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:
int i=rno[0];
...since it's not a downcast.
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 i2 = 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();
Or the Byte constructor:
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.
Answer from T.J. Crowder on Stack OverflowYour 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:
int i=rno[0];
...since it's not a downcast.
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 i2 = 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();
Or the Byte constructor:
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.
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.
java - How to Convert Int to Unsigned Byte and Back - Stack Overflow
Java: Convert byte to integer - Stack Overflow
java - Convert Byte[] to int - Stack Overflow
types - Convert a byte array to integer in Java and vice versa - Stack Overflow
A byte is always signed in Java. You may get its unsigned value by binary-anding it with 0xFF, though:
int i = 234;
byte b = (byte) i;
System.out.println(b); // -22
int i2 = b & 0xFF;
System.out.println(i2); // 234
Java 8 provides Byte.toUnsignedInt to convert byte to int by unsigned conversion. In Oracle's JDK this is simply implemented as return ((int) x) & 0xff; because HotSpot already understands how to optimize this pattern, but it could be intrinsified on other VMs. More importantly, no prior knowledge is needed to understand what a call to toUnsignedInt(foo) does.
In total, Java 8 provides methods to convert byte and short to unsigned int and long, and int to unsigned long. A method to convert byte to unsigned short was deliberately omitted because the JVM only provides arithmetic on int and long anyway.
To convert an int back to a byte, just use a cast: (byte)someInt. The resulting narrowing primitive conversion will discard all but the last 8 bits.
You can use ByteBuffer for this:
ByteBuffer buffer = ByteBuffer.wrap(myArray);
buffer.order(ByteOrder.LITTLE_ENDIAN); // if you want little-endian
int result = buffer.getShort();
See also Convert 4 bytes to int.
In Java, Bytes are signed, which means a byte's value can be negative, and when that happens, @MattBall's original solution won't work.
For example, if the binary form of the bytes array are like this:
1000 1101 1000 1101
then myArray[0] is 1000 1101 and myArray[1] is 1000 1101, the decimal value of byte 1000 1101 is -115 instead of 141(= 2^7 + 2^3 + 2^2 + 2^0)
if we use
int result = (myArray[0] << 8) + myArray[1]
the value would be -16191 which is WRONG.
The reason why its wrong is that when we interpret a 2-byte array into integer, all the bytes are unsigned, so when translating, we should map the signed bytes to unsigned integer:
((myArray[0] & 0xff) << 8) + (myArray[1] & 0xff)
the result is 36237, use a calculator or ByteBuffer to check if its correct(I have done it, and yes, it's correct).
Thank you for the answers but I found that the problem comes from somewhere else, not from the conversion methode,
there is an other version of the methode
public int convertirOctetEnEntier(byte[] b){
int MASK = 0xFF;
int result = 0;
result = b[0] & MASK;
result = result + ((b[1] & MASK) << 8);
result = result + ((b[2] & MASK) << 16);
result = result + ((b[3] & MASK) << 24);
return result;
}
OK, so you have a big-endian byte array that you are converting to an int. I don't think it matters what the value of each byte is; you just need to stick them all together:
public int convertByteToInt(byte[] b)
{
int value= 0;
for(int i=0; i<b.length; i++)
value = (value << 8) | b[i];
return value;
}
If I'm missing something here, let me know.
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(); // { 0, 1 }
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: <<, &, |
// when operators of equal precedence (here bitwise OR) appear in the same expression, they are evaluated from left to right
int fromByteArray(byte[] bytes) {
return bytes[0] << 24 | (bytes[1] & 0xFF) << 16 | (bytes[2] & 0xFF) << 8 | (bytes[3] & 0xFF);
}
// packing an array of 4 bytes to an int, big endian, clean code
int fromByteArray(byte[] bytes) {
return ((bytes[0] & 0xFF) << 24) |
((bytes[1] & 0xFF) << 16) |
((bytes[2] & 0xFF) << 8 ) |
((bytes[3] & 0xFF) << 0 );
}
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.
This is the standard way to do that transformation. If you want to get rid of the checkstyle complaints, try defining a constant, it could help:
public final static int MASK = 0xff;
BTW - keep in mind, that it is still a custom conversion. byte is a signed datatype so a byte can never hold the value 255. A byte can store the bit pattern 1111 1111 but this represents the integer value -1.
So in fact you're doing bit operations - and bit operations always require some magic numbers.
BTW-2 : Yes, there is a Byte.MAX_VALUE constant but this is - because byte is signed - defined as 27-1 (= 127). So it won't help in your case. You need a byte constant for -1.
Ignore checkstyle. 0xFF is not a magic number. If you define a constant for it, the constant is a magic constant, which is much less understandable than 0xFF itself. Every programmer educated in the recent centuries should be more familiar with 0xFF than with his girlfriend, if any.
should we write code like this?
for(int i = Math.ZERO; ... )
I am converting an arbitrary ascii string, which is incidently a number padded with zeros, for example "1" as "0000000001" then to bytes then back in python. Of course the value can be "0000004231" etc. also. It is always numeric and within range of signed 32bit value padded by zeros.
When I tell python that it is bytes and I want it in int, it converts it to a nice large random looking number. Then I can convert it back to original value later using to_bytes() function.
In [74]: value = int.from_bytes(bytes(format(1, '010d'),'ascii'), byteorder='little') In [75]: value.to_bytes(10,byteorder=sys.byteorder) Out[75]: b'0000000001' In [76]: value Out[76]: 232284873704446901628976 In [77]:
How can I achieve same with java?
Note: I need the number to be padded with zeros and 10 characters long. It is a number in range of 32bit signed int with padding to fixed 10 character length
Note 2 : I already tried this. I get [B@7852e922 in the testBytesvariable and not 232284873704446901628976 which I expect