如何将InputStream复制到AsynchronousFileChannel
我希望从(Tomcat ) InputStream中读取，并使用AsynchronousFileChannel异步地将(大型)内容复制到文件中。我可以在with a regular FileChannel上读到关于the missing transferTo的文章。但是如果我使用Java 7 AsyncFileChannel，我总是会得到BufferOverflowException。
try (AsynchronousFileChannel output = AsynchronousFileChannel.open(path, StandardOpenOption.CREATE, StandardOpenOption.WRITE);
output.lock(); // need to lock, this is one key reason to use channel
ReadableByteChannel input = Channels.newChannel(inputStream); // servlet InputStream
ByteBuffer buf = ByteBuffer.allocate(4096);
int position = 0;
int count;
Future<Integer> lastWrite = null;
while ((count = input.read(buf)) >= 0 || buf.position() > 0) {
logger.info("read {} bytes", count);
buf.flip();
output.write(buf, position);
if (count > 0) position += count;
buf.compact();
}
if (lastWrite != null) lastWrite.get(10, TimeUnit.SECONDS);然后当我跑的时候
14:12:30.597 [http-bio-9090-exec-3] INFO c.b.p.c.BlobUploadServlet - read 4096 bytes
14:12:30.597 [http-bio-9090-exec-3] INFO c.b.p.c.BlobUploadServlet - read 0 bytes
... many more with 0 bytes read ...
14:12:30.597 [http-bio-9090-exec-3] INFO c.b.p.c.BlobUploadServlet - read 3253 bytes
14:12:30.605 [http-bio-9090-exec-3] ERROR c.b.p.c.BlobUploadServlet - null
java.nio.BufferOverflowException: null
at java.nio.HeapByteBuffer.put(HeapByteBuffer.java:183) ~[na:1.7.0_17]
at java.nio.channels.Channels$ReadableByteChannelImpl.read(Channels.java:393) ~[na:1.7.0_17]我怎样才能修复BufferOverflow？另外，当读取0字节时，挂起循环并等待的正确方法是什么？
回答 1
Stack Overflow用户
发布于 2015-06-17 14:14:48
对原海报来说太晚了，但无论如何。
我试图复制您的问题(但使用略有不同的示例，我使用通道复制了大型文件)：
public static void main(String[] args) throws IOException, InterruptedException, ExecutionException {
final InputStream inputStream = new FileInputStream("/home/me/Store/largefile");
final ReadableByteChannel inputChannel = Channels.newChannel(inputStream);
final AsynchronousFileChannel outputChannel = AsynchronousFileChannel.open(
FileSystems.getDefault().getPath(
"/home/me/Store/output"),
StandardOpenOption.CREATE, StandardOpenOption.WRITE);
outputChannel.lock();
final ByteBuffer buffer = ByteBuffer.allocate(4096);
int position = 0;
int recievedBytes = 0;
Future<Integer> lastWrite = null;
while ((recievedBytes = inputChannel.read(buffer)) >= 0
|| buffer.position() != 0) {
System.out.println("Recieved bytes: " + recievedBytes);
System.out.println("Buffer position: " + buffer.position());
buffer.flip();
lastWrite = outputChannel.write(buffer, position);
// do extra work while asynchronous channel is writing bytes to disk,
// in perfect case more extra work can be done, not just simple calculations
position += recievedBytes;
// extra work is done, we should wait, because we use only one buffer which can be still busy
if (lastWrite != null) lastWrite.get();
buffer.compact();
}
outputChannel.close();
inputChannel.close();
inputStream.close();
}在循环的每一次迭代中，我们从输入流中读取一块数据，然后将这个块“推送”到输出流中。当前的线程不会等待写作的完成，它会继续进行，所以我们可以做额外的工作。但是在新的迭代之前，我们应该等待写作的完成。试着评论一下
if (lastWrite != null) lastWrite.get();
你会得到
java.nio.BufferOverflowException。
您的代码给了我一个技巧，让我使用未来来处理最后一次写操作。但你错过了最后一次手术。
此外，我还省略了片段中的一些额外的调优(只是为了简单起见，在处理文件时不需要额外的调优)。
https://stackoverflow.com/questions/23922280
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