Bytes in Java range from -128 to 127. If over 127 they overflow. So I assigned 128 to a byte variable to know how overflow work, but the compiler shows a error message: "Type mismatch: cannot convert from int to byte".
Why do I get a compiler error instead of it overflowing?
javac will let you assign values within range to bytes. When you go out of range, the value needs to be explicitly cast.
byte b = (byte)128;
Would work with the compiler. To see overflow you might try
byte b = 127;
b += 1;
System.out.println(b);
-128
The nuts and bolts can be found in the JLS. Java will do arithmetic as int, or long.
byte a = 1;
byte b = 2;
//byte c = a + b; //fails because the operation is performed as an int.
byte c = (byte)(a + b);
A byte is an 8-bit signed integer value that can hold values between -128 and 127.
You can cast 128 into a byte 'byte b = (byte)128;', but then b will just equal -128 because of how twos-complement numbers work.
Integer constants like 1, -32, 128, or 7654321 have int type in Java. Assigning an int value to a byte variable is simply not allowed. This is why you get the error cannot convert from int to byte. To get around it, you have to convert the int value to byte with a type cast, b=(byte)128.
The real question should be, why do you not get an error for this code?
byte b = 100; // !!!! ASSIGNING int TO byte !!!!
You don't get an error here because of this a special "assignment conversion" rule that only applies for compile time constant expressions:
A narrowing primitive conversion may be used if the type of the variable is byte, short, or char, and the value of the constant expression is representable in the type of the variable.
Type
byte B = (byte)128;
instead of byte B = 128;
I was learning wrapper class, here I learned how to convert int to Interger wrapper class.
But I want to convert int to Byte using Byte wrapper class.
I have tried
int a =10;
Byte c = Byte; //(Not getting suggestion in eclipse)
for example, I know how to convert int to Interger refer code below.
int a =10;
Integer b = Integer.valueOf(a);
You could add an additional cast to the integer like this:
int a = 10;
Byte c = Byte.valueOf((byte)a);
Try it online.
Btw, going from a primitive to an Boxed value is done implicitly by the compiler. So the code above doesn't necessary need the .valueOf:
int a = 10;
Byte c = (byte)a;
Try it online.
But, when you're explicitly casting values, always keep in mind that it can hold unexpected results when doing it wrong. In this case for example, the size of a a byte is smaller than an integer (the range of a byte is [-128, 127] and of an int is [-2147483648, 2147483647]), so something like this wouldn't give the expected result:
int a = 200;
Byte c = Byte.valueOf((byte)a); // Results in -56
Try it online.
This is also the reason why you usually can't go to a smaller type, since it might not fit, but you can go to a larger one without needing the cast:
byte a = 10;
Integer c = Integer.valueOf(a);
Try it online.
int x=30;
Byte c = (byte)x; // Java has auto boxing for wrapper class Byte to byte and opposite automatically
You can cast the same way you normally do with Integer wrapper.
int a =10;
Byte b = Byte.valueOf((byte)a);
//b = 10
An int is 4 bytes. So you can't convert an integer to a byte without losing some data.
int a =1000;
Byte b = Byte.valueOf((byte)a);
//b = -24
You can, however, convert an int to an array of bytes.
Java - Convert int to Byte Array of 4 Bytes?
Same you can do for Byte too:
1. autoboxing
Byte b = (byte) a;
2. valueOf(byte b)
Byte b = Byte.valueOf((byte)a)
Returns a Byte instance representing the specified byte value. If a new Byte instance is not required, this method should generally be used in preference to the constructor Byte(byte), as this method is likely to yield significantly better space and time performance since all byte values are cached.
From JLS 5.0-2. Conversions In Various Contexts so you can't perform 5.1.3. Narrowing Primitive Conversion without cast operator.
// Casting conversion (5.4) of a float literal to
// type int. Without the cast operator, this would
// be a compile-time error, because this is a
// narrowing conversion (5.1.3):
int i = (int)12.5f;
Recently,while I was going through typecasting concept in java, I have seen that type casting of larger variable to smaller variable results in the modulo of larger variable by the range of smaller variable.Can anyone please explain this in detail why this is the case and is it true for any explicit type conversion?.
class conversion {
public static void main(String args[])
{
double a = 295.04;
int b = 300;
byte c = (byte) a;
byte d = (byte) b;
System.out.println(c + " " + d);
}
}
The above code gives the answer of d as 44 since 300 modulo 256 is 44.Please explain why this is the case and also what happens to the value of c?
It is a design decision that goes all the way back the the C programming language and possibly to C's antecedents too.
What happens when you convert from a larger integer type to a smaller integer type is that the top bits are lopped off.
Why? Originally (and currently) because that is what hardware integer instructions support.
The "other" logical way to do this (i.e. NOT the way that Java defines integer narrowing) would be to convert to that largest (or smallest) value representable in the smaller type; e.g.
// equivalent to real thin in real java
// b = (byte) (Math.max(Math.min(i, 127), -128))
would give +127 as the value of b. Incidentally, this is what happens when you convert a floating-point value to an integer value, and the value is too large. That is what is happening in your c example.
You also said:
The above code gives the answer of d as 44 since 300 modulo 256 is 44.
In fact, the correct calculation would be:
int d = ((b + 128) % 256) - 128;
That is because the range of the Java byte type is -128 to +127.
For completeness, the above behavior only happens in Java when the larger type is an integer type. If the larger type is a floating point type and the smaller one is an integer type, then a source value that is too large or too small (or an infinity) gets converted to the largest or smallest possible integer value for the target type; e.g.
double x = 1.0e200;
int i = (int) x; // assigns 'Integer.MAX_VALUE' to 'i'
And a NaN is converted to zero.
Reference:
Java 17 Language Specification: §5.1.3
I need to convert a number into an unsigned byte. The number is always less than or equal to 255, and so it will fit in one byte.
I also need to convert that byte back into that number. How would I do that in Java? I've tried several ways and none work. Here's what I'm trying to do now:
int size = 5;
// Convert size int to binary
String sizeStr = Integer.toString(size);
byte binaryByte = Byte.valueOf(sizeStr);
and now to convert that byte back into the number:
Byte test = new Byte(binaryByte);
int msgSize = test.intValue();
Clearly, this does not work. For some reason, it always converts the number into 65. Any suggestions?
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.
If you just need to convert an expected 8-bit value from a signed int to an unsigned value, you can use simple bit shifting:
int signed = -119; // 11111111 11111111 11111111 10001001
/**
* Use unsigned right shift operator to drop unset bits in positions 8-31
*/
int psuedoUnsigned = (signed << 24) >>> 24; // 00000000 00000000 00000000 10001001 -> 137 base 10
/**
* Convert back to signed by using the sign-extension properties of the right shift operator
*/
int backToSigned = (psuedoUnsigned << 24) >> 24; // back to original bit pattern
http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html
If using something other than int as the base type, you'll obviously need to adjust the shift amount: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
Also, bear in mind that you can't use byte type, doing so will result in a signed value as mentioned by other answerers. The smallest primitive type you could use to represent an 8-bit unsigned value would be a short.
Except char, every other numerical data type in Java are signed.
As said in a previous answer, you can get the unsigned value by performing an and operation with 0xFF. In this answer, I'm going to explain how it happens.
int i = 234;
byte b = (byte) i;
System.out.println(b); // -22
int i2 = b & 0xFF;
// This is like casting b to int and perform and operation with 0xFF
System.out.println(i2); // 234
If your machine is 32-bit, then the int data type needs 32-bits to store values. byte needs only 8-bits.
The int variable i is represented in the memory as follows (as a 32-bit integer).
0{24}11101010
Then the byte variable b is represented as:
11101010
As bytes are signed, this value represent -22. (Search for 2's complement to learn more about how to represent negative integers in memory)
Then if you cast is to int it will still be -22 because casting preserves the sign of a number.
1{24}11101010
The the casted 32-bit value of b perform and operation with 0xFF.
1{24}11101010 & 0{24}11111111
=0{24}11101010
Then you get 234 as the answer.
The solution works fine (thanks!), but if you want to avoid casting and leave the low level work to the JDK, you can use a DataOutputStream to write your int's and a DataInputStream to read them back in. They are automatically treated as unsigned bytes then:
For converting int's to binary bytes;
ByteArrayOutputStream bos = new ByteArrayOutputStream();
DataOutputStream dos = new DataOutputStream(bos);
int val = 250;
dos.write(byteVal);
...
dos.flush();
Reading them back in:
// important to use a (non-Unicode!) encoding like US_ASCII or ISO-8859-1,
// i.e., one that uses one byte per character
ByteArrayInputStream bis = new ByteArrayInputStream(
bos.toString("ISO-8859-1").getBytes("ISO-8859-1"));
DataInputStream dis = new DataInputStream(bis);
int byteVal = dis.readUnsignedByte();
Esp. useful for handling binary data formats (e.g. flat message formats, etc.)
The Integer.toString(size) call converts into the char representation of your integer, i.e. the char '5'. The ASCII representation of that character is the value 65.
You need to parse the string back to an integer value first, e.g. by using Integer.parseInt, to get back the original int value.
As a bottom line, for a signed/unsigned conversion, it is best to leave String out of the picture and use bit manipulation as #JB suggests.
Even though it's too late, I'd like to give my input on this as it might clarify why the solution given by JB Nizet works. I stumbled upon this little problem working on a byte parser and to string conversion myself.
When you copy from a bigger size integral type to a smaller size integral type as this java doc says this happens:
https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.3
A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value, this may cause the sign of the resulting value to differ from the sign of the input value.
You can be sure that a byte is an integral type as this java doc says
https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
byte: The byte data type is an 8-bit signed two's complement integer.
So in the case of casting an integer(32 bits) to a byte(8 bits), you just copy the last (least significant 8 bits) of that integer to the given byte variable.
int a = 128;
byte b = (byte)a; // Last 8 bits gets copied
System.out.println(b); // -128
Second part of the story involves how Java unary and binary operators promote operands.
https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.2
Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:
If either operand is of type double, the other is converted to double.
Otherwise, if either operand is of type float, the other is converted to float.
Otherwise, if either operand is of type long, the other is converted to long.
Otherwise, both operands are converted to type int.
Rest assured, if you are working with integral type int and/or lower it'll be promoted to an int.
// byte b(0x80) gets promoted to int (0xFF80) by the & operator and then
// 0xFF80 & 0xFF (0xFF translates to 0x00FF) bitwise operation yields
// 0x0080
a = b & 0xFF;
System.out.println(a); // 128
I scratched my head around this too :). There is a good answer for this here by rgettman.
Bitwise operators in java only for integer and long?
If you want to use the primitive wrapper classes, this will work, but all java types are signed by default.
public static void main(String[] args) {
Integer i=5;
Byte b = Byte.valueOf(i+""); //converts i to String and calls Byte.valueOf()
System.out.println(b);
System.out.println(Integer.valueOf(b));
}
In terms of readability, I favor Guava's:
UnsignedBytes.checkedCast(long) to convert a signed number to an unsigned byte.
UnsignedBytes.toInt(byte) to convert an unsigned byte to a signed int.
Handling bytes and unsigned integers with BigInteger:
byte[] b = ... // your integer in big-endian
BigInteger ui = new BigInteger(b) // let BigInteger do the work
int i = ui.intValue() // unsigned value assigned to i
in java 7
public class Main {
public static void main(String[] args) {
byte b = -2;
int i = 0 ;
i = ( b & 0b1111_1111 ) ;
System.err.println(i);
}
}
result : 254
I have tested it and understood it.
In Java, the byte is signed, so 234 in one signed byte is -22, in binary, it is "11101010", signed bit has a "1", so with negative's presentation 2's complement, it becomes -22.
And operate with 0xFF, cast 234 to 2 byte signed(32 bit), keep all bit unchanged.
I use String to solve this:
int a = 14206;
byte[] b = String.valueOf(a).getBytes();
String c = new String(b);
System.out.println(Integer.valueOf(c));
and output is 14206.
I need to perform the following operation on a byte (2*x*x)+x where x is a single byte. Can i perform that operation directly as we will do for int. If no how can we perform above operation.
Have you tried the following?
byte x =
int f = 2 * x * x + x;
As an exercise I suggest you print out the results of every possibility byte value and see if you get the expected value. There is only 256 possible byte values.
Yes.
Java integer arithmetic is two-complement; that means that (as long there are enough bits to write down the values) lengthning or shortening the field does not affect the result.
NOTE1: Check for overflows. If the result is not in the 128 / -127 range it will not fit in a byte (or 255 / 0 for unsigned).
NOTE2: Float and double are not part of integer arithmetic.
So apparently you believe that the operator exists because the compiler allows it but you don't believe that the JVM will compute the result correctly. Why?
And how hard is this to test? Bytes only have 256 values each.
There can be two different cases:
byte a = 2; byte b = 2; byte c = a*b
it is correct but
byte a = 22; byte b = 22; byte c = a*b
it is not correct.
because while assigning value to a and b we given value less than 127 but after multiplication the value is longer than one byte that is why we need a bigger datatype to store value of c.
for that case you can use int c = a*b;