Bit Packing (Bytes into Int)
I have two bytes and im trying to pack their bits into an int.
I have it working fine now:
int result =((b2 < 0 ? (b2 + 256) : b2) << 8) | ((b1 < 0 ? (b1 + 256) : b1) << 0);
Is there a way to do it without checking if the byte is negative?
I dont want the value of the byte converted to an int. I want the bits
copy and pasted in.
For example, this:
int result =b2 << 8 | b1;
wont work for this reason:
byte= 1010,1010 = -86
int for -86 = 11111111111111111111111110101010
int = 1010,1010 = 170
publicclass ShiftTest{
publicstaticvoid main(String[] args){
new ShiftTest();
}
public ShiftTest(){
int num = 682;
byte b1 = (byte)num;
byte b2 = (byte)(num >>> 8);
int result =((b2 < 0 ? (b2 + 256) : b2) << 8) |
((b1 < 0 ? (b1 + 256) : b1) << 0);
System.out.println("Result: " + result +" , " + Integer.toBinaryString(result));
System.out.println();
System.out.println("Num Bits: " + Integer.toBinaryString(num));
System.out.println("B1: " + (b1) +" , " + ((int)b1) +" , " + Integer.toBinaryString((int)b1));
System.out.println("B2: " + (b2) +" , " + ((int)b2) +" , " + Integer.toBinaryString((int)b2));
}
}
If you want to pack bytes b1 and b2 in an int simply do this:int i= ((b1&0xff)<<8)|(b2&0xff);kind regards,Jos
(b && 0xff) << 8 I think.Too lazy to try it.- Edit -Yeah, what Jos said.- - -
And the other way:
byte b1 = (byte)(num & 0xff);
byte b2 = (byte)((num >>> 8) & 0xff);
I'm surprised you didn't know this. The 0xFF trick must have been posted here a million times.
Jverd wrote:
> Yeah, what Jos said.
Indeed; would you be so kind to repeat that once more for the record?
> Yeah, what Jos said.
Thank you; one more for the road please?
> Yeah, what Jos said.
You're great; thanks a bunch ;-)
kind regards,
Jos (< what he says ;-)
Ah, yes.int result2 = ((b2 & 0xff) << 8) | (b1 & 0xff);is certainly the more correct way to do this, lol.D'ohp.s.: and as always thanksexcept for Jos that modest b@stard, lol.
>> byte b1 = (byte)(num & 0xff);
>> byte b2 = (byte)((num >>> 8) & 0xff);
Mr Evil, btw this isnt needed.
When int is cast to a byte it automatically just truncates the bits
to the last 8.
So its just
byte b1 = int;
byte b2 = int >>> 8
(where in this case >>> and >> are interchangeable really)
The problem was that for bitshifting the byte is cast up to an int -
unhappily if its a negative number because of twos complement.
> Mr Evil, btw this isnt needed.
> When int is cast to a byte it automatically just truncates the bits
> to the last 8.
>
> So its just
> byte b1 = int;
Nope, the compiler will whine about a possible loss of precision. You
have to narrow cast your int to a byte explicitly:
byte b1= (byte)i;
byte b2= (byte)(i>>8);
modest regards,
Jos ;-)
> Mr Evil, btw this isnt needed.
> When int is cast to a byte it automatically just
> truncates the bits
> to the last 8.
Sorry, you're right. I was thinking that negative int values would yield negative bytes, but the only int values that would not turn into negative bytes by merely truncating, are all out of a byte's range, so it would be pointless.
> > Mr Evil, btw this isnt needed. > > When int is cast to a byte it automatically just> > truncates the bits> > to the last 8.> > Sorry, you're right.No he isn't. See my reply #7.kind regards,Jos
I am right regarding what I was talking about to Mr Evil you Jos
are right about what you were arguing:
Your:
byte b1= (byte)i;
byte b2= (byte)(i>>8);
is much different than his:
>> byte b1 = (byte)(num & 0xff);
>> byte b2 = (byte)((num >>> 8) & 0xff);
You were telling me you indeed need the (byte) cast.
Meanwhile what I was saying to him is that he didnt need the & oxff.
There was some argument crossfire there, lol.
> I am right regarding what I was talking about to Mr Evil you Jos
> are right about what you were arguing:
>
> Your:
> byte b1= (byte)i;
> byte b2= (byte)(i>>8);
>
> is much different than his:
> >> byte b1 = (byte)(num & 0xff);
> >> byte b2 = (byte)((num >>> 8) & 0xff);
>
> You were telling me you indeed need the (byte) cast. Meanwhile what
> I was saying to him is that he didnt need the & oxff.
>
> There was some argument crossfire there, lol.
Ok, agreed but your example was still wrong; doing a:
byte b1= i;
makes the compiler whine, i.e. you still need the explicit cast.
kind regards,
Jos
>> Ok, agreed but your example was still wrong; doing anotice that in my code example in post 1 i did the explicit cast.In my fervor, When I deleted his &oxff i deleted the cast.: P
> >> Ok, agreed but your example was still wrong; > notice that in my code example in post 1 i did the explicit cast.> In my fervor, When I deleted his &oxff i deleted the cast.Ok, let's call it a day.kind regards,Jos
> Ok, let's call it a day.Haha, I was just teasing.I just got home from work to do a full counter of dishes so im glad you called it - i have an excuse to crash now.Except for the small matter of dinner...
