如何指定表示特定函数的arg类型的类型

Question

我有一个函数(下面称为runLater(，它接受两个参数：

1. 任意函数
2. 一组参数任意函数

像这样：

function runLater(aFunction, aFunctionsParams) {
// store for later use
}

如何键入runLater函数，以便当我将函数作为第一个参数传入时，第二个参数仅限于该函数的参数类型？

function logNameAndAge(name: string, age: number) {...}
runLater(logNameAndAge, ['hoff', 42]) // ok, the parameter types match up
runLater(logNameAndAge, [false, 'oops']) // no ok, someFunction has [string, number] as paramters

Answer 1

使用Parameters<T>实用程序类型轻松实现

type func = (...args: any) => any
function runLater<T extends func>(aFunction: T, aFunctionsParams: Parameters<T>) {
// store for later use
}
function logNameAndAge(name: string, age: number) {}
// ok:
runLater(logNameAndAge, ['hoff', 42])
// errors:
//   Type 'boolean' is not assignable to type 'string'.(2322)
//   Type 'string' is not assignable to type 'number'.(2322)
runLater(logNameAndAge, [false, 'oops'])