问题

可以像这样做吗？这将非常有帮助。使用泛型的方法签名：

class Foo<Parent extends Foo> {
    public bar<U extends Parameters<Parent.bar>>(x: U) {
    }
}

class Foo<Parent extends Foo> {
    public bar<U extends Parameters<Parent.bar>>(x: U) {
    }
}

答案1

得分: 1

以下是您要翻译的代码部分：

class Foo<Parent extends { bar: (...a: any[]) => any }> {
    public bar<U extends Parameters<Parent['bar']>>(x: U) {
    }
}

Playground链接

您还可能对这个将参数展开到bar函数的版本感兴趣：

class Foo<Parent extends { bar: (...a: any[]) => any }> {
    public bar(...x: Parameters<Parent['bar']>) {
    }
}

Playground链接

<details>
<summary>英文:</summary>

You can&#39;t if you want `Parent` to extend `Foo`, since that would cause a circular dependency in the type, but you can do it if you constrain `Parent` to something else with a `bar`:

```ts
class Foo&lt;Parent extends { bar: (...a: any[]) =&gt; any}&gt; {
    public bar&lt;U extends  Parameters&lt;Parent[&#39;bar&#39;]&gt;&gt;(x: U) {
    }
}

Playground Link

You might also be interested in this version that spreads the arguments back to the bar function:

class Foo<Parent extends { bar: (...a: any[]) => any}> {
    public bar(...x: Parameters<Parent['bar']>) {
    }
}

Playground Link