为什么 class 属性的类型缩小检查在分配给变量（别名条件）时不起作用？

Question

如何在别名条件表达式中缩小 class 的属性类型？

Short：在 class 方法中，我想进行类型缩小检查，例如 this._end === null && this._head === null，但我想首先分配结果在类型缩小 if 子句中使用它之前检查变量。它没有按预期工作。

不短：假设我们有一个队列 class:

class Queue {
  private _length: number = 0;
  private _head: Node | null = null;
  private _end: Node | null = null;

  public enqueue(node: Node): boolean {
    this._length += 1;

    const isQueueEmpty = this._end === null;
    if (isQueueEmpty) {
      this._head = node;
      this._end = node;
      return true;
    }

    this._end.setLink(node); // TypeScript doesn't understand that this._end passed the null check 
                             // in the if statement and writes that this._end can be null. 
    this._end = node;
    return true;
  }
}

如果你直接在 if 条件中的 enqueue 函数中写一个检查，那么 TypeScript 理解 this._end 不是 null:

public enqueue(node: Node): boolean {
  this._length += 1;

  if (this._end === null) {
    this._head = node;
    this._end = node;
    return true;
  }

  this._end.setLink(node); // Now TypeScript understand that this._end passed the null check
  this._end = node;
  return true;
}

但代码的描述性较差。

此外，一般来说，我需要说的是 class 的两个属性都通过了 null 检查。具体来说：this._head 和 this._end.

我写了下面的方法：

type QueuePointerKeys = '_head' | '_end';

private _isQueueEmpty(): this is this & { [K in QueuePointerKeys]: null } {
  return this._end === null;
}

而且它不起作用。如果放在enqueue方法中，那么：

public enqueue(node: Node): boolean {
  this._length += 1;

  if (this._isQueueEmpty()) {
    this._head = node;     // TypeScript writes that a variable of type null cannot be assigned the type Node
    this._end = node;      // TypeScript writes that a variable of type null cannot be assigned the type Node
    return true;
  }

  this._end.setLink(node); // TypeScript doesn't understand that this._end passed the null check
                           // in the if statement and writes that this._end can be null.
  this._end = node;
  return true;
}

如何使用 class 方法（例如 _isQueueEmpty 进行类型缩小？这类似于 this._end === null && this._head === null 直接用作 if 语句的条件？

Answer 1

Control Flow Analysis of Aliased Conditions and Discriminants 是最近才添加的，它有局限性，包括：

Narrowing through indirect references occurs only when the conditional expression or discriminant property access is declared in a const variable declaration with no type annotation, and the reference being narrowed is a const variable, a readonly property, or a parameter for which there are no assignments in the function body.

由于 this._end 不是只读的属性，它不会为您工作。我通过将 this._end 分配给本地 const 变量，然后将其用于条件以及 setLink 调用来确认这是您的问题：

interface Node {
  setLink(node: Node): void
}

class Queue {
  private _length: number = 0;
  private _head: Node | null = null;
  private _end: Node | null = null;

  public enqueue(node: Node): boolean {
    this._length += 1;

    const end = this._end
    const isQueueEmpty = end === null;

    if (isQueueEmpty) {
      this._head = node;
      this._end = node;
      return true;
    }

    end.setLink(node);
    this._end = node;
    return true;
  }

}

事实上，有一个解决您的具体案例的未决问题：Control flow analysis of aliased conditions is not able to narrow object properties #46412, as well as Allow non-readonly properties to be used in aliased conditional expressions #44972。

希望他们能尽快找到解决这些问题的方法

为什么 class 属性的类型缩小检查在分配给变量（别名条件）时不起作用？

Why doesn't a type narrowing check of a class property not work when assigned to a variable (aliased conditions)?

javascript

typescript

typeguards

type-narrowing

typescript4.0

为什么 class 属性 的类型缩小检查在分配给变量（别名条件）时不起作用？

Why doesn't a type narrowing check of a class property not work when assigned to a variable (aliased conditions)?

javascript

typescript

typeguards

type-narrowing

typescript4.0

为什么 class 属性的类型缩小检查在分配给变量（别名条件）时不起作用？