小编hjf*_*yer的帖子

Object.hashCode()的默认行为是返回对象的"地址",以便当且仅当a == b时,a.hashCode()== b.hashCode().如果超类已经定义了hashCode(),我如何在用户定义的类中获得此行为？例如:

class A {
  public int hashCode() {
    return 0;
  }
}

class B extends A {
  public int hashCode() {
    // Now I want to return a unique hashcode for each object.
    // In pythonic terms, it'd look something like:
    return Object.hashCode(this);
  }
}

想法？

我遇到了any在 Typescript 中似乎无法避免的情况。这是一个反映我正在尝试做的事情的示例：

type NativeFn<A, B> = {
    kind: 'native'
    fn: (a: A) => B
}

type ComposeFn<A, B, C> = {
    kind: 'compose'
    f: Fn<B, C>
    g: Fn<A, B>
}

type Fn<A, B> = NativeFn<A, B>
    | ComposeFn<A, any, B>  // <=== HERE

function evalfn<A, B>(fn: Fn<A, B>, arg: A): B {
    switch (fn.kind) {
        case 'native': return fn.fn(arg)
        case 'compose': {
            // intermediate has type "any", which is a drag.
            const intermediate = evalfn(fn.g, arg)
            return …