TS 中的递归类型推断

作者：xie392地址：https://v.douyin.com/ieQtfv3e/更新时间：2024-12-21

案例代码

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type Curried<T, K> = T extends []
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    ? () => K
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    : T extends [infer AGR]
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    ? (params: AGR) => K
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    : T extends [infer AGR, ...infer REST]
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    ? (params: AGR) => Curried<REST, K>
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    : never
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declare function curry<T extends any[], K>(fn: (...args: T) => K): Curried<T, K>
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function sum(a: string, b: number, c:object) {
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    return a + b
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}
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const curryFn = curry(sum)
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curryFn('sss')(333)({a:1})