Java one byte to integer

Convert a byte array to integer in Java and vice versa

I want to store some data into byte arrays in Java. Basically just numbers which can take up to 2 Bytes per number. I’d like to know how I can convert an integer into a 2 byte long byte array and vice versa. I found a lot of solutions googling but most of them don’t explain what happens in the code. There’s a lot of shifting stuff I don’t really understand so I would appreciate a basic explanation.

How much do you understand about bit shifting? It sounds like the question is really «what does bit shifting do» more than about the conversion to byte arrays, really — if you actually want to understand how the conversion would work.

(Just to clarify, I’m fine with either question, but it’s worth making it clear which question you really want answered. You’re likely to get an answer which is more useful to you that way.)

Okay i got your point! Thanks for the remark. I know what bit shifting does i just didn’t understand what its used for in converting byte arrays yet.

@prekageo and Jeff Mercado Thanks for your two answers. prekageo gave a good explanation of how this is done, nice link! That makes it a lot clearer to me. And Jeff Mercados solution solved the problem i had.

8 Answers 8

Use the classes found in the java.nio namespace, in particular, the ByteBuffer . It can do all the work for you.

byte[] arr = < 0x00, 0x01 >; ByteBuffer wrapped = ByteBuffer.wrap(arr); // big-endian by default short num = wrapped.getShort(); // 1 ByteBuffer dbuf = ByteBuffer.allocate(2); dbuf.putShort(num); byte[] bytes = dbuf.array(); //

Is it too expensive if the byte array contains only 1 or 2 integer? Not sure about the cost constructing a ByteBuffer .

How often are you working with binary data in 2-4 byte chunks? Really? A sane implementation would either work with it in BUFSIZ chunks (typically 4kb) or use other IO libraries that hides this detail. There’s an entire library within the framework that’s dedicated to help you work on buffers of data. You do a disservice to yourself and other maintainers of your code when you implement common operations without good reason (be it perf or other critical operation). These buffers are merely wrappers that operates on arrays, nothing more.

@JaveneCPPMcGowan There is no direct instantiation present in this answer. If you mean the factory methods wrap and allocate , they don’t return an instance of the abstract class ByteBuffer .

Not a solution for 3 byte stride. We can get Char , Short , Int . I suppose I could pad to 4 bytes and discard the 4th each time, but I would rather not.

byte[] toByteArray(int value) < return ByteBuffer.allocate(4).putInt(value).array(); >byte[] toByteArray(int value) < return new byte[] < (byte)(value >> 24), (byte)(value >> 16), (byte)(value >> 8), (byte)value >; > int fromByteArray(byte[] bytes) < return ByteBuffer.wrap(bytes).getInt(); >// packing an array of 4 bytes to an int, big endian, minimal parentheses // operator precedence: // packing an array of 4 bytes to an int, big endian, clean code int fromByteArray(byte[] bytes)

When packing signed bytes into an int, each byte needs to be masked off because it is sign-extended to 32 bits (rather than zero-extended) due to the arithmetic promotion rule (described in JLS, Conversions and Promotions).

There’s an interesting puzzle related to this described in Java Puzzlers («A Big Delight in Every Byte») by Joshua Bloch and Neal Gafter . When comparing a byte value to an int value, the byte is sign-extended to an int and then this value is compared to the other int

byte[] bytes = (…) if (bytes[0] == 0xFF) < // dead code, bytes[0] is in the range [-128,127] and thus never equal to 255 >

Note that all numeric types are signed in Java with exception to char being a 16-bit unsigned integer type.

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Converting from byte to int in Java

Your array is of byte primitives, but you’re trying to call a method on them.

You don’t need to do anything explicit to convert a byte to an int , just:

Note that the default behavior of byte -to- int conversion is to preserve the sign of the value (remember byte is a signed type in Java). So for instance:

byte b1 = -100; int i1 = b1; System.out.println(i1); // -100 

If you were thinking of the byte as unsigned (156) rather than signed (-100), as of Java 8 there’s Byte.toUnsignedInt :

byte b2 = -100; // Or `= (byte)156;` int = Byte.toUnsignedInt(b2); System.out.println(i2); // 156 

Prior to Java 8, to get the equivalent value in the int you’d need to mask off the sign bits:

byte b2 = -100; // Or `= (byte)156;` int i2 = (b2 & 0xFF); System.out.println(i2); // 156 

Just for completeness #1: If you did want to use the various methods of Byte for some reason (you don’t need to here), you could use a boxing conversion:

Byte b = rno[0]; // Boxing conversion converts `byte` to `Byte` int i = b.intValue(); 

Byte b = new Byte(rno[0]); int i = b.intValue(); 

But again, you don’t need that here.

Just for completeness #2: If it were a downcast (e.g., if you were trying to convert an int to a byte ), all you need is a cast:

int i; byte b; i = 5; b = (byte)i; 

This assures the compiler that you know it’s a downcast, so you don’t get the «Possible loss of precision» error.

Why is boxing conversion superior to Byte constructor in this case? to my understanding the boxing conversion ends up using the Byte constructor in the end any way? is my understanding incorrect?

@user3772575: Yes. Boxing will use Byte.valueOf , not new Byte() , to avoid allocating a new instance if one already exists for that byte value; more here. (The spec doesn’t actually say that, but that’s what it does. You can tell by looking at the bytecode.)

Last I checked in Java 8 Byte class provides a static method called toUnsignedInt(byte). This should help converting easier and code more readable.

byte b = (byte)0xC8; int v1 = b; // v1 is -56 (0xFFFFFFC8) int v2 = b & 0xFF // v2 is 200 (0x000000C8) 

Most of the time v2 is the way you really need.

This is the correct answer, and v2 is almost always what you want. Implicit conversion of byte to int without the & 0xff is a source of bugs in Java.

This isn’t really due to «bugs in Java» implicit conversion. It’s due to developer misunderstanding. Java’s byte primitive only supports values between -128 to 127. If someone tries to stuff an unsigned value (like 0xC8) into that, it can roll over with a bunch of leading ones (due to the twos complement notation). Similarly, for this question, a full bit space can have a sign bit set and a numeric promotion will do similarly.

Java is actually doing numeric conversion under the hood, according to the JLS, which means that both operands are converted to int primitives, anyway.

Not bugs in Java, bugs in user code written in Java 🙂 Most programmers don’t realize bytes are signed in Java, and frankly it is a stupid default behavior that they treat bytes as signed in the Java spec, but they did it for consistency with the other integral types, since none of them have unsigned versions. The same problem comes up with conversion of shorts to ints, since people also generally don’t realize shorts are signed. (It’s easier to trigger sign extension bugs with bytes though, because of the very limited range of values.) People do expect ints and longs to be signed though.

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how to convert the byte array into an integer array

I am working on a project in which I am receiving image data as an array of bytes (1-byte per pixel). Each byte represents a greyscale integer (0-255). In order to perform a number of functions on the data, I need to convert the byte array into an integer array. please help me..

2 Answers 2

Anything wrong with the simple approach?

public static int[] convertToIntArray(byte[] input) < int[] ret = new int[input.length]; for (int i = 0; i < input.length; i++) < ret[i] = input[i] & 0xff; // Range 0 to 255, not -128 to 127 >return ret; > 

EDIT: If you want a range of -128 to 127:

public static int[] convertToIntArray(byte[] input) < int[] ret = new int[input.length]; for (int i = 0; i < input.length; i++) < ret[i] = input[i]; >return ret; > 

this head bow is for the speed of you answering the questions, i thought to answer the question, in between seconds i alarmed as you answered

Leave the bit masking away which will sign extend the byte. And yes I agree with Damodar.. almost eerie — maybe google finally created an AI that can pass the Turing Test and tries to test it on SO? :p

It depends on the use of the resulted int array, but normally, converting a byte array or ByteBuffer into an integer array means «wraping» 4 bytes into 1 integer, e.g. for bitmaps, so in that case I would suggest the following conversion:

IntBuffer ib = ByteBuffer.wrap(input).order(ByteOrder.BIG_ENDIAN).asIntBuffer(); int[] ret = new int[ib.capacity()]; ib.get(ret); return ret; 

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Convert Byte to Int in Java

1. Convert Byte to Int Using the Byte Wrapper Class and Casting in Java
2. Using Byte to the Unsigned Byte Conversion

A byte data type is defined as an 8-bit signed two’s complement integer, whereas an int data type can be classified as a 32-bit signed two’s complement integer. Unless there is concern about memory, an integer is the default data type used to store integral values. In this article, we’re going to discuss how you can convert a byte into an int data type.

Convert Byte to Int Using the Byte Wrapper Class and Casting in Java

A byte holds 0 as the default value and its range varies from -128 = (-2^7) to 127 = (2^7 -1) . An integer holds a default value of 0, and its range varies from -2^31 to 2^31-1. The Wrapper class for byte is Byte, whereas for int is Integer.

If the byte value exceeds their range, the compiler automatically promotes it to an int variable. In Java, there is only a signed integer type (no unsigned concept); this means that all the integer primitive types, including the byte, are signed. However, the range is 0 to 255 for an unsigned byte data type.

The whole point of byte-to-int conversion is to preserve the sign of the value. Here in the code, we have a byte type variable b , which we’ll convert into an int type variable i . Hence, it’s not downcast. We can directly assign the byte to the int data type.

Secondly, we have a Wrapper class method intValue() that returns the value of byte as an int after widening the primitive conversion as we’re storing a smaller data type into a larger one.

If we take the byte as unsigned, then we have the Byte.toUnsignedInt() method that converts the argument passed to an int using the unsigned conversion.

public class JavaByteToInt   public static void main(String args[])  byte b = -98;  System.out.println("byte b : = "+b);  int i = b;  System.out.println("int i : "+i);   Byte byt = new Byte(b);  int i1 = byt.intValue();  int i2 = Byte.toUnsignedInt(byt);  System.out.println("int i1 : "+i1);  System.out.println("int i2 : "+i2);   > > 

byte b : = -98 int i : -98 int i1 : -98 int i2 : 158 

Using Byte to the Unsigned Byte Conversion

The range for an unsigned byte data type goes from 0 to 255; however, Java doesn’t have unsigned bytes. What you can do is cast the byte into an int to make an unsigned byte, and mask (bitwise) the new int with a 0xff to get the last 8 bits or prevent the sign extension.

When a byte value is used with the & operator, it automatically casts the byte to an integer. The Hex value of 0x000000FF is denoted as 0xFF .

public class JavaByteToInt   public static void main(String args[])  byte b = -127;  int i = (b & 0xFF);  System.out.println(i);   > > 

Rupam Saini is an android developer, who also works sometimes as a web developer., He likes to read books and write about various things.

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