我正在尝试注释作为函数参数的对象的属性。
特别是当鼠标悬停在函数定义上时,我希望 options.length
解释出现在 vscode 中。
/**
* Memoize a function
* @param {(...fnArgs: T[]) => U} fn The function to memoize
* @param {Object} options Memoization options
* @param {number} options.max The max size of the LRU cache
* @param {number | undefined} options.length The response will be cached
* by the first N args, based on this value. Trailing args will still be
* passed to the underlying function but will be ignored during memoization.
*/
export const memo = <T, U>(
fn: (...fnArgs: T[]) => U,
{ max, length }: { max: number; length?: number }
) => {
const cachedArgs: T[][] = []
const cachedValues: U[] = []
const get = (args: T[]): U | undefined => {
const index = cachedArgs.findIndex(x => argsAreEqual(x, args, length))
if (index === -1) return
onUsed(index)
return cachedValues[index]
}
const set = (args: T[], value: U) => {
cachedArgs.push(args)
cachedValues.push(value)
}
const onUsed = (index: number) => {
moveToEnd(index, cachedArgs)
moveToEnd(index, cachedValues)
}
const prune = () => {
if (cachedArgs.length >= max) {
cachedArgs.shift()
cachedValues.shift()
}
}
return (...args: T[]) => {
let value = get(args)
if (value) return value
prune()
value = fn(...args)
set(args, value)
return value
}
}
将鼠标悬停在类型签名上时,我得到以下信息
@param fn — The function to memoize
@param options — Memoization options
我已尽力解决copy from the docs,但它没有显示options.length
的解释。
我希望它显示 options.length
参数如何工作的解释。我该怎么做?
额外的问题,不确定如何使泛型与 jsdoc 一起工作……非常感谢 @template
的帮助!
答案 0 :(得分:1)
此行为似乎是一个已知错误,即解构参数不会显示其 JSDoc 注释:microsoft/TypeScript#24746。我不太确定为什么,但根据 a comment,给 @param
类型的 Object
等同于给它类型的 any
,你可以得到通过使用不同的类型来实现您想要的行为。在下面,我将 Object
更改为 {}
:
/**
* @param {(...fnArgs: T[]) => U} fn
* @param {{}} options Memoization options
* @param {number} options.max The max size of the LRU cache
* @param {number | undefined} options.length The response will be cached
* by the first N args, based on this value. Trailing args will still be
* passed to the underlying function but will be ignored during memoization.
*/
export const memo = <T, U>(
fn: (...fnArgs: T[]) => U,
{ max, length }: { max: number; length?: number }
) => { }
并且您的 IntelliSense 开始工作,至少在将鼠标悬停在该功能上时。 (我认为当您将鼠标悬停在代码中名为 max
或 length
的实际标识符上时,不可能获得评论)。观察:
/* IntelliSense shows:
const memo: <T, U>(fn: (...fnArgs: T[]) => U, { max, length }: {
max: number;
length?: number | undefined;
}) => void
@param fn
@param options — Memoization options
@param options.max — The max size of the LRU cache
@param options.length — The response will be cached by the first N args,
based on this value. Trailing args will still be passed to the
underlying function but will be ignored during memoization.
*/
请注意,这似乎只适用于 TypeScript 代码(如 .ts
或 .tsx
),而不适用于 JavaScript 代码。如果您在 JavaScript 中执行此操作,编译器将抱怨除 Object
或 object
之外的任何内容。
至于你的额外问题,我认为它应该像在 JavaScript 中一样在 TypeScript 中工作:
/**
* @template T the arg type of fn
* @template U the return type of fn
* @param {(...fnArgs: T[]) => U} fn
显示给我
/* IntelliSense shows:
const memo: <T, U>(fn: (...fnArgs: T[]) => U, { max, length }: {
max: number;
length?: number | undefined;
}) => void
@template — T the arg type of fn
@template — U the return type of fn
@param fn
*/