Bitwise operators in Java acting on two bytes

Question

This should be simple but I'm having trouble changing some C code that uses bitwise operators and shifts into Java.

In C I have:

unsigned short Color(byte r, byte g, byte b)
{
  return( ((unsigned short)g & 0x1F )<<10 | ((unsigned short)b & 0x1F)<<5 | (unsigned short)r & 0x1F);
}

This function, Color, returns 16bits where 15 of them represent three 5bit values for red, green, and blue color channels. I'm trying to do something similar in Java except I'm having trouble with data types.

In java I have:

int r=someval, g=anotherval, b=yetanotherval;

And I want to convert these integers to a 2byte data type and use the same bitwise/shift operations as above.

What is the best way to approach this? I was thinking of a 2byte array but that seems unnecessary.

edit

So I gave it a try in Java with shorts but there still seems to be an issue. I started simple with:

short r=someval, g=anotherval, b=yetanotherval;
short colorData = g & 0x001F;

But the compiler complains: "cannot convert from int to short"

Note, that casting to (unsigned short) in C does not have a sense at all, because arithmetic/bitwise operation arguments automaically promoted to [unsigned] int, and result of "&" will be int too. — Vovanium
– Vovanium, Commented Nov 11, 2010 at 14:55

unwind · Accepted Answer · 2010-11-11 14:49:55Z

2

You have the short type in Java as well, but it's always signed, there is no unsigned in Java.

Therefore, it might be better to use an int, so you don't need to use the sign bit to hold actual color data, which can be a bit (no pun intended!) confusing.

This should work as a starting point, at least:

int Color(int r, int g, int b)
{
  return ((g & 0x1F) << 10) | ((b & 0x1F) << 5) | (r & 0x1F);
}

I made the arguments int as well, for simplicity's sake.

Update: It sounds (and seems) as if you can actually use short, since you only need 15 bits for the color format. Then let's do that:

short Color(short r, short g, short b)
{
  return ((g & 0x1F) << 10) | ((b & 0x1F) << 5) | (r & 0x1F);
}

I haven't tried compiling the above, but it's likely that you'll get conversion errors and will need to add explicit short casts throughout to make it compile.

edited Nov 11, 2010 at 14:49

answered Nov 11, 2010 at 14:40

unwind

402k64 gold badges492 silver badges620 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Anon Over a year ago

Thanks, I'm actually not using the left most bit since I only need 15 of the 16 bits. Would shorts work for me then?

codymanix · Accepted Answer · 2010-11-11 14:49:18Z

2

In java, all calculation operations, even with only types smaller than int, always result to an int, so you have to explicitly cast it back to short:

short colorData = (short)(g & 0x001F);

answered Nov 11, 2010 at 14:49

codymanix

29.7k21 gold badges98 silver badges155 bronze badges

Comments

Jörn Horstmann · Accepted Answer · 2010-11-11 15:51:01Z

2

In your example

short r=someval, g=anotherval, b=yetanotherval;
short colorData = g & 0x001F;

Both g and the hex literal get automatically promoted to int because the logical and operator always works on and returns integers. You can of course cast the result back to short.

short colorData = (short)(g & 0x001F);

The same should work for the char datatype, which is in fact the only unsigned integer type on the JVM. I would recommend against using the char type in this way, unless this is an isolated performance/memory sensitive part of your program, since using char for arithmetic might seem strange and counterintuitive for other maintainers of the source.

answered Nov 11, 2010 at 15:51

Jörn Horstmann

34.1k11 gold badges77 silver badges122 bronze badges

Comments

Jules · Accepted Answer · 2010-11-11 14:41:31Z

1

char: The char data type is a single 16-bit Unicode character. It has a minimum value of '\u0000' (or 0) and a maximum value of '\uffff' (or 65,535 inclusive).

http://download.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

The Java char datatype is 2 bytes by default.

answered Nov 11, 2010 at 14:41

Jules

1,3718 silver badges19 bronze badges

2 Comments

Anon Over a year ago

I gave char a try earlier but the compiler complains "cannot convert from char to int" whenever I try to do a bitwise 'and' on it with 0x001F.

Jules Over a year ago

This is getting a bit dirty now, but you could try specifying your hex like this "'\u001F'". This is the literal for char-values. But looking at your other comment, use the short if you only need 15 bit anyway.

Flavio · Accepted Answer · 2010-11-11 14:42:39Z

1

Since you are using only 15 bits, the sign bit on the short data type will be no problem.

answered Nov 11, 2010 at 14:42

Flavio

12k3 gold badges37 silver badges41 bronze badges

Collectives™ on Stack Overflow

Bitwise operators in Java acting on two bytes

5 Answers 5

1 Comment

Comments

Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

1 Comment

Comments

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related