使用Retrofit rxjava concatWith时堆栈溢出

Bhuvan 发表于 Dev

布万

我想使用rxjava Observable处理翻新中的分页。我听了另一个问题的建议。

我有100多个页面需要获取，但是链在第20页左右失败，并且在logcat中使用以下日志停止了对可观察对象的任何进一步订阅

04-04 04:12:11.766    2951-3012/com.example.app I/dalvikvm﹕ threadid=28: stack overflow on call to Ljava/util/concurrent/atomic/AtomicLongFieldUpdater$CASUpdater;.compareAndSet:ZLJJ
04-04 04:12:11.766    2951-3012/com.example.app I/dalvikvm﹕ method requires 56+20+32=108 bytes, fp is 0x94b52350 (80 left)
04-04 04:12:11.766    2951-3012/com.example.app I/dalvikvm﹕ expanding stack end (0x94b52300 to 0x94b52000)
04-04 04:12:11.766    2951-3012/com.example.app I/dalvikvm﹕ Shrank stack (to 0x94b52300, curFrame is 0x94b548dc)

有人知道为什么会发生这种情况吗？

更新：我知道这是由于递归而发生的，但是有没有更优雅的方式来处理带有翻新和rxjava的分页？

par

因此，鉴于我回答了您引用的原始问题，我可能也应该尝试回答这种情况。:)

这是我最初的寻呼答案的另一种有效（且可能更简单）的替代方法，因为我已经在军械库中开发了更多的Rx技巧。:)（以java8 lambda样式的伪代码完成）：

Observable.range(Integer.MAX_VALUE)
    // Get each page in order.
    .concatMap(page -> getResults(page))
    // Take every result up to and including the one where the next page index is null.
    .takeUntil(result -> result.next == null)
    // Add each output to a list builder. I'm using Guava's ImmutableList, but you could
    // just as easily use a regular ArrayList and avoid having to map afterwards. I just
    // personally prefer outputting an immutable data structure, and using the builder
    // is natural for this.
    //
    // Also, if you wanted to have the observable stream the full output at each page,
    // you could use collect instead of reduce. Note it has a slightly different syntax. 
    .reduce(
        ImmutableList.<ResponseObject>builder(),
        (builder, response) -> builder.addAll(response.results))
    // Convert list builder to one List<ResponseObject> of all the things.
    .map(builder -> builder.build())
    .subscribe(results -> { /* Do something with results. */ });