我建立了一个小的类型级别的库来简化形式形式的函数对的编写
function a(...): T | undefined;
function b(...): T;
其中b是从a派生的,如果返回undefined,则抛出异常。基本思路如下:
export enum FailMode { CanFail, CanNotFail }
export const canFail = FailMode.CanFail;
export const canNotFail = FailMode.CanNotFail
export type Failure<F extends FailMode> = F extends FailMode.CanFail ? undefined : never;
export type Maybe<T, F extends FailMode> = T | Failure<F>;
export function failure<F extends FailMode>(mode: F): Failure<F> {
if (mode === canFail) return undefined as Failure<F>;
throw new Error("failure");
}
现在,我们可以将a
和b
合并为一个函数,该函数需要一个额外的参数来区分类型:
function upperCaseIfYouCan<F extends FailMode>(x: string | undefined, mode: F): Maybe<string, F> {
if (x === undefined)
return failure<F>(mode);
return x.toUpperCase();
}
// both of these now work
let y: string = upperCaseIfYouCan("foo", canNotFail); // can throw, but will never return undefined
let z: string | undefined = upperCaseIfYouCan("foo", canFail); // cannot throw, but can return undefined
现在,在大多数情况下,我需要canNotFail
变体,并且我想知道是否有一种方法可以使它成为“默认”,因为我不必传递canNotFail
参数在这种情况下,则可以执行以下操作:
let y: string = upperCaseIfYouCan("foo"); // can throw
我已经确定这不能通过默认参数来实现,因为默认参数的类型必须无法与F
统一。有没有办法以不同的方式实现这一目标,例如,像upperCaseIfYouCan
这样的函数定义起来同样容易?
答案 0 :(得分:1)
编译器对您指定默认参数并不满意,因为有人总是可以在调用时随手手动指定通用类型参数,例如:upperCaseIfYouCan<FailMode.CanFail>("");
。如果您不希望发生这种情况,则可以使用type assertion来抑制该错误,此外,还应为F
类型参数提供自己的default,以便没有明显的选择对于F
,编译器将选择您的默认值,而不是完整的FailMode
:
function upperCaseIfYouCan<F extends FailMode = FailMode.CanNotFail>(
x: string | undefined,
mode: F = canNotFail as F
): Maybe<string, F> {
if (x === undefined)
return failure<F>(mode);
return x.toUpperCase();
}
这应该适合您的用例:
let str = upperCaseIfYouCan("foo", canNotFail); // string
let strOrUndef = upperCaseIfYouCan("foo", canFail); // string | undefined
// this is the behavior you want
str = upperCaseIfYouCan("foo"); // string
另一方面,您可以使用overloads处理两种不同的调用函数的方法。尽管泛型在概念上更优雅,但在这种情况下重载可能更直观:
function upperCaseIfYouCan(x: string | undefined, mode: FailMode.CanFail): string | undefined;
function upperCaseIfYouCan(x: string | undefined, mode?: FailMode.CanNotFail): string;
function upperCaseIfYouCan(
x: string | undefined,
mode: FailMode = FailMode.CanNotFail
) {
if (x === undefined) return failure(mode);
return x.toUpperCase();
}
let str = upperCaseIfYouCan("foo", canNotFail); // string
let strOrUndef = upperCaseIfYouCan("foo", canFail); // string | undefined
str = upperCaseIfYouCan("foo"); // string
实现中没有比带有类型断言的版本更安全的方法,但是至少不能轻易以错误的方式调用重载(没有F
可以手动指定)。
好的,希望其中之一能给您一些指导。祝你好运!