I expected that, when a java character is stored as "UTF-16", each character uses 2 bytes, so "hello" should consume 10 bytes, but this code:
String h = "hello";
System.out.println(new String(h.getBytes("UTF-16"), "UTF-16").length());
System.out.println(new String(h.getBytes("UTF-8"), "UTF-8").getBytes("UTF-16").length);
Will print "5 12"
My question:
(1) I expected that the first println should get "10" as I mentioned. But why 5?
(2) For the second println, I am trying to getBytes for it first as "UTF-8" then as "UTF-16". I suppose it should also be 10. But actually it's 12.
I'm using MAC and my region is HongKong. Would you help to explain what's happening in the program, and how "5 12" actually came out?
Thanks a lot!
(1) I expected that the first println should get "10" as I mentioned. But why 5?
You take a 5 character string, encode it as bytes using UTF-16 encoding.
Then you create a new string by decoding the bytes (correctly) from UTF-16, which gives you a new string consisting of your original 5 characters again.
(2) For the second println, I am trying to getBytes for it first as "UTF-8" then as "UTF-16". I suppose it should also be 10. But actually it's 12.
This part of the code:
new String(h.getBytes("UTF-8"), "UTF-8")
is actually a no-op. It is just a rather expensive way to copy a string. You encode the string to bytes using UTF-8 as the encoding scheme, and then you create a new string by decoding the UTF-8 encoded bytes.
So effectively, you are doing this:
"hello".getBytes("UTF-16").length
The reason for the extra 2 bytes is that UTF-16 encoding puts a BOM (byte order mark) as the first (2 byte) code unit.
For more information, read the Unicode FAQs on "UTF-8, UTF-16, UTF-32 & BOM".
I expected that the first println should get "10" as I mentioned. But why 5?
You are calling length() on the String, not on the byte[]. So this will give you the length of the String in characters (at least as long as we are staying in the Unicode Basic Multilingual Plane -- this unfortunately breaks down when you have characters that need variable-length encoding even in UTF-16).
Once you have a String, it does not matter what encoding was used to create it. length is always given in terms of characters.
If you converted this into a byte[] using UTF-16, you might rightfully have expected 10 (for the five characters times two bytes each) -- that it actually ends up being 12 is due to a Byte Order Mark being included.
Related
When an I/O stream manages 8-bit bytes of raw binary data, it is
called a byte stream. And, when the I/O stream manages 16-bit Unicode
characters, it is called a character stream.
Byte stream is clear. It uses 8-bit bytes. So if I were to write a character that uses 3 bytes it would only write its last 8 bits! Thus making incorrect output.
So that is why we use character streams. Say I want to write Latin Capital Letter Ạ. I would need 3 bytes for storing in UTF-8. But say I also want to store 'normal' A. Now it would take 1 byte to store.
Are you seeing pattern? We can't know how much bytes it will take for writing any of these characters until we convert them. So my question is why is it said that character streams manage 16-bit Unicode characters? When in case where I wrote Ạ that takes 3 bytes it didn't cut it to last 16-bits like byte streams cut last 8-bits. What does that quote even mean then?
In Java, a String is composed of a sequence of 16-bit chars, representing text stored in the UTF-16 encoding.
A Charset is an object that describes how to convert Unicode characters to a sequence of bytes. UTF-8 is an example of a charset.
A character stream like Writer, when it outputs to a thing that contains bytes -- a file, or a byte output stream like OutputStream -- uses a Charset to convert Strings to simple byte sequences for output. (Technically, it converts the UTF-16 chars to Unicode characters and then converts those to byte sequences with the Charset.) A Reader, when reading from a byte source, does the reverse conversion.
In UTF-16, Ạ is represented as the 16-bit char 0x1EA1. It takes only 16 bits in UTF-16, not 24 bits as in UTF-8.
If you converted it to bytes with the UTF-8 encoding, as here:
ByteArrayOutputStream baos = new ByteArrayOutputStream();
Writer writer = new OutputStreamWriter(baos, StandardCharsets.UTF_8);
writer.write("Ạ");
writer.close();
return baos.toByteArray();
Then you would get the 3 byte sequence 0xE1 0xBA 0xA1 as expected.
In Java, a character (char) is always 16 bits, as can be seen from its max value - 65535. This is why the quote is not wrong. 16 bit is indeed a character.
"How can all the Unicode characters be stored in just 16 bits?" you might ask. This is done in Java using the UTF-16 encoding. Here's how it works (in very simplified terms):
Every Unicode code point in the Basic Multilingual Plane is encoded in 16 bits. (Yes 16 bit is enough for that) Every code point outside of the BMP is encoded with a pair of 16 bit characters, called surrogate pairs.
"Ạ" (U+1EA0) is inside the BMP, so can be encoded with 16 bits.
You said:
Say I want to write Latin Capital Letter Ạ. I would need 3 bytes for storing in UTF-8. But say I also want to store 'normal' A. Now it would take 1 byte to store!
That does not make the quote incorrect. The stream still "manages 16-bit characters", because that's what you will give it with Java code. When you call println on a PrintStream, you are giving it a String, which is a bunch of chars under the hood, which is a bunch of 16-bits. So it is really managing a stream of 16-bit characters. It's just that it outputs them in a different encoding.
It's probably worth mentioning what happens when you try to print a character that is not in the BMP. This would still not make the quote incorrect. The quote does not say "code point". It says "character" which would refer to the upper/lower surrogates of the surrogate pair that you are printing.
While programming I encountered a weird behavior of Strings that are converted to bytes and then back to Strings again. Some chars are converted wrongly, and therefore the hashCode of the String is also changed. The length of the Strings remain the same.
The problem seems to occur with chars from 55296 - 57343 (U+D800 to U+DFFF). Other chars work fine. Is it because they are surrogates?
String string = new String(new char[] { 56000 });
System.out.println((int)string.charAt(0));
System.out.println((int)new String(string.getBytes(StandardCharsets.UTF_8), StandardCharsets.UTF_8).charAt(0));
The console output is:
56000
63
What is going on here? Is this a java bug, or am I misunderstanding something?
That's because these values are not characters but surrogates. Two of these values form a surrogate pair that in turn represents one character. If you have just one low or high surrogate value this is an invalid encoding and not a character.
Since this is an invalid encoding, it is replaced by a "?" character when you convert it to UTF-8.
You can read more about it for example here https://en.wikipedia.org/wiki/UTF-16
I been doing some coding with String in Java8,Java 11 but this question is based on Java 8. I have this little snippet.
final char e = (char)200;//È
I just thought that the characters between 0.255[Ascii+extended Ascii] would always fit in a byte just because 2^8=256 but this seems not to be true i have try on the website https://mothereff.in/byte-counter and states that the character is taking 2 bytes can somebody please explain to me.
Another question in a lot of post states that Java is UTF-16 but in my machine running Windows 7 is returning UTF-8 in this snippet.
String csn = Charset.defaultCharset().name();
Is this platform depent?
Other questions i have try this snippet.
final List<Charset>charsets = Arrays.asList(StandardCharsets.ISO_8859_1,StandardCharsets.US_ASCII,StandardCharsets.UTF_16,StandardCharsets.UTF_8);
charsets.forEach(a->print(a,"È"));
System.out.println("getBytes");
System.out.println(Arrays.toString("È".getBytes()));
charsets.forEach(a->System.out.println(a+" "+Arrays.toString(sb.toString().getBytes(a))));
private void print(final Charset set,final CharSequence sb){
byte[] array = new byte[4];
set.newEncoder()
.encode(CharBuffer.wrap(sb), ByteBuffer.wrap(array), true);
final String buildedString = new String(array,set);
System.out.println(set+" "+Arrays.toString(array)+" "+buildedString+"<<>>"+buildedString.length());
}
And prints
run:
ISO-8859-1 [-56, 0, 0, 0] È//PERFECT USING 1 BYTE WHICH IS -56
US-ASCII [0, 0, 0, 0] //DONT GET IT SEE THIS ITEM FOR LATER
UTF-16 [-2, -1, 0, -56] È<<>>1 //WHAT IS -2,-1 BYTE USED FOR? I HAVE TRY WITH OTHER EXAMPLES AND THEY ALWAYS APPEAR AM I LOSING TWO BYTES HERE??
UTF-8 [-61, -120, 0, 0] 2 È //SEEMS TO MY CHARACTER NEEDS TWO BYTES?? I THOUGHT THAT CODE=200 WOULD REQUIRE ONLY ONE
getBytes
[-61, -120]//OK MY UTF-8 REPRESENTATION
ISO-8859-1 [-56]//OK
US-ASCII [63]//OK BUT WHY WHEN I ENCODE IN ASCCI DOESNT GET ANY BYTE ENCODED?
UTF-16 [-2, -1, 0, -56]//AGAIN WHAT ARE -2,-1 IN THE LEADING BYTES?
UTF-8 [-61, -120]//OK
I have try
System.out.println(new String(new byte[]{-1,-2},"UTF-16"));//SIMPLE "" I AM WASTING THIS 2 BYTES??
In resume.
Why UTF-16 always has two leading bytes are they wasted? new byte[]{-1,-2}
Why when i encode "È" i dont get any bytes in ASCCI Charset but when i do È.getBytes(StandardCharsets.US_ASCII) i get {63}?
Java uses UTF-16 but in my case UTF-8 is platform depend??
Sorry if this post is confussing
Environment
Windows 7 64 Bits Netbeans 8.2 with Java 1.8.0_121
First question
For your first question: those bytes are the BOM code and they specify the byte order (whether the least or most significant comes first) of multibyte encoding such as UTF-16.
Second question
Every ASCII character can be encoded as a single byte in UTF-8. But ASCII is not an 8-bit encoding, it uses 7 bits for every character. And in fact, all Unicode character with code points >= 128 require at least two bytes. (The reason is that you need a way to distinguish between 200 and a multibyte code point whose first byte happens to be 200. UTF-8 solves this by using the bytes >= 128 to represent multibyte codepoints.)
'È' is not an ASCII character, so it cannot be represented in ASCII. This explains the second output: 63 is ASCII for the character '?'. And indeed, the Javadoc for the getBytes(Charset) method specifies that unmappable input is mapped to "the default replacement byte array", in this case '?'. On the other hand, to obtain the first ASCII byte array you used the CharsetEncoder directly, which is a more low-level API and does not perform such automatic replacements. (When you would have checked the result of the encode method, you would have found it to have returned a CoderResult instance representing an error.)
Third question
Java 8 Strings use UTF-16 internally, but when communicating with other software, different encodings may be expected, such as UTF-8. The Charset.defaultCharset() method returns the default character set of the virtual machine, which depends on the locale and character set of the operating system, not on the encoding used internally by Java strings.
Let's back up a bit…
Java's text datatypes use the UTF-16 character encoding of the Unicode character set. (As do, VB4/5/6/A/Script, JavaScript, .NET, ….) You can see this in the various operations you do with the string API: indexing, length, ….
Libraries support converting between the text datatypes and byte arrays using various encodings. Some of them are categorized as "Extended ASCII", but stating that is a very poor substitute for naming the character encoding actually being used.
Some operating systems allow the user to designate a default character encoding. (Most users don't know or care, though.) Java attempts to pick this up. It is only useful when the program understands that input from the user is that character encoding or that output should be. This century, users dealing in text files prefer to use a specific encoding, communicate them unchanged across systems, don't appreciate lossy conversions and therefore don't have any use for this concept. From a program's point of view, it is never what you want unless it is exactly what you want.
Where a conversion would be lossy, you have the choice of a replacement character (such a '?'), omitting it, or throwing an exception.
A character encoding is a map between a codepoint (integer) of a character set and one or more code units, according to the definition of the encoding. A code unit is a fixed size and the number of code units needed for a codepoint, might vary by codepoint.
In libraries, it is not generally useful to have an array of code units so they take the further step of converting to/from an array of bytes. byte values do range from -128 to 127, however, that's the Java interpretation as two's complement 8-bit integers. As the bytes are understood to be encoding text, the values would be interpret according to the rules of the character encoding.
Because some Unicode encodings, have code units more than one byte long, byte order becomes important. So, at the byte array level, there is UTF-16 Big Endian and UTF-16 Little Endian. When communicating a text file or stream, you would send the bytes and well as having a shared knowledge of the encoding. This "metadata" is required for understanding. So, UTF-16BE or UTF-16LE, for example. To make that a bit easier, Unicode allows some metadata beginning of the file or stream to indicate the byte order. It is called the byte-order mark (BOM) So, the external metadata can share the encoding (say, UTF-16), while the internal metadata shares the byte order. Unicode allows the BOM to be present even when byte order is not relevant, such as UTF-8. So, if the understanding is that the bytes are text encoded with any Unicode encoding and a BOM is present, then it's a very simple matter to figure out which Unicode encoding it is and what the byte order is, if relavent.
1) You are seeing the BOM in some of your Unicode encoding outputs.
2) È is not in the ASCII character set. What would want to happen in this case? I often prefer an exception.
3) The system you were using, for your account, at the time of your tests, may have had UTF-8 as the default character encoding, Is that important to the way you want and have encoded your text files on that system?
Below code will print the length of byte store to below String which contain double byte Japanese character. Per my understanding, output of this program should be 2, however it is coming as 3. Why it this the case?
String j = "大";
System.out.println(j.getBytes().length);
If this will be always the case, then should I assume below:
1, for single byte character, output of program will be always 1
2, for double byte character, output of program will be always 3
UTF 8 characters byte length can be between 1 to 4 bytes. So your code is printing whatever is the correct byte length for the input japanese character.
I believe the code point for that character is 0x5927, which when represented as UTF-8 is the three bytes E5 A4 A7. (Not all non-ASCII characters take 3 bytes in UTF-8, only those with code points in the range of 0x0800 and 0xFFFF.)
.getBytes() method uses the default system encoding (in case of Linux it's usually UTF-8).
Since you mentioned "one-byte" and "two-byte Japanese characters", I guess you want to use SJIS encoding. You do it this way:
String j = "大";
System.out.println(j.getBytes("SJIS").length);
prints 2.
As a guideline, never use .getBytes without specifying an encoding and never use any other method or class that uses the default system encoding. You'll run your code on a different computer and it will stop working.
I used RandomAccessFile to read a byte from a text file.
public static void readFile(RandomAccessFile fr) {
byte[] cbuff = new byte[1];
fr.read(cbuff,0,1);
System.out.println(new String(cbuff));
}
Why am I seeing one full character being read by this?
A char represents a character in Java (*). It is 2 bytes large (or 16 bits).
That doesn't necessarily mean that every representation of a character is 2 bytes long. In fact many character encodings only reserve 1 byte for every character (or use 1 byte for the most common characters).
When you call the String(byte[]) constructor you ask Java to convert the byte[] to a String using the platform's default charset(**). Since the platform default charset is usually a 1-byte encoding such as ISO-8859-1 or a variable-length encoding such as UTF-8, it can easily convert that 1 byte to a single character.
If you run that code on a platform that uses UTF-16 (or UTF-32 or UCS-2 or UCS-4 or ...) as the platform default encoding, then you will not get a valid result (you'll get a String containing the Unicode Replacement Character instead).
That's one of the reasons why you should not depend on the platform default encoding: when converting between byte[] and char[]/String or between InputStream and Reader or between OutputStream and Writer, you should always specify which encoding you want to use. If you don't, then your code will be platform-dependent.
(*) that's not entirely true: a char represents a UTF-16 code unit. Either one or two UTF-16 code units represent a Unicode code point. A Unicode code point usually represents a character, but sometimes multiple Unicode code points are used to make up a single character. But the approximation above is close enough to discuss the topic at hand.
(**) Note that on Android the default character set is always UTF-8 and starting with Java 18 the Java platform itself also switched to this default (but it can still be configured to act the legacy way)
Java stores all it's "chars" internally as two bytes. However, when they become strings etc, the number of bytes will depend on your encoding.
Some characters (ASCII) are single byte, but many others are multi-byte.
Java supports Unicode, thus according to:
Java Character Docs
The max value supported is "\uFFFF" (hex FFFF, dec 65535), or 11111111 11111111 binary (two bytes).
The constructor String(byte[] bytes) takes the bytes from the buffer and encodes them to characters.
It uses the platform default charset to encode bytes to characters. If you know, your file contains text, that is encoded in a different charset, you can use the String(byte[] bytes, String charsetName) to use the correct encoding (from bytes to characters).
In ASCII text file each character is just one byte
Looks like your file contains ASCII characters, which are encoded in just 1 byte. If text file was containing non-ASCII character, e.g. 2-byte UTF-8, then you get just the first byte, not whole character.
There are some great answers here but I wanted to point out the jvm is free to store a char value in any size space >= 2 bytes.
On many architectures there is a penalty for performing unaligned memory access so a char might easily be padded to 4 bytes. A volatile char might even be padded to the size of the CPU cache line to prevent false sharing. https://en.wikipedia.org/wiki/False_sharing
It might be non-intuitive to new Java programmers that a character array or a string is NOT simply multiple characters. You should learn and think about strings and arrays distinctly from "multiple characters".
I also want to point out that java characters are often misused. People don't realize they are writing code that won't properly handle codepoints over 16 bits in length.
Java allocates 2 of 2 bytes for character as it follows UTF-16. It occupies minimum 2 bytes while storing a character, and maximum of 4 bytes. There is no 1 byte or 3 bytes of storage for character.
The Java char is 2 bytes. But the file encoding may be different.
So first you should know what encoding your file uses. For example, the file could be UTF-8 or ASCII encoded, then you will retrieve the right chars by reading one byte at a time.
If the encoding of the file is UTF-16, it may still show you the correct char if your UTF-16 is little endian. For example, the little endian UTF-16 for A is [65, 0]. Then when you read the first byte, it returns 65. After padding with 0 for the second byte, you will get A.