Possible Loss of Precision

[Rob Sweeny]

Hi All,

I've Included two source files that I would think would do the same thing, but they dont

Using Compound Assignment Operators I have no compilation errors.
But when I convert the code to strict Assignment Operators I get "Possible Loss of Precision"

1. How do I change the 2nd source code so it doesn't give me these errors
2. Why is there a difference. Why doesn't the Compound Assignment Operators give me the same errors.


No Compilation Errors:

import java.io.*;

public class AssignmentOperators{
	public static void main(String[] args) {
		// Compound Assingment Operators
		byte a;
		a = 10;
		System.out.println(a += 3); // 13
		a = 15;
		System.out.println(a -= 3); // 12
		a = 20;
		System.out.println(a *= 3); // 60
		a = 25;
		System.out.println(a /= 3); // 8
		a = 30;
		System.out.println(a %= 3); // 0
	}
}

Compliation Errors:

import java.io.*;

public class AssignmentOperators{
	public static void main(String[] args) {
		// Compound Assingment Operators
		byte a;
		a = 10;
		System.out.println(a = a + 3); // 13
		a = 15;
		System.out.println(a = a - 3); // 12
		a = 20;
		System.out.println(a = a * 3); // 60
		a = 25;
		System.out.println(a = a / 3); // 8
		a = 30;
		System.out.println(a = a % 3); // 0
	}
}

Compilation Error:

src\com\scjaexam\tutorial\pirates\AssignmentOperators.java:11: possible loss of
precision
found : int
required: byte
_________System.out.println(a = a + 3); // 13
_____________________________^


Thanks!

[W. Joe Smith]

When you are doing this:

a = a + whatever

    

The addition operation is returning an int. You are trying to assign an int to a byte, which is giving you a warning. You can either switch a's declaration to be an int, or explicity cast all of your additions to byte, essentially telling the compiler you are aware there is a possible loss of precision but you want to do it anyway.

[Rob Sweeny]

How do i "explicity cast all of your additions to byte" ?

System.out.println(a = a + 3b); // 13

Doesn't work.

[W. Joe Smith]

You explicitly cast the additions like this:

 

a = (byte)(a + whatever)

   

This is telling the compiler that you are taking the result of your addition, no matter what it is, and changing it to a byte. If any of your additions should go beyond the bounds of a byte, though, you will get unexpected results.

I would suggest doing a Google search for explicit casting in java, as it has many other uses.

[Campbell Ritchie]

W. Joe Smith wrote: . . . I would suggest [a] search for explicit casting in java . . .

You will also find something about it in the Java™ Language Specification, but it is by no means easy reading.