Byte not showing addition result in Java

Question

Why am I not getting the output since byte value is from -128 to 127 and the result is 30 ?

class test15 {          
   public static void main(String ...s) {
      byte a = 10;
      byte b = 20;
      byte c = a+b;
      System.out.println(c);
   }    
}

Answer 1

CompileTime vs RunTime

a,b are bytes and bytes doesn't have addition operator. At run time, since a,b are not final and integer addition takes place which result is an int . And you are trying to assign it to a byte. Which results a compilation error.

byte c = (byte)a+b;

And as a side note. try to change your code as

 final byte a = 10;
 final byte b = 20;
 byte c = a + b;

That compiles fine because of narrowing conversion and it is possible, if all the operands are constant values in an expression.

Answer 2

You must cast to byte . Since a and b are of type byte, Java will convert them to an int , and then the result is an int .

byte c = (byte) (a+b);

Answer 3

Compile time error, if I change this

byte c = a + b;

to

byte c = (byte) (a + b);

I get the output

30

Answer 4

Change

byte c = a + b;

to

byte c = (byte)(a + b);

It works for me.

Answer 5

Java converts the result of arithmetic operations to int type by default. Try this:

public static void main(String... s)
{

    byte a = 10;
    byte b = 20;
    byte c = (byte) (a + b);

System.out.println(c);
}

That (byte) changes the int value returned by (a+b) into a byte value. Normally, this cannot be done, as the range of int is much greater than the range of byte, so a loss of precision would occur. Using casting forces the conversion, regardless of precision.

Answer 6

There is no instruction set to perform operation on a byte type. Rather the instruction set for int type is used for the operation on boolean , byte , char , and short types. http://docs.oracle.com/javase/specs/jvms/se7/html/jvms-2.html#jvms-2.11.1 Also A data type of higher size (occupying more memory) cannot be assigned to a data type of lower size. This is not done implicitly by the JVM and requires explicit casting; a casting operation to be performed by the programmer

The result will be an Integer. For this you must do

byte a = 10;
byte b = 20;
byte c = (byte) (a + b); //u will have to typecast int to byte 

System.out.println(c);

Answer 7

Here's whats happening

byte a = 10;

variable a of byte type created 10 assigned to it.

byte b = 20;

variable b of byte type created 10 assigned to it.

byte c = a+b;

a+b is automatically promoted to integer, since now result of summation is integer you cannot assign integer to byte.