Question

在打字稿中，我可以执行以下操作：

const foo = <T>({ someArg }: { someArg: T }) => someArg

foo({ someArg: 4 }) // returns a number
foo({ someArg: "cool" }) // returns a string

我该如何使用jsdoc / checkjs进行等效处理？我以为它会像这样：

/**
 * @template T
 *
 * @param {object} props
 * @param {T} props.someArg
 * @returns {T}
 */
const foo = ({ someArg }) => someArg

foo({ someArg: 4 })
foo({ someArg: 'cool' })

但是我收到一个看起来像这样的错误：

src/components/example.js:10:7 - error TS2322: Type 'number' is not assignable to type 'T'.

10 foo({ someArg: 4 })
         ~~~~~~~

  src/components/example.js:5:4
    5  * @param {T} props.someArg
         ~~~~~~~~~~~~~~~~~~~~~~~~
    6  * @returns {T}
      ~~~
    The expected type comes from property 'someArg' which is declared here on type '{ someArg: T; }'

src/components/example.js:11:7 - error TS2322: Type 'string' is not assignable to type 'T'.

11 foo({ someArg: 'cool' })
         ~~~~~~~

  src/components/example.js:5:4
    5  * @param {T} props.someArg
         ~~~~~~~~~~~~~~~~~~~~~~~~
    6  * @returns {T}
      ~~~
    The expected type comes from property 'someArg' which is declared here on type '{ someArg: T; }'
      ~~~~~~~

不破坏args似乎可以：

/**
 * @template T
 *
 * @param {object} props
 * @param {T} props.someArg
 * @returns {T}
 */
const foo = props => props.someArg

更新-我认为可能是这个问题： Generic types from JSDoc aren't really generic

如何在JSDoc / Typescript中将@template与结构化参数一起使用

0 个答案: