Java NIO server receives random string [duplicate] - java

I'm writing a web application in Google app Engine. It allows people to basically edit html code that gets stored as an .html file in the blobstore.
I'm using fetchData to return a byte[] of all the characters in the file. I'm trying to print to an html in order for the user to edit the html code. Everything works great!
Here's my only problem now:
The byte array is having some issues when converting back to a string. Smart quotes and a couple of characters are coming out looking funky. (?'s or japanese symbols etc.) Specifically it's several bytes I'm seeing that have negative values which are causing the problem.
The smart quotes are coming back as -108 and -109 in the byte array. Why is this and how can I decode the negative bytes to show the correct character encoding?

The byte array contains characters in a special encoding (that you should know). The way to convert it to a String is:
String decoded = new String(bytes, "UTF-8"); // example for one encoding type
By The Way - the raw bytes appear may appear as negative decimals just because the java datatype byte is signed, it covers the range from -128 to 127.
-109 = 0x93: Control Code "Set Transmit State"
The value (-109) is a non-printable control character in UNICODE. So UTF-8 is not the correct encoding for that character stream.
0x93 in "Windows-1252" is the "smart quote" that you're looking for, so the Java name of that encoding is "Cp1252". The next line provides a test code:
System.out.println(new String(new byte[]{-109}, "Cp1252"));

Java 7 and above
You can also pass your desired encoding to the String constructor as a Charset constant from StandardCharsets. This may be safer than passing the encoding as a String, as suggested in the other answers.
For example, for UTF-8 encoding
String bytesAsString = new String(bytes, StandardCharsets.UTF_8);

You can try this.
String s = new String(bytearray);

public class Main {
/**
* Example method for converting a byte to a String.
*/
public void convertByteToString() {
byte b = 65;
//Using the static toString method of the Byte class
System.out.println(Byte.toString(b));
//Using simple concatenation with an empty String
System.out.println(b + "");
//Creating a byte array and passing it to the String constructor
System.out.println(new String(new byte[] {b}));
}
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
new Main().convertByteToString();
}
}
Output
65
65
A

public static String readFile(String fn) throws IOException
{
File f = new File(fn);
byte[] buffer = new byte[(int)f.length()];
FileInputStream is = new FileInputStream(fn);
is.read(buffer);
is.close();
return new String(buffer, "UTF-8"); // use desired encoding
}

I suggest Arrays.toString(byte_array);
It depends on your purpose. For example, I wanted to save a byte array exactly like the format you can see at time of debug that is something like this : [1, 2, 3] If you want to save exactly same value without converting the bytes to character format, Arrays.toString (byte_array) does this,. But if you want to save characters instead of bytes, you should use String s = new String(byte_array). In this case, s is equal to equivalent of [1, 2, 3] in format of character.

The previous answer from Andreas_D is good. I'm just going to add that wherever you are displaying the output there will be a font and a character encoding and it may not support some characters.
To work out whether it is Java or your display that is a problem, do this:
for(int i=0;i<str.length();i++) {
char ch = str.charAt(i);
System.out.println(i+" : "+ch+" "+Integer.toHexString(ch)+((ch=='\ufffd') ? " Unknown character" : ""));
}
Java will have mapped any characters it cannot understand to 0xfffd the official character for unknown characters. If you see a '?' in the output, but it is not mapped to 0xfffd, it is your display font or encoding that is the problem, not Java.

Related

How can I save a String Byte without losing information?

I'm developing a JPEG decoder(I'm in the Huffman phase) and I want to write BinaryString's into a file.
For example, let's say we've this:
String huff = "00010010100010101000100100";
I've tried to convert it to an integer spliting it by 8 and saving it integer represantation, as I can't write bits:
huff.split("(?<=\\G.{8})"))
int val = Integer.parseInt(str, 2);
out.write(val); //writes to a FileOutputStream
The problem is that, in my example, if I try to save "00010010" it converts it to 18 (10010), and I need the 0's.
And finally, when I read :
int enter;
String code = "";
while((enter =in.read())!=-1) {
code+=Integer.toBinaryString(enter);
}
I got :
Code = 10010
instead of:
Code = 00010010
Also I've tried to convert it to bitset and then to Byte[] but I've the same problem.
Your example is that you have the string "10010" and you want the string "00010010". That is, you need to left-pad this string with zeroes. Note that since you're joining the results of many calls to Integer.toBinaryString in a loop, you need to left-pad these strings inside the loop, before concatenating them.
while((enter = in.read()) != -1) {
String binary = Integer.toBinaryString(enter);
// left-pad to length 8
binary = ("00000000" + binary).substring(binary.length());
code += binary;
}
You might want to look at the UTF-8 algorithm, since it does exactly what you want. It stores massive amounts of data while discarding zeros, keeping relevant data, and encoding it to take up less disk space.
Works with: Java version 7+
import java.nio.charset.StandardCharsets;
import java.util.Formatter;
public class UTF8EncodeDecode {
public static byte[] utf8encode(int codepoint) {
return new String(new int[]{codepoint}, 0, 1).getBytes(StandardCharsets.UTF_8);
}
public static int utf8decode(byte[] bytes) {
return new String(bytes, StandardCharsets.UTF_8).codePointAt(0);
}
public static void main(String[] args) {
System.out.printf("%-7s %-43s %7s\t%s\t%7s%n",
"Char", "Name", "Unicode", "UTF-8 encoded", "Decoded");
for (int codepoint : new int[]{0x0041, 0x00F6, 0x0416, 0x20AC, 0x1D11E}) {
byte[] encoded = utf8encode(codepoint);
Formatter formatter = new Formatter();
for (byte b : encoded) {
formatter.format("%02X ", b);
}
String encodedHex = formatter.toString();
int decoded = utf8decode(encoded);
System.out.printf("%-7c %-43s U+%04X\t%-12s\tU+%04X%n",
codepoint, Character.getName(codepoint), codepoint, encodedHex, decoded);
}
}
}
https://rosettacode.org/wiki/UTF-8_encode_and_decode#Java
UTF-8 is a variable width character encoding capable of encoding all 1,112,064[nb 1] valid code points in Unicode using one to four 8-bit bytes.[nb 2] The encoding is defined by the Unicode Standard, and was originally designed by Ken Thompson and Rob Pike.[1][2] The name is derived from Unicode (or Universal Coded Character Set) Transformation Format – 8-bit.[3]
It was designed for backward compatibility with ASCII. Code points with lower numerical values, which tend to occur more frequently, are encoded using fewer bytes. The first 128 characters of Unicode, which correspond one-to-one with ASCII, are encoded using a single byte with the same binary value as ASCII, so that valid ASCII text is valid UTF-8-encoded Unicode as well. Since ASCII bytes do not occur when encoding non-ASCII code points into UTF-8, UTF-8 is safe to use within most programming and document languages that interpret certain ASCII characters in a special way, such as "/" (slash) in filenames, "\" (backslash) in escape sequences, and "%" in printf.
https://en.wikipedia.org/wiki/UTF-8
Binary 11110000 10010000 10001101 10001000 becomes F0 90 8D 88 in UTF-8. Since you are storing it as text, you go from having to store 32 characters to storing 8. And because it's a well known and well designed encoding, you can reverse it easily. All the math is done for you.
Your example of 00010010100010101000100100 (or rather 00000001 0010100 0101010 00100100) converts to *$ (two unprintable characters on my machine). That's the UTF-8 encoding of the binary. I had mistakenly used a different site that was using the data I put in as decimal instead of binary.
https://onlineutf8tools.com/convert-binary-to-utf8
For a really good explanation of UTF-8 and how it can apply to the answer:
https://hackaday.com/2013/09/27/utf-8-the-most-elegant-hack/
Edit:
I took this question as a way to reduce the amount of characters needed to store values, which is a type of encoding. UTF-8 is a type of encoding. Used in a "non-standard" way, the OP can use UTF-8 to encode their strings of 0's & 1's in a much shorter format. That's how this answer is relevant.
If you concatenate the characters, you can go from 4x 8 bits (32 bits) to 8x 8 bits (64 bits) easily and encode a value as large as 9,223,372,036,854,775,807.

RandomAccesFile and UTF8 line

I use a RandomAccessFile object to read an UTF-8 French file. I use the readLine method.
My Groovy code below:
while ((line = randomAccess.readLine())) {
def utfLine = new String(line.getBytes('UTF-8'), 'UTF-8')
++count
long nextRecordPos = randomAccess.getFilePointer()
compareNextRecords(utfLine, randomAccess)
randomAccess.seek(nextRecordPos)
}
My problem is utfLine and line are the same: the accented characters stay like é instead of é. No conversion is done.
First of all, this line of code does absolutely nothing. The data is the same. Remove it:
def utfLine = new String(line.getBytes('UTF-8'), 'UTF-8')
According to the Javadoc, RandomAccessFile.readLine() is not aware of character encodings. It reads bytes until it encounters "\r" or "\n" or "\r\n". ASCII byte values are put into the returned string in the normal way. But byte values between 128 and 255 are put into the string literally without interpreting it as being in a character encoding (or you could say this is the raw/verbatim encoding).
There is no method or constructor to set the character encoding in a RandomAccessFile. But it's still valuable to use readLine() because it takes care of parsing for a newline sequence and allocating memory.
The easiest solution in your situation is to manually convert the fake "line" into bytes by reversing what readLine() did, then decode the bytes into a real string with awareness of character encoding. I don't know how to write code in Groovy, so I'll give the answer in Java:
String fakeLine = randomAccess.readLine();
byte[] bytes = new byte[fakeLine.length()];
for (int i = 0; i < fakeLine.length(); i++)
bytes[i] = (byte)fakeLine.charAt(i);
String realLine = new String(bytes, "UTF-8");

UTF-8 and charcters in Java and Eclipse IDE

public static void main(String[] args) throws UnsupportedEncodingException {
String str = "अ";
byte[] bytes = str.getBytes("UTF-8");
for (byte b : bytes) {
System.out.print(b + "\t");
}
String hindi = new String(bytes, "UTF-8");
System.out.println("\nHindi = " + hindi);
System.out.println((int) 'अ');
}
OUTPUT:
-32 -92 -123
Hindi = अ
2309
I need explanation on those three outputs. Especially the last one.
Also, I copy paste this characterअ from a web page. How do I type it manually in Eclipse IDE? For example, ALT + 65 will give 'A' but ALT + 2309 does not give me 'अ' (I copy paste this again).
The first print:
See public byte[] getBytes(Charset charset):
Encodes this String into a sequence of bytes using the given charset,
storing the result into a new byte array.
The second print:
See public String(byte[] bytes,
Charset charset):
Constructs a new String by decoding the specified array of bytes using
the specified charset.
The third print:
See this link:
You're printing the decimal code of it, which is 2309.
The links provided above should help you to understand the output you're getting in each case.
Ideally you need to type letter's associated unicode number "\uNNNN" in Java in any IDE.
Further for single character, straight approach would be:
char c = '\uNNNN';
System.out.println((int) c);
Update: List of Unicode ranges for Devanagri http://www.unicode.org/charts/PDF/U0900.pdf
To type in Hindi in notepad/IDE etc, search for software which which maps keyboard keys to specific Hindi letter and related grammaticle punctuations.

How to convert 1s and 0s to String?

Please have a look at the following machine code
‎0111001101110100011100100110010101110011011100110110010101100100
This means something. I need to convert this to string. When I use Integer.parseInt() with the above as the string and 2 as the radix(to convert it to bytes), it gives number format exception.
And I believe I have to seperate this into sets of 8 pieces (like ‎01110011 , 10111010, etc). Am I correct?
Please help me to convert this correctly to string.
Thanks
final String s =
"0111001101110100011100100110010101110011011100110110010101100100";
final StringBuilder b = new StringBuilder();
for (int i = 0; i < s.length(); i+=8)
b.append((char)Integer.parseInt(s.substring(i,i+8),2));
System.out.println(b);
prints "stressed"
A shorter way of reading large integers is to use BigInteger
final String s = "0111001101110100011100100110010101110011011100110110010101100100";
System.out.println(new String(new BigInteger('0'+s, 2).toByteArray(), 0));
prints
stressed
It depends on the encoding of the String.
An ASCII coded string uses 1 byte for each character while a unicode coded string takes 2 bytes for each character. There are many other types of encodings. The binary layout differs for each encoding.
So you need to find the encoding that was used to write this string to binary format

Java: String to byte array conversion

I am getting some unexpected results from what I thought was a simple test. After running the following:
byte [] bytes = {(byte)0x40, (byte)0xE2, (byte)0x56, (byte)0xFF, (byte)0xAD, (byte)0xDC};
String s = new String(bytes, Charset.forName("UTF-8"));
byte[] bytes2 = s.getBytes(Charset.forName("UTF-8"));
bytes2 is a 14 element long array nothing like the original (bytes). Is there a way to do this sort of conversion and retain the original decomposition to bytes?
Is there a way to do this sort of conversion and retain the original decomposition to bytes?
Well that doesn't look like valid UTF-8 to me, so I'm not surprised it didn't round-trip.
If you want to convert arbitrary binary data to text in a reversible way, use base64, e.g. via this public domain encoder/decoder.
This should do:
public class Main
{
/*
* This method converts a String to an array of bytes
*/
public void convertStringToByteArray()
{
String stringToConvert = "This String is 76 characters long and will be converted to an array of bytes";
byte[] theByteArray = stringToConvert.getBytes();
System.out.println(theByteArray.length);
}
/**
* #param args the command line arguments
*/
public static void main(String[] args)
{
new Main().convertStringToByteArray();
}
}
Two things:
The byte sequence does not appear to be valid UTF-8
$ python
>>> '\x40\xe2\x56\xff\xad\xdc'.decode('utf8')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib64/python2.7/encodings/utf_8.py", line 16, in decode
return codecs.utf_8_decode(input, errors, True)
UnicodeDecodeError: 'utf8' codec can't decode byte 0xe2 in position 1: invalid continuation byte
Even if it were valid UTF-8, decoding and then encoding can result in different bytes due to things like precombined characters and other Unicode features.
If you want to encode arbitrary binary data in a string in a way where you are guaranteed to get the same bytes back when you decode them, your best bet is something like base64.

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