How does Java form TCP packets from byte arrays?

Question

While SocketIn/OutputStreams in Java seem stream based to an API user, TCP packets are packet based. One can write only one byte or an array of bytes. But one could also write more bytes than a TCP packet could carry.

So how does Java form TCP packets from the write methods?

Creates calling write(singleByte) 4 times 4 TCP packets? Or does java join the bytes together?

Does java join and split bytes to form the TCP packets?

The OS is a lot better at handling those things than Java. No need to duplicate functionality by rewriting all that code into the JDK. — Kayaman, Aug 04 '14 at 18:16
@Kayaman So this is handled by the os? Then how do most os's handle it?And why the downvote? — MinecraftShamrock, Aug 04 '14 at 18:34
I didn't downvote you. Most OSes handle it *effectively* so we don't have to care about it. — Kayaman, Aug 04 '14 at 18:58

score 1 · Accepted Answer · answered Aug 04 '14 at 19:19

1

No, Java, or more precisely the Java Virtual Machine, has no reason to get into these details.

What the JVM does, is that it opens a native Socket just like any native program would. It then allows the Java code to interact with the native socket through the various Java API. This leaves the networking details to the operating system's network stack (TCP/IP implementation).

answered Aug 04 '14 at 19:19

Malt

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Is it even possible for native sockets to know whether the code would write a single byte more or not? – MinecraftShamrock Aug 04 '14 at 20:33
I was going to to write that it's impossible to know whether the underlying socket buffer was full, before calling write(), but then I googled and found this: http://stackoverflow.com/a/1805234/3199595 – Malt Aug 04 '14 at 20:46

How does Java form TCP packets from byte arrays?

1 Answers1