Convert Int (last 8 bits) to Byte

[gonzalioz]

Hi all,

I need to convert a int to a byte and I am having a really really hard time. I tried shifting operators, casting all kinds of weird stuff I found on the internet. But nothing works. Everytime my Integer gets to 128 or above and I cast it to a byte it will turn negative.

My testing code:

int i1 = 234;		

byte attempt1 = (byte) i1;	
byte attempt2  = (byte) (i1 >> 8);
byte attempt3  = (byte) ((i1 << 24) >> 24);
		
System.out.println("i1: " + i1 + "  |  " + Integer.toBinaryString(i1));
System.out.println("attempt1: " + attempt1);
System.out.println("attempt2: " + attempt2);
System.out.println("attempt3: " + attempt3);

output:

i1: 234  |  11101010
attempt1: -22
attempt2: 0
attempt3: -22

Please please can somebody help me? I am pretty desperated.

Thnx a lot in advance.

[JosAH]

A simple cast will do the job:

int i= ...;
byte b= (byte)i;

kind regards,

Jos

[Toll]

Bytes will never go higher than 127. There are no unsigned variables in Java, if that's what you're after; if you want a number higher than 127, you have to either use a char (16bit) or an int (32bit).

EDIT: There's also long, of course (64bit) but an int will generally suffice.

[gonzalioz]

Thanks. I was confused with .NET byte. In .NET byte is unsigned by default :).

Thank you!