typescript - 如何从其字符串表示中获取类型

Question

如何使 typescript 根据 expectedType 的值派生 assertTypof 的泛型参数

也就是说，我想在不指定 number 两次的情况下应用下面的函数

playable example

type TypeLiteral = "string" | "number" | "boolean" | "symbol" | "undefined" | "object" | "function"

// I know its bad to return generic, but dont know how to make without it
function assertTypeof<T>(expectedType: TypeLiteral, target: any): T {
    const actualType = typeof target
    if (actualType === expectedType) return target

    throw new Error(`Expected ${expectedType}, but got ${actualType}`)
}


const n1 = assertTypeof<number>('number', 1) // fine
const n2 = assertTypeof<number>('number', {}) // error

Answer 1

您可以在接口(interface)中编码字符串 -> 类型映射，并使用 indexed access type operator作为 assertTypeof 的返回类型:

interface TypeMap {
    string: string,
    number: number,
    boolean: boolean,
    symbol: Symbol,
    undefined: undefined,
    object: object,
    function: Function
};

function assertTypeof<N extends keyof TypeMap>(expectedType: N, target: any)
 : TypeMap[N] {
    const actualType = typeof target
    if (actualType === expectedType) return target

    throw new Error(`Expected ${expectedType}, but got ${actualType}`)
}

const n1 = assertTypeof('number', 1) // fine
const n2 = assertTypeof('number', {}) // runtime error