Type promotion in Java

Question

I have a problem with the below Java statements:

byte b = 10;
byte r = (byte) (b * b); // Giving correct result
byte r = (byte) b * b; // Giving error " POSSIBLE LOSS OF PRECISION"

Why is it mandatory to give parentheses to b * b?

Answer 1

(byte) b * b casts the value of the first b to byte (which is redundant since it was already byte), and multiples it by the value of the second b. Multiplying two bytes promotes them to int first, since there is no * operator for bytes. Therefore the result is int, and cannot be assigned to a byte variable.

On the other hand, (byte)(b * b) casts the int multiplication result to byte, which can be assigned to a byte variable.

This is covered in the JLS in 5.6.2. Binary Numeric Promotion:

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order:

1. If any operand is of a reference type, it is subjected to unboxing conversion (§5.1.8).

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.

Answer 2

Casting problem

byte r = (byte) (b * b);

It casts the (byte) type to the result of (b * b)

byte r = (byte) b * b;

It casts the (byte) type to the first b only, therefore it will become ((byte) b) * b

Answer 3

By the precedence rule you are casting only the first b to byte instead of the whole result.

And Java follow some rules, as you can see here

All integer values (byte, short and int) in an arithmetic operations (+, −, *, /, %) are converted to int type before the arithmetic operation in performed. However, if one of the values in an arithmetic operation (+, −, *, /, %) is long, then all values are converted to long type before the arithmetic operation in performed.

So, by just casting the first b you are doing this:

byte = byte * integer

Hence:

byte = integer

Thus, raised error.

Answer 4

Variables of type byte must be [-128,127], that is why the compiler must not accept any operation, b*b;b+b;b-b;b/b without a cast on the result of operation, like: (byte)(b*b).

In the code below, when you change the result type to int it compiles.

 byte b=10;
    byte c=(byte)b*b;      //incompatible but correct when c is int
    byte d=((byte)b)*b;      //incompatible but correct when d is int
    byte r=(byte)(b*b); 
    byte t= b*b;             //incompatible but correct when t is int
    byte e=(byte)b/b;      //incompatible but correct when e is int
    byte f=((byte)b)/b;      //incompatible but correct when f is int
    byte o=(byte)(b/b); 
    byte w=b/b;             //incompatible but correct when w is int
    byte g=(byte)b+b;      //incompatible but correct when g is int
    byte p=((byte)b)+b;      //incompatible but correct when p is int
    byte q=(byte)(b+b);
    byte x=b+b;             //incompatible but correct when x is int
    byte h=(byte)b-b;      //incompatible but correct when h is int
    byte v=((byte)b)-b;      //incompatible but correct when v is int
    byte s=(byte)(b-b);
    byte y=b-b;               //incompatible but correct when y is int
    byte k=(byte)b;
    byte u=b;   

NOTE

As @andy Truner pointed out in comments, when b is final, all the previous instructions compile!!, except for the following set up:

   final byte fn=-120;
    byte z=fn-b;               //this does not compile even when both final, because the result would be -130, out of the byte type interval!!