Parse string with java 8 streams

Question

I have ip address string like "192.168.1.1". Is it possible to parse string using java 8 streams and get primitive byte array as a result. Currently i have code

Arrays.stream(address.split("\\."))
      .map(UnsignedBytes::parseUnsignedByte)
      .toArray(Byte[]::new);

where is UnsignedBytes is guava class which return byte, but this code return Byte[] not byte[]

Update: Yes, i read about why ByteStream class is absent. My question is about, is it possible to get byte array using java-8 streams, without overhead such as create intermediate list.

The only way I could possibly conceive of doing that would be to write a custom collector. That could work, but it might be tricky. — Louis Wasserman, Jul 03 '16 at 21:08
You might collect to a list and use guava's `Bytes.toArray` to get a primitive array. — Hank D, Jul 03 '16 at 21:27
Possible duplicate of [In Java 8, is there a ByteStream class?](http://stackoverflow.com/questions/32459683/in-java-8-is-there-a-bytestream-class) — user140547, Jul 03 '16 at 21:27
see the final `collect` in this answer: http://stackoverflow.com/a/36613808/3487617 — Hank D, Jul 03 '16 at 21:38
Frankly, this is an interesting intellectual effort, but this is the kind of situation where a plain old simple for loop is much simpler and readable. And you won't parallelize that stream, so it doesn't even have that advantage. — JB Nizet, Jul 03 '16 at 21:53

score 3 · Answer 1 · answered Jul 03 '16 at 22:13

3

You could just use Java's InetAddress class.

 InetAddress i = InetAddress.getByName("192.168.1.1");
 byte[] bytes = i.getAddress();

answered Jul 03 '16 at 22:13

Robert

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I like this answer, but the javadoc says *The result is in network byte order: the highest order byte of the address is in `getAddress()[0]`*. I think you need to reverse the byte array to suit OP's need. – Bohemian Jul 03 '16 at 22:42
@Bohemian Not sure what order the OP wants. The code above has 192 in `bytes[0]`, 168 in `bytes[1]`, and so on. Which makes sense to me. – Robert Jul 04 '16 at 02:08
I must have been thinking bits (where 0 is the "last" bit). This is fine! – Bohemian Jul 04 '16 at 04:14

score 1 · Answer 2 · edited May 23 '17 at 12:32

1

You can use Pattern.splitAsStream(CharSequence) use a ByteArrayOutputStream in a collector:

byte[] bytes = Pattern.compile("\\.")
        .splitAsStream(address)
        .mapToInt(Integer::parseInt)
        .collect(
                ByteArrayOutputStream::new,
                ByteArrayOutputStream::write,
                (baos1, baos2) -> baos1.write(baos2.toByteArray(), 0, baos2.size())
        )
        .toByteArray();

Collect code adapted from https://stackoverflow.com/a/32470838/3255152.

edited May 23 '17 at 12:32

Community

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answered Jul 05 '16 at 17:41

mfulton26

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I like this answer better because it fits in the theme of the question, and would be useful outside of just parsing IP addresses. – lscoughlin Aug 17 '18 at 14:58

Parse string with java 8 streams

2 Answers2