Convert byte[] to String using binary encoding

Question

I want to translate each byte from a byte[] into a char, then put those chars on a String. This is the so-called "binary" encoding of some databases. So far, the best I could find is this huge boilerplate:

byte[] bytes = ...;
char[] chars = new char[bytes.length];
for (int i = 0; i < bytes.length; ++i) {
    chars[i] = (char) (bytes[i] & 0xFF);
}
String s = new String(chars);

Is there another option from Java SE or perhaps from Apache Commons? I wish I could have something like this:

final Charset BINARY_CS = Charset.forName("BINARY");
String s = new String(bytes, BINARY_CS);

But I'm not willing to write a Charset and their codecs (yet). Is there such a ready binary Charset in JRE or in Apache Commons?

How is that "huge boilerplate"? Just wrap it in a method that takes a byte array and returns a string. — ColinD
– ColinD, Commented Dec 7, 2011 at 17:13
Not entirely sure of your problem. Won't ISO 8859-1 (Latin-1) do the job? It is an 8-bit single byte encoding... — Dunes
– Dunes, Commented Dec 7, 2011 at 17:28
@ColinD That wrapping was done before. But I had to code this method a few times in distinct projects with no shared library between them. And I don't want to build a library only for this. That's why we use stuff like java.util, java.text, java.lang and Apache Commons. — fernacolo
– fernacolo, Commented Dec 7, 2011 at 17:34

Peter Lawrey · Accepted Answer · 2011-12-07 17:25:06Z

9

You could use the ASCII encoding for 7-bit characters

String s = "Hello World!";
byte[] b = s.getBytes("ASCII");
System.out.println(new String(b, "ASCII"));

or 8-bit ascii

String s = "Hello World! \u00ff";
byte[] b = s.getBytes("ISO-8859-1");
System.out.println(new String(b, "ISO-8859-1"));

EDIT

System.out.println("ASCII => " + Charset.forName("ASCII"));
System.out.println("US-ASCII => " + Charset.forName("US-ASCII"));
System.out.println("ISO-8859-1 => " + Charset.forName("ISO-8859-1"));

prints

ASCII => US-ASCII
US-ASCII => US-ASCII
ISO-8859-1 => ISO-8859-1

edited Dec 7, 2011 at 17:25

answered Dec 7, 2011 at 17:11

Peter Lawrey

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13 Comments

fernacolo Over a year ago

Thanks, but there will exist some 8-bit characteres.

jtahlborn Over a year ago

other way around, "ASCII" is an alias for "US-ASCII". obviously, both will work, i'm just saying that this is the "official" name java uses.

ColinD Over a year ago

@Peter Lawrey: From the article you linked: "US-ASCII is the Internet Assigned Numbers Authority (IANA) preferred charset name for ASCII." Also, I believe the charset that's required to be present in all Java implementations is "US-ASCII".

fernacolo Over a year ago

ISO-8859-1 did the trick. Very interesting. I thought that it would map some bytes to 0x7F, because not all byte values have meaning in this encoding (according to en.wikipedia.org/wiki/ISO/IEC_8859-1).

ColinD Over a year ago

The standard charset identifiers are listed in the Charset Javadoc: docs.oracle.com/javase/6/docs/api/java/nio/charset/Charset.html

|

Chris Aldrich · Accepted Answer · 2011-12-07 21:42:00Z

1

You could skip the step of a char array and putting in String and even use a StringBuilder (or StringBuffer if you are worried about multi-threading). My example shows StringBuilder.

byte[] bytes = ...;
StringBuilder sb = new StringBuilder(bytes.length);
for (int i = 0; i < bytes.length; i++) {
  sb.append((char) (bytes[i] & 0xFF));
}

return sb.toString();

I know it doesn't answer your other question. Just seeking to help with simplifying the "boilerplate" code.

edited Dec 7, 2011 at 21:42

answered Dec 7, 2011 at 17:13

Chris Aldrich

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2 Comments

ColinD Over a year ago

There is no good reason not to use StringBuilder instead of StringBuffer if you're using it as a local variable like in your example.

Chris Aldrich Over a year ago

@ColinD Modified to StringBuilder. You are right there. Used to using StringBuffer as that was all we had before Java 5. Also, we have a multi-threaded app, so StringBuffer works well for us. But ++ to your point.

Andrew Rasmussen · Accepted Answer · 2011-12-07 17:12:58Z

0

There is a String constructor that takes an array of bytes and a string specifying the format of the bytes:

String s = new String(bytes, "UTF-8");   // if the charset is UTF-8
String s = new String(bytes, "ASCII");   // if the charset is ASCII

answered Dec 7, 2011 at 17:12

Andrew Rasmussen

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3 Comments

fernacolo Over a year ago

UTF-8 will translate some multi-byte characters in single-char characteres, so it won't work. ASCII only handles 7-bit characteres, and there will exist some 7-bit characters.

Andrew Rasmussen Over a year ago

Why would this get a downvote? I told you that the String constructor does exactly what you want it to do. Sorry for not doing your research for you as to what charset to use...

fernacolo Over a year ago

This downvote was not from me. By the way, thanks for trying to answer.

havexz · Accepted Answer · 2011-12-07 17:15:58Z

0

You can use base64 encoding. There is an implementation done by apache

http://commons.apache.org/codec/

Base 64 http://commons.apache.org/codec/apidocs/org/apache/commons/codec/binary/Base64.html

answered Dec 7, 2011 at 17:15

havexz

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