1

Typescript 3.7 引入了 assertions in control flow analysis :

class A {
    point: Point | null = null;
    validatePoint(): asserts this is { point: Point} {
        if (!this.point) throw new Error()
    }

    update(p: Point | null) {
        this.validatePoint()
        // now ts is sure that this.point can't be null
        doSomething(this.point)
    }
}

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