What will be output of following program?

public class Test{
      public static void main(String[] args){
            byte b=127;
            b++;
            b++;
            System.out.println(b);
      }
}

Shouldn't we consider 0 as a value after 128?

c. -127 is correct answer. As b++ is b+=1 and += is implicit cast, means it cast the result from integer to byte automatically. Now 129(32 bit , as it was integer) is converted into 8 bit (byte can store 8 bit).
129 -> 10000001 , As we know that byte are signed and left most digit represent the sign (0 means positive and 1 means negative), here the left most value is 1 (in 10000001) so negative of it will be the answer , to find the negative we do 2's complement and add 1 to it
10000001 -----> 01111110 (2's complement) +1 ---> 01111111 = 127
and as we already know the left most bit was 1 so it is -127

hope this helps :)

C is the correct answer
Byte range is from -127 to 128
b=127
b++= 128
after 128 it will back to -127 again..
b++= -127

Answer is -127 because Increment operators has implicit cast.
b++;
is equivalent to
b = (byte) (b + 1);

but, on the other hand,
b = b + 1;
is a simple arithmetic operation and need type caste from int to byte.

result of arithmetic operation is int so it needs to type conversion, and here it stores in short so compilation error.....understand

can anyone help me,with detailed explanation..

the solution is not relevant.