打字稿，数组包含错误：“字符串”类型的参数不能分配给“A”类型的参数 | “B”'

Question

我所遇到的问题可以归结为以下错误的打字稿代码，

let arr = ["A","B"] as const // has to be const/readonly
function isAorB(str: string): str is "A" | "B" {
  return arr.includes(str);
}

Argument of type 'string' is not assignable to parameter of type '"A" | "B"'

聊天中提出了一些解决方案，但在某种程度上都是粗糙的，并且“采取额外的步骤”，真的没有更好的方法吗？

1. 选项 1：缩小

   str

的类型

let arr = ["A", "B"] as const;

function isAorB(str: string): str is "A" | "B" {
  return arr.includes(str as "A" | "B");
}

This works because you’re telling TypeScript, "I know str can only be "A" or "B" for this comparison."

这是错误的！

2. 选项 2：使用

   Set

const arrSet = new Set(["A", "B"] as const);

function isAorB(str: string): str is "A" | "B" {
  return arrSet.has(str);
}

这是一个“矫枉过正”，包含可以在不创建新对象的情况下工作！！

3. 选项 3：将 arr 转换为更宽松的类型

let arr = ["A", "B"] as const;

function isAorB(str: string): str is "A" | "B" {
  return (arr as readonly string[]).includes(str);
}

If you want to avoid modifying the function, you can cast arr to readonly string[], which loosens the type of includes, This approach sacrifices some type safety but resolves the error.

我同意聊天，这是一个奇怪的选择。

有没有办法在不欺骗编译器或创建新对象的情况下实现这个功能？

Answer 1

输入数组来包含字符串怎么样？

const arr: readonly string[] = ["A","B"];
function isAorB(str: string): str is "A" | "B" {
    return arr.includes(str);
}