如何在泛型函数上使用正确的类型函数？

Question

我正在尝试修复 @typescript-eslint/ban-types lint 错误：

warning Don't use Function as a type @typescript-eslint/ban-types

我同意 Function 是一个不安全的类型，应该替换为适当的函数类型，例如 () => void.

但是我的代码有点复杂，我不知道如何在这里避免它，因为我正在使用 typeof that returns a Function.

这是我的代码的简化版本。完整版是here.

function start<T>(args: unknown[], couldBeAFunction: T) {
  if (typeof couldBeAFunction === 'function') {
    runFunction(args, couldBeAFunction); // couldBeAFunction now is a Function
  }
}

function runFunction<T extends Function>(args: unknown[], target: T) {
  target.apply(args, args);
}

我不能用 runFunction<T extends (...args: unknown[]) => unknown> 替换 runFunction<T extends Function>，否则我会在 runFunction 调用时收到以下错误：

Argument of type 'T & Function' is not assignable to parameter of type '(...args: unknown[]) => unknown'. Type 'Function' provides no match for the signature '(...args: unknown[]): unknown'.ts(2345)

如何解决此 lint 错误？

Answer 1

您可以使用类型保护：

type Fn = (...args: unknown[]) => unknown

const isFn = (fn: unknown): fn is Fn => typeof fn === 'function'

function start<T>(args: unknown[], couldBeAFunction: T) {
  if (isFn(couldBeAFunction)) {
    runFunction(args, couldBeAFunction); // couldBeAFunction now is a Function
  }
}

function runFunction<T extends Fn>(args: unknown[], target: T) {
  target.apply(args, args);
}

Playground

如何在泛型函数上使用正确的类型函数？

How to use a proper type Function on a generic function?

typescript

type-safety