打字稿通用与返回类型不兼容

时间:2019-04-16 11:42:46

标签: typescript generics typescript-typings typescript-generics

我遇到打字稿泛型的问题:

function isString(a: any): a is string {
    return typeof a === 'string'
}

function concat<T extends string | number>(a: T, b: T): T {
    if (isString(a) && isString(b)) {
        return a.concat(b)
    }
    return a + b
}

游乐场网址:https://www.typescriptlang.org/play/index.html#src=function%20isString(a%3A%20any)%3A%20a%20is%20string%20%7B%0D%0A%20%20%20%20return%20typeof%20a%20%3D%3D%3D%20'字符串'%0D%0A%7D%0D%0A%0D%0Afunction%20concat%3CT%20扩展%20string%20%7C%20number%3E(a%3A%20T %2C%20b%3A%20T)%3A%20T%20%7B%0D%0A%20%20%20%20if%20(isString(a)%20%26%26%20isString(b))%20 %7B%0D%0A%20%20%20%20%20%20%20%20return%20a.concat(b)%0D%0A%20%20%20%20%20%7D%0D%0A%20% 20%20%20return%20a%20%2B%20b%0D%0A%7D%0D%0A

键入似乎合适,但我有一些错误。关于打字稿泛型似乎有些困惑,但是我发现的答案都没有帮助我解决这个基本用例。

1 个答案:

答案 0 :(得分:2)

TypeScript不narrow generic types via control flow。因此,即使已知a的类型为stringT的类型也会顽固地保持T。使代码按原样编译的唯一方法是使用type assertions来放置编译器(如注释中所述):

function concat<T extends string | number>(a: T, b: T): T {
    if (isString(a) && isString(b)) {
        return a.concat(b) as T; // assert as T
    }
    return (a as number) + (b as number) as T; // assert as numbers and T
}

警告:使用类型声明时,需要非常小心,不要对编译器撒谎。您可以从以下情况中看到我们拥有的:

// string literal types
const oops1 = concat("a", "b");
// type "a" | "b" at compile time, but "ab" at runtime

// numeric literal types
const oops2 = concat(5, 6); 
// type 5 | 6 at compile time, but 11 at runtime

// string | number types
let notSure = Math.random() < 0.5 ? "a" : 1 
const oops3 = concat(notSure, 100); // no error
// I bet you didn't want concat() to possibly accept string + number

最大的问题是T extends string | numberprompt编译器推断为string literal typenumeric literal typeToverloads。当您将诸如"a"之类的字符串文字作为参数传递时,T将缩小为"a",这意味着T仅是字符串"a"而没有其他价值。我想你不想那样。

您要执行的功能是传统上(无论如何在TS2.8之前)使用conditional types完成的功能:

function concat(a: string, b: string): string;
function concat(a: number, b: number): number;
function concat(a: string | number, b: string | number): string | number {
    if (isString(a) && isString(b)) {
        return a.concat(b);
    }
    return (a as number) + (b as number);
}

现在这些示例将按照您的预期进行操作:

const oops1 = concat("a", "b"); // string
const oops2 = concat(5, 6); // number
let notSure = Math.random() < 0.5 ? "a" : 1 
const oops3 = concat(notSure, 100); // error, notSure not allowed

使用泛型和{{3}},您可以获得相同的行为,但这可能不值得:

type StringOrNumber<T extends string | number> =
    [T] extends [string] ? string :
    [T] extends [number] ? number : never

function concat<T extends string | number>(
    a: T,
    b: StringOrNumber<T>
): StringOrNumber<T> {
    if (isString(a) && isString(b)) {
        return a.concat(b) as any;
    }
    return (a as number) + (b as number) as any;
}

const oops1 = concat("a", "b"); // string
const oops2 = concat(5, 6); // number
let notSure = Math.random() < 0.5 ? "a" : 1
const oops3 = concat(notSure, 100); // error

无论如何,希望能有所帮助。祝你好运!