CODEWITHSUNDEEP
javacasting
Euriloco
Euriloco

Reputation: 253

Type promotion Java

I`m learning Java with the Herbert Schildt book's: Java a Beginner's Guide. In that book appears this code:

// A promotion surprise!
class PromDemo{
    public static void main(String args[]){
        byte b;
        int i;
        b = 10;
        i = b * b;      // OK, no cast needed

        b = 10;
        b = (byte) (b * b);     // cast needed!!

        System.out.println("i and b: " + i + " " + b);
    }
}

I don't understand why I must use (byte) in the line:

b = (byte) (b * b);     // cast needed!!

b was defined as a byte and the result of b * b is 100 which is right value for a byte (-128...127).

Thank you.

Upvotes: 4

Views: 398

Answers (2)

yas
yas

Reputation: 3620

The JLS (5.6.2. Binary Numeric Promotion) gives rules about combining numeric types with a binary operator, such as the multiplication operator (*):

  • If either of the operands is of type double, the other one will be converted to a double.
  • Otherwise, if either of the operands is of type float, the other one will be converted to a float.
  • Otherwise, if either of the operands is of type long, the other one will be converted to a long.
  • Otherwise, both operands will be converted to an int.

The last point applies to your situation, the bytes are converted to ints and then multiplied.

Upvotes: 7

Thomas Uhrig
Thomas Uhrig

Reputation: 31605

In Java, byte and short will always be promoted to int, when you have a calculation like this:

byte b = 10;
b = (byte) (b * b);

So you actually multiply an integer with an integer, which will return an integer. Since you cannot assign an integer to a byte, you need the cast.

This is called "automatic type promotion" if you would like to Google it (to find e.g. https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.2)

Upvotes: 3

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