Java default Integer value is int

Question

My code looks like this

import java.util.Scanner;


public class StudentGrades {

public static void main(String[] argv)
{
    Scanner keyboard = new Scanner(System.in);
    byte q1 = keyboard.nextByte() * 10;
}
}

It gives me an error Type mismatch: cannot convert from int to byte.

Why would Java store a literal operand that is small enough to fit in a byte, into an int type? Do literals get stored in variables/registers before the ALU performs arithmetic operations?

Answer 1

The problem is the line byte q1 = keyboard.nextByte() * 10;. There are no arithmetic operations on byte or short. The value of keyboard.nextByte() is casted up to an int prior to multiplication with 10, which is also an int. The result of the multiplication is an int, which can not be stored into q1 if it's defined as a byte.

Possible solutions would be to cast the result of the multiplication to a byte using (byte) (keyboard.nextByte() * 10) or by changing the type of q1 to int.

Unfortunately, I'm not finding a reference to this in the tutorials. However, you can find an explanation in the Java Language Specification. Sections 5.6.2 Binary Numeric Promotion and 15.18.2 Additive Operators (+ and -) for Numeric Types define this behavior.