getBulk memory leak?

Java SDK
1

I encapsulated getBulk method like this:

public Map<String, Object> getBulk(Collection<String> ids) {
	Observable<Map<String, Object>> asyncResult = Observable.from(ids)
			.flatMap(new Func1<String, Observable<BinaryDocument>>() {
			  @Override
			  public Observable<BinaryDocument> call(String id) {
				return bucket.async().get(id, BinaryDocument.class);
			  }
			})
			.toMap(new Func1<BinaryDocument, String>() {
					 @Override
					 public String call(BinaryDocument bd) {
					   return bd.id();
					 }
				   },
				new Func1<BinaryDocument, Object>() {
				  @Override
				  public Object call(BinaryDocument bd) {
					return bd.content();
				  }
				});

		Map<String, Object> syncResult = asyncResult.toBlocking().single();
		return syncResult;
}

When I used getBulk, I got the following log:

2017-04-13 20:05:11,835 cb-io-5-3 com.couchbase.client.deps.io.netty.util.ResourceLeakDetector [ERROR]: LEAK: ByteBuf.release() was not called before it’s garbage-collected. See Netty.docs: Reference counted objects for more information

I’m quite sure I haven’t call ReferenceCountUtil.release method in getBulk, since I wrote get method like this, the above log wouldn’t happen.

public byte[] get(String key) {
	BinaryDocument res = bucket.get(BinaryDocument.create(key));
	if (res == null)
		return null;
	byte[] data = new byte[res.content().capacity()];
	for (int i = 0; i < res.content().capacity(); ++i) {
		data[i] = res.content().getByte(i);
	}
	ReferenceCountUtil.release(res.content()); // quite important !!
	return data;
}

The problem is I don’t know how to add release call within getBulk method.
Please help me to write a safer getBulk without memory leak reported by log.
Thanks.

2

By the way, my java sdk version is 2.3.2

3

Yeah if you use the binary document its up to you to release the buffers, otherwise you leak memory and netty complains.

In your code you could turn it into a byte[] in your toMap function and store it as a byte[] instead of a ByteBuf? The other option is to pass it along as a ByteBuf but then make sure release the buffer once used.

4

Ah, you mean I process the BinaryDocument just the same way as I did in get method ?
That is, convert BinaryDocument to byte[] in toMap.call function and release bd.content at the same time.

5

If it’s not BinaryDocument, is it necessary to release manually ?

6

No, only with the BinaryDocument. All the others handle this for you. The BinaryDocument is designed to give you lowest level of access but it also means more duties :slight_smile:

7

Now I got. Thank you. :smile: