![函数返回值]( "函数返回值")
本文介绍了TypeScript typeof 函数返回值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
承认我有这样的功能
const createPerson = () =>({ firstName: 'John', lastName: 'Doe' })
如何在声明 createPerson 之前不声明接口或类型，获取返回值类型?
像这样:
type Person = typeof createPerson()
示例场景
我有一个 Redux 容器，它将状态和分派动作映射到组件的 props.
containers/Counter.tsx
import { CounterState } from 'reducers/counter'//... 这里我还定义了 MappedState 和 mapStateToProps//我想移除的接口接口映射调度{增量:() =>任何}//并获取该函数的返回值类型const mapDispatchToProps =(调度:调度):映射调度=>({增量:() =>调度(增量)})//在这里导出它而不是 MappedDispatch导出类型 MappedProps = MappedState &映射调度导出默认连接(mapStateToProps，mapDispatchToProps)(计数器)
components/Counter.tsx
import { MappedProps } from 'containers/Counter'导出默认值(道具:MappedProps)=>(<div><h1>计数器</h1><p>{props.value}</p><button onClick={props.increment}>+</button>
)
我希望能够导出 mapDispatchToProps 的类型而无需创建 MappedDispatch 接口.
我在这里减少了代码，但它让我输入了两次相同的东西.
解决方案
原帖
打字稿
我创建了一个允许变通方法的小库，直到将完全声明性的方式添加到 TypeScript 中:
https://npmjs.com/package/returnof
还在 Github 上创建了一个问题，要求 泛型类型推断，这将允许以完全声明的方式来做到这一点:
https://github.com/Microsoft/TypeScript/issues/14400
2018 年 2 月更新
TypeScript 2.8
TypeScript 2.8 引入了一个新的静态类型 ReturnType，它允许实现:
https://github.com/Microsoft/TypeScript/pull/21496
您现在可以以完全声明的方式轻松获取函数的返回类型:
const createPerson = () =>({firstName: '约翰',姓氏:'母鹿'})type Person = ReturnType
Admit I have a function like this
const createPerson = () => ({ firstName: 'John', lastName: 'Doe' })
How can I, without declaring an interface or a type before declaring createPerson, get the return value type?
Something like this:
type Person = typeof createPerson()
Example Scenario
I have a Redux container that maps state and dispatch actions to props of a component.
containers/Counter.tsx
import { CounterState } from 'reducers/counter'
// ... Here I also defined MappedState and mapStateToProps
// The interface I would like to remove
interface MappedDispatch {
increment: () => any
}
// And get the return value type of this function
const mapDispatchToProps =
(dispatch: Dispatch<State>): MappedDispatch => ({
increment: () => dispatch(increment)
})
// To export it here instead of MappedDispatch
export type MappedProps = MappedState & MappedDispatch
export default connect(mapStateToProps, mapDispatchToProps)(Counter)
components/Counter.tsx
import { MappedProps } from 'containers/Counter'
export default (props: MappedProps) => (
<div>
<h1>Counter</h1>
<p>{props.value}</p>
<button onClick={props.increment}>+</button>
</div>
)
I want to be able to export the type of mapDispatchToProps without having to create MappedDispatch interface.
I reduced the code here, but it makes me type the same thing two times.
解决方案
Original Post
TypeScript < 2.8
I created a little library that permits a workaround, until a fully declarative way is added to TypeScript:
https://npmjs.com/package/returnof
Also created an issue on Github, asking for Generic Types Inference, that would permit a fully declarative way to do this:
https://github.com/Microsoft/TypeScript/issues/14400
Update February 2018
TypeScript 2.8
TypeScript 2.8 introduced a new static type ReturnType which permits to achieve that:
https://github.com/Microsoft/TypeScript/pull/21496
You can now easily get the return type of a function in a fully declarative way:
const createPerson = () => ({
firstName: 'John',
lastName: 'Doe'
})
type Person = ReturnType<typeof createPerson>
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