Byte incrementing operation in Java

Question

What happens when we try to increment a byte variable using the increment operator and also by the addition operator.

public class A {
    public static void main(final String args[]) {
        byte b = 1;

        b++;

        b = b + 1;
    }
}

Please give me the source where we can find such small things unleashed? Please help me out.

"Please give me the source where we can find such small things unleashed" - The Java Tutorial, any good Java text book ... or if you are good at reading formal documents, the Java Language Specification — Stephen C, Nov 19 '12 at 11:20
@PeterLawrey.. Why? The 3rd statement will in fact not compile. Cannot cast form an `int` to a `byte`. — Rohit Jain, Nov 19 '12 at 11:26

score 13 · Accepted Answer · edited Jun 20 '19 at 10:08

13

The difference is that there is an implicit casting in the ++ operator from int to byte, whereas, you would have to do that explicitly in case if you use b = b + 1

b = b + 1;  // Will not compile. Cannot cast from int to byte

You need an explicit cast:

b = (byte) (b + 1);

Whereas b++ will work fine. The ++ operator automatically casts the value b + 1, which is an int to a byte.

This is clearly listed in JLS - §15.26.2 Compound Assignment Operators : -

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once

Please note that operation b + 1 will give you a result of type int. So, that's why you need an explicit cast in your second assignment.

edited Jun 20 '19 at 10:08

TechDog

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answered Nov 19 '12 at 11:22

Rohit Jain

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But when I use an int it still shows me error??? – Praveen Nov 19 '12 at 11:25
Sorry Rohit.....When i use the explicit cast that u've said, still gives me error. – Praveen Nov 19 '12 at 11:28
@Praveen.. It should not give error AFAIK. Can you please post your current code? – Rohit Jain Nov 19 '12 at 11:29
public class A3{ public static void main(String str[]){ byte b=1; b++; //line 2 System.out.println(b); b=(byte)b+1; //line 3 System.out.println(b); }} – Praveen Nov 19 '12 at 11:32
So what happens if I write b = b + (byte)1? Just Curious!!! – Vishal Nov 19 '12 at 11:35
@Praveen.. Note the bracket. `(byte)b + 1` should be `(byte)(b + 1)` – Rohit Jain Nov 19 '12 at 11:36
@Vishal.. That would also not work. Because, even though you converted `1` to a byte value, the result of operation between two byte values is an `int`. And then again the same reason. – Rohit Jain Nov 19 '12 at 11:37
NO it is not compiling Vishal – Praveen Nov 19 '12 at 11:38
Yeah. That would not compile, as per the reason I posted in my last to last comment. – Rohit Jain Nov 19 '12 at 11:38
@RohitJain I'l compile it and see now – Praveen Nov 19 '12 at 11:38
@RohitJain Now I am little confuse... Anyway I will try to find it later. Thanks – Vishal Nov 19 '12 at 11:39
@RohitJain Hurrah thanx............It s working – Praveen Nov 19 '12 at 11:40
1

@Vishal.. Actually, the type of the result you get of an expression is at least an `int`. Now, as you add a higher type, the type of result is that higher type. Operation between: - `byte and byte is int`, `int and byte is int`, `int and long is long`, and so on., – Rohit Jain Nov 19 '12 at 11:40
@RohitJain where do u find these smaller things....Give me some tips to do good programming??? – Praveen Nov 19 '12 at 11:41
@Praveen.. All comes from experience my bro. Don't worry much if you don't know about them now. Everyday you keep on adding new stuffs to your knowledge bag, and thus learning many things. Just Enjoy Coding, and forget everything :) – Rohit Jain Nov 19 '12 at 11:42
@RohitJain it is an information to me... Hope I could remember it.. Thanks. – Vishal Nov 19 '12 at 11:44
Yep bro...Thnx for ur help. Can i get ur mail id? – Praveen Nov 19 '12 at 11:44
@RohitJain Can i get ur mail id? – Praveen Nov 19 '12 at 11:46
@Praveen.. That's in my profile information. And You're welcome :) – Rohit Jain Nov 19 '12 at 11:46
@RohitJain Thank u bro...Keep Sharing – Praveen Nov 19 '12 at 12:00
@Praveen.. Sure. You can anytime mail me whenever you have any problem. Else, SO is always there. Be in touch with SO to learn better programming, and yes, Enjoy Coding. :) – Rohit Jain Nov 19 '12 at 12:04

Petar Minchev · Answer 2 · 2012-11-19T11:25:34.877

3

What happens? Actually b = b + 1 won't compile.

You must explicitly convert it to byte, because b + 1 evaluates to int. And it is not guaranteed that an int can fit into a byte.

b = (byte)(b + 1);

edited Nov 19 '12 at 11:25

answered Nov 19 '12 at 11:18

Petar Minchev

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Well, b = b + 1 will not compile. – Rohit Jain Nov 19 '12 at 11:19
Why is it happening so Rohit? – Praveen Nov 19 '12 at 11:21
@Praveen.. See my answer below. – Rohit Jain Nov 19 '12 at 11:24

score 0 · Answer 3 · answered Nov 19 '12 at 11:18

0

The variable will be incremented twice

answered Nov 19 '12 at 11:18

Blank Chisui

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score 0 · Answer 4 · answered Nov 19 '12 at 11:19

0

b++; and b=b+1; are equivalent and will result in the same bytecode.

b will be equal to 3 before the main method is finished.

Edit: actually they are not equivalent: b=b+1; is wrong and should be b=(byte) (b+1); [cast to a byte, otherwise it is an int]

answered Nov 19 '12 at 11:19

Burkhard

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But when I execute the code I'm getting a compile time error??? – Praveen Nov 19 '12 at 11:20

Byte incrementing operation in Java

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