I read that we should use Reader/Writer for reading/writing character data and InputStream/OutputStream for reading/writing binary data. Also, in java characters are 2 bytes. I am wondering how the following program works. It reads characters from standard input stores them in a single byte and prints them out. How are two byte characters fitting into one byte here?
http://www.cafeaulait.org/course/week10/06.html
The comment explains it pretty clearly:
// Notice that although a byte is read, an int
// with value between 0 and 255 is returned.
// Then this is converted to an ISO Latin-1 char
// in the same range before being printed.
So basically, this assumes that the incoming byte represents a character in ISO-8859-1.
If you use a console with a different encoding, or perhaps provide a character which isn't in ISO-8859-1, you'll end up with problems.
Basically, this is not good code.
Java stores characters as 2 bytes, but for normal ASCII characters the actual data fits in one byte. So as long as you can assume the file being read there is ASCII then that will work fine, as the actual numeric value of the character fits in a single byte.
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.
I defined a unicode character as a byte array:
private static final byte[] UNICODE_MEXT_LINE = Charsets.UTF_8.encode("\u0085").array();
At the moment byte array length is 3, is it safe to assume the length of the array is always 3 across platforms?
Thank you
It's safe to assume that that particular character will always be three bytes long, regardless of platform.
But unicode characters in UTF-8 can be one byte, two bytes, three bytes or even four bytes long, so no, you can't assume that if you convert any character to UTF-8 then it'll come out as three bytes.
That particular character will always be 3 bytes in length, but others will be different. Unicode characters are anywhere from 1-4 bytes long. The 8 in 'UTF-8' just means that it uses 8-bit code units.
The Wikipedia page on UTF-8 provides a pretty good overview of how that works. Basically, the first bits of the first byte tell you how many bytes long that character will be. For instance, if the first bit of the first byte is a 0 as in 01111111, then that means this character is only one byte long (in utf-8, these are the ascii characters). If the first bits are 110 as in 11011111, then that tells you that this character will be two bytes long. The chart in the Wikipedia page provides a good illustration of this.
There's also this question, which has some good answers as well.
We read and write binary files using the java primitive 'byte' like fileInputStream.read(byte) etc. In some more example we see byte[] = String.getBytes(). A byte is just 8-bit value. Why we use byte[] to read binaries? What does a byte value contains after reading from file or string ?
We read and write binary files using the java primitive 'byte' like fileInputStream.read(byte) etc.
Because the operating system models files as sequences of bytes (or more precisely, as octets). The byte type is the most natural representation of an octet in Java.
Why we use byte[] to read binaries?
Same answer as before. Though, in reality, you can also read binary files in other ways as well; e.g. using DataInputStream.
What does a byte value contains after reading from file or string ?
In the first case, the byte that was in the file.
In the second case, you don't "read" bytes from a String. Rather, when you call the String.getBytes() you get the bytes that comprise the String's characters when they are encoded in a particular character-set. If you use the no-args getBytes() method you will get the JVM's default character-set / encoding. You can also supply an argument to choose a different encoding.
Java makes a clear distinction between bytes (8 bit) quantities and characters. Conceptually, Java characters are Unicode code points, and strings and similar representations of text are sequences of characters ... not sequences of bytes.
(Unfortunately, there is a "wrinkle" in the implementation. When Java was designed, the Unicode character space fitted into a 16 bits; i.e. there were <= 65536 recognized code points. Java was designed to match this ... and the char type was defined as a 16 bit unsigned integral type. But then Unicode was expanded to > 65536 code points, and Java was left with the awkward problem that some Unicode code points could not be represented using one char values. Instead, they are represented by a pair of char values ... a so-called surrogate pair ... and Java strings are effectively represented in UTF-16. For most common characters / character-sets, this doesn't matter. But if you need to deal with unusual characters / character-sets, the correct way to deal with Strings is to use the "codepoint" methods.)
The String is built upon bytes. The bytes are built upon bits. The bits are "physically" stored on the drive.
So instead of reading data from drive bit by bit it is read in larger portions which are bytes.
So the byte[] contains raw data. Raw data are equal to that what is stored on drive.
You eventually alaways read raw data, then you can apply a formatter what turns that bytes into characters and eventually into letters dispalyed on the screed if that is a txt file. If you dead with image out will read bytes that store the information about color instaed of character.
Because the smallest storage unit is byte.
Title is pretty self-explanatory. In a lot of the JRE javadocs I see the phrases "stream of bytes" and "stream of characters" all over the place.
But aren't they the same thing? Or are they slightly different (e.g. interpreted differently) in Java-land? Thanks in advance.
In Java, a byte is not the same thing as a char. Therefore a byte stream is different from a character stream. Bytes are intended for arbitrary binary data; characters are specifically for data representing the building blocks of strings.
but if a char is only 1 byte in width
Except that it's not.
As per the JLS §4.2.1 a char is a number in the range:
from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535
But a byte is a number in the range
from -128 to 127, inclusive
Stream of byte is just plain byte, like how you would see it when you open a file in HEX Editor.
Character is different from just plain byte. ASCII encoding uses exactly 1 byte per character, but that is not true for many other encoding. For example, UTF-8 encoding may use from 1 to 4 bytes to encode a single character. Stream of character is designed to abstract away the underlying encoding, and produce char of one type of encoding (in Java, char and String uses UTF-16 encoding).
As a rule of thumb:
When you are dealing with text, you must use stream of character to decode the byte into character with the appropriate encoding.
When you are dealing with binary data or mixed of binary and text, you must use stream of byte, since it doesn't make sense otherwise. If a sequence of byte represents a String in certain encoding, then you can always pick those bytes out and use String(byte[] bytes, Charset charset) constructor to get back the String.
They are different. char is a 2-byte datatype in Java: byte is a 1-byte datatype.
Edit: char is also an unsigned type, while byte is not.
Generally it is better off to talk about streams in terms of their sizes, rather than what they carry. Stream of bytes is more intuitive than streams of chars, because streams of chars make us have to double check that a char is indeed a byte, not a unicode char, or anything fancy.
A char is a representation, which can be represented by a byte, but a byte is always going to be a byte. All world will burn when bytes will stop being 8 bits.
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.