TypeScript const 断言：如何使用 Array.prototype.includes？

Question

我正在尝试使用元素数组作为联合类型，这在 TS 3.4 中使用 const 断言变得很容易，所以我可以这样做：

const CAPITAL_LETTERS = ['A', 'B', 'C', ..., 'Z'] as const;
type CapitalLetter = typeof CAPITAL_LETTERS[string];

现在我想测试一个字符串是否为大写字母，但以下失败并显示“不可分配给类型的参数”：

let str: string;
...
CAPITAL_LETTERS.includes(str);

有没有比将CAPITAL_LETTERS 转换为unknown 然后再转换为Array<string> 更好的方法来解决这个问题？

Answer 1

使用 lodash

const CAPITAL_LETTERS = ['A', 'B', 'C', 'Z'] as const;
_.includes(CAPITAL_LETTERS, 'A');

Answer 2

这是一个适用于字符串和字符串文字的解决方案，使用 TypeScript 4.1 Template Literal Types 不会破坏其他任何内容，并且在条件下使用时为了方便起见还缩小了类型：

declare global {
    interface ReadonlyArray<T> {
        includes<S, R extends `${Extract<S, string>}`>(
            this: ReadonlyArray<R>,
            searchElement: S,
            fromIndex?: number
        ): searchElement is R & S;
    }
}

最初由 noppa 在a TypeScript github issue related to this 发布。

Answer 3

另一种解决方法是使用类型保护

https://www.typescriptlang.org/docs/handbook/advanced-types.html#user-defined-type-guards

const myConstArray = ["foo", "bar", "baz"] as const

function myFunc(x: string) {
    //Argument of type 'string' is not assignable to parameter of type '"foo" | "bar" | "baz"'.
    if (myConstArray.includes(x)) {
        //Hey, a string could totally be one of those values! What gives, TS?
    }
}

//get the string union type
type TMyConstArrayValue = typeof myConstArray[number]

//Make a type guard
//Here the "x is TMyConstArrayValue" tells TS that if this fn returns true then x is of that type
function isInMyConstArray(x: string): x is TMyConstArrayValue {
    return myConstArray.includes(x as TMyConstArrayValue)

    //Note the cast here, we're doing something TS things is unsafe but being explicit about it
    //I like to this of type guards as saying to TS:
    //"I promise that if this fn returns true then the variable is of the following type"
}

function myFunc2(x: string) {
    if (isInMyConstArray(x)) {
        //x is now "foo" | "bar" | "baz" as originally intended!
    }
}

虽然您必须引入另一个“不必要的”功能，但最终看起来干净且工作完美。在你的情况下，你会添加

const CAPITAL_LETTERS = ['A', 'B', 'C', ..., 'Z'] as const;
type CapitalLetter = typeof CAPITAL_LETTERS[string];
function isCapitalLetter(x: string): x is CapitalLetter {
    return CAPITAL_LETTERS.includes(x as CapitalLetter)
}

let str: string;
isCapitalLetter(str) //Now you have your comparison

//Not any more verbose than writing .includes inline
if(isCapitalLetter(str)){
  //now str is of type CapitalLetter
}

Answer 4

The standard library signature for Array<T>.includes(u) 假定要检查的值与数组元素T 的类型相同或更窄。但在您的情况下，您正在做相反的事情，检查一个 wider 类型的值。事实上，只有在T 和U 之间没有重叠时（即T & U 是never），你才会说Array<T>.includes<U>(x: U) 是一个错误并且必须被禁止。

现在，如果您不打算经常对includes() 进行这种“相反”的使用，并且希望运行时效应为零，那么您应该通过类型断言将CAPITAL_LETTERS 扩大到ReadonlyArray<string>：

(CAPITAL_LETTERS as ReadonlyArray<string>).includes(str); // okay

另一方面，如果您认为includes() 的这种使用应该在没有类型断言的情况下被接受，并且您希望它在您的所有代码中发生，您可以merge in a custom declaration：

// global augmentation needed if your code is in a module
// if your code is not in a module, get rid of "declare global":
declare global { 
  interface ReadonlyArray<T> {
    includes<U>(x: U & ((T & U) extends never ? never : unknown)): boolean;
  }
}

这将使数组（嗯，一个只读数组，但这就是您在此示例中所拥有的）将允许.includes() 的任何参数，只要数组元素类型和参数之间存在一些重叠类型。由于string & CapitalLetter 不是never，它会允许调用。不过，它仍然会禁止CAPITAL_LETTERS.includes(123)。

好的，希望对您有所帮助；祝你好运！