Question

我有这种奇怪的类型CompletableFuture<CompletableFuture<byte[]>>，但我想要CompletableFuture<byte[]>。这可能吗？

public Future<byte[]> convert(byte[] htmlBytes) {
    PhantomPdfMessage htmlMessage = new PhantomPdfMessage();
    htmlMessage.setId(UUID.randomUUID());
    htmlMessage.setTimestamp(new Date());
    htmlMessage.setEncodedContent(Base64.getEncoder().encodeToString(htmlBytes));

    CompletableFuture<CompletableFuture<byte[]>> thenApply = CompletableFuture.supplyAsync(this::getPhantom, threadPool).thenApply(
        worker -> worker.convert(htmlMessage).thenApply(
            pdfMessage -> Base64.getDecoder().decode(pdfMessage.getEncodedContent())
        )
    );

}

Answer 1

其文档中有bug，但CompletableFuture#thenCompose系列方法相当于flatMap。它的声明还应该给你一些线索

public <U> CompletableFuture<U> thenCompose(Function<? super T,? extends CompletionStage<U>> fn)

thenCompose获取接收者CompletableFuture的结果（称之为 1 ）并将其传递给您提供的Function，其必须返回自己的CompletableFuture {1}}（称之为 2 ）。当 2 完成时，CompletableFuture返回的thenCompose（称为 3 ）将会完成。

在你的例子中

CompletableFuture<Worker> one = CompletableFuture.supplyAsync(this::getPhantom, threadPool);
CompletableFuture<PdfMessage /* whatever */> two = one.thenCompose(worker -> worker.convert(htmlMessage));
CompletableFuture<byte[]> result = two.thenApply(pdfMessage -> Base64.getDecoder().decode(pdfMessage.getEncodedContent()));

什么是CompletableFuture相当于flatMap？

1 个答案: