Inconsistent "Required type: byte provided: int" in Java

Question

I stored integers in byte arrays but suddenly i got a "Required type: byte provided: int" error and some lines above not. So i tried to find out what different was, tests below:

    byte b;
    int integer = 12;
    final int finalInteger = 12;
    final int finalIntegerLO = 128; // 1000 0000

    b = integer;               //Required type byte provided int
    b = finalInteger;          //OK
    b = finalIntegerLO;        //Required type byte provided int

I guess having a final int with no '1' in the 2^7 place is Ok? It gave me an idea what happens if you combine it with bitwise operators and now it makes much less sense to me..

    b = finalIntegerLO & 0xf;  //OK

Is now ok.. but

    b = integer & 0xf;         //Required type byte provided int

not??

Can someone explain me why it acts so different?

Answer 1

Let's break every line

---

case 1:

b = integer

Here we can see we are trying to convert int into a byte, so compiler asks as to explicitly typecast like so. (value of integer may exceed byte range.)

b = (byte) integer;

---

case 2:

b = finalInteger;

This case is slightly different since finalInteger is a constant whose value is fixed and the compiler can tell beforehand whether it lies within the range of byte or not. If it lies within the range compiler is smart enough to convert int to byte without us to explicitly typecast it.

---

case 3:

b = finalIntegerLO;

Range of byte is -128 to 127 clearly we cannot convert int to a byte and compiler sees that finalIntegerLO is a constant so it is impossible to carry out this conversion and we see an error

To remove this error we can explicitly typecase (DONT DO IT THOUGH) which will give use as b = -128

b = (byte) finalIntegerLO;

---

case 4:

b = finalIntegerLO & 0xf;

Here finalIntegerLO and 0xf both are constants and compiler can determine what will be the result it's 0 which is within the range of byte.

---

case 5:

b = integer & 0xf;

Here integer value can be changed before the execution of this line so the compiler is not sure if the result is within the range of int or not, so it asks us to explicitly typecast like so.

b = (byte) (integer & 0xf);

Again like case 3 you may get an unexpected result.

Answer 2

The error that is received while executing b = integer is "incompatible types: possible lossy conversion from int to byte". When this statement is being executed by the compiler, an implicit type conversion from a higher data type(int) to a lower data type(byte) is being carried out. This is not possible as there might be loss of information/precision with such conversions and hence, java tried to avoid them. An alternative would be to force/coerce the conversion using an explicit type conversion as follows:

byte b;
int integer = 12;
b = (byte) integer;

Moving on,

final int finalInteger = 12;
final int finalIntegerLO = 128;

b = finalInteger works because the final keyword ensures that the variable finalInteger does not change its value. Since, we are already telling our Java compiler that we cannot change the value of finalInteger variable and because finalInteger stores a value that is in the range of values that can be stored in a byte variable (-128 to 127), the conversion from int to byte is possible. b = finalIntegerLO does not work because the value of finalIntegerLO exceeds the maximum value for a byte variable, i.e, -128 to 127.