Question

I'm trying to understand the pipe operator of observable API:

 export declare class Observable<T> implements Subscribable<T> {
.......
    pipe<A>(op1: OperatorFunction<T, A>): Observable<A>;
    pipe<A, B>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>): Observable<B>;
    pipe<A, B, C>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>, op3: OperatorFunction<B, C>): Observable<C>;
    pipe<A, B, C, D>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>, op3: OperatorFunction<B, C>, op4: OperatorFunction<C, D>): Observable<D>;
    pipe<A, B, C, D, E>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>, op3: OperatorFunction<B, C>, op4: OperatorFunction<C, D>, op5: OperatorFunction<D, E>): Observable<E>;
    pipe<A, B, C, D, E, F>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>, op3: OperatorFunction<B, C>, op4: OperatorFunction<C, D>, op5: OperatorFunction<D, E>, op6: OperatorFunction<E, F>): Observable<F>;
    pipe<A, B, C, D, E, F, G>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>, op3: OperatorFunction<B, C>, op4: OperatorFunction<C, D>, op5: OperatorFunction<D, E>, op6: OperatorFunction<E, F>, op7: OperatorFunction<F, G>): Observable<G>;
    pipe<A, B, C, D, E, F, G, H>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>, op3: OperatorFunction<B, C>, op4: OperatorFunction<C, D>, op5: OperatorFunction<D, E>, op6: OperatorFunction<E, F>, op7: OperatorFunction<F, G>, op8: OperatorFunction<G, H>): Observable<H>;
    pipe<A, B, C, D, E, F, G, H, I>(op1: OperatorFunction<T, A>, op2: OperatorFunction<A, B>, op3: OperatorFunction<B, C>, op4: OperatorFunction<C, D>, op5: OperatorFunction<D, E>, op6: OperatorFunction<E, F>, op7: OperatorFunction<F, G>, op8: OperatorFunction<G, H>, op9: OperatorFunction<H, I>): Observable<I>;
.......
}

As you could see, all of pipe overload methods receive and OperationFunction type.

Lets check the filter operator API:

export declare function filter<T, S extends T>(predicate: (value: T, index: number) => value is S, thisArg?: any): OperatorFunction<T, S>;
export declare function filter<T>(predicate: (value: T, index: number) => boolean, thisArg?: any): MonoTypeOperatorFunction<T>;

As you can see, you have an overload method that returns an OperationFunction and another that returns MonoTypeOperatorFunction.

I have the following questions:

RxJS6 operators like filter, could only be used on pipe method?
Why in filter we have a overload method that return different types but the pipe only receives one?

Answer 1

RxJS6运算符（例如过滤器）只能用于管道方法吗？

管道方法是 pure high-order 函数。所以不，它们可以用作任何其他方法。区别在于，如果在pipe闭包之外调用它们，则会丢失类型为intellisense。

为什么在过滤器中我们有一个重载方法返回不同的类型但是管道只收到一个？

随身携带一粒盐

流中的大多数运算符/用例都涉及转换，例如将值 x 从域 X 转换为值 y 来自域 Y 。 OperatorFunction<X,Y>接口的定义抽象了这个概念。

对于filter运算符，我们有一个特殊的情况OperatorFunction<X,Y>，其中满足条件 X为Y而Y为X ，因为操作从域 X 中获取值 x 并将其转换为来自同一域 X 中的值 y 。这个特殊情况再次由MonoTypeOperatorFunction<X>接口抽象，该接口的确是OperatorFunction<X,X>的{{3}}。

我的猜测是MonoTypeOperatorFunction界面的创建有两个目标：

抽象先前解释的概念

pipe(...fn: OperatorFunction<X,Y>)

保留 OperatorFunction的合同

RxJS6: Why observable Pipe operator only receives OperatorFunction and not MonoTypeOperatorFunction

1 个答案: