Question

我想知道这两种方法之间的行为是否存在差异，或者只是风格问题：

private Single<JsonObject> foo() {
     return Single.just(new JsonObject()).flatMap(next -> Single.just(next));
}

private Single<JsonObject> bar() {
     return Single.just(new JsonObject()).map(next -> next);
}

Answer 1

There is no difference in behavior as both are pointless operations. The first simply repeats wrapping the object into a Single, while the second maps it to itself. You would never have a reason to do either.

Read up on 'flatMap()' and 'map()': the first turns each value into an observable of different values, the second turns each value into a different value.

Answer 2

您可以自己代表 flatMap 运算符，就像其他两个运算符 map 和 merge 的序列一样。

地图会将您的源项目转换为可观察的项目，并根据地图内部的函数发出一个值。此时，合并将有助于将您的每个新可观察物（而不是源物）发出的所有物品放在一起。

那本书https://www.manning.com/books/rxjava-for-android-developers

上有一个很好的例证

map with merge together

为简化此代码，引入了 flatMap 运算符

only flatMap

rxjava：flatmap和map之间的区别

2 个答案: