Java simple boolean[] to byte conversion

Question

I have an array of 8 booleans which I want to simply convert to a byte. Is there a simple way to do this? Or do I have to use for loop?

Personally I'd prefer a simple up to two lines solution if it exists.

Thanks for help.

EDIT: Possible duplicate is just one boolean to a byte, I have an array.

ANOTHER EDIT: I get a byte from a udp packet then I set the first bit (boolean) to false, then I would need to get a byte out of that again.

That question is just one boolean to byte, I have a whole array. — grizeldi
– grizeldi, Commented Aug 17, 2015 at 10:29
So iterate over array. We might be more helpful if you tell us what have you tried so far. — Kiki
– Kiki, Commented Aug 17, 2015 at 10:30
Yes, that's a way, but I'm looking for a simpler solution if it exists. — grizeldi
– grizeldi, Commented Aug 17, 2015 at 10:31
oneliner solution is to make method that converts boolean array to byte array - method will take something like 6 to 10 lines, but then you will use it just like byte[] arr = booleanToByteArray(booleanArrayHere); — itwasntme
– itwasntme, Commented Aug 17, 2015 at 10:41

Eran · Accepted Answer · 2015-08-17 10:37:57Z

8

I think a loop is better, but if you must have a one liner :

byte b = (byte)((bool[0]?1<<7:0) + (bool[1]?1<<6:0) + (bool[2]?1<<5:0) + 
                (bool[3]?1<<4:0) + (bool[4]?1<<3:0) + (bool[5]?1<<2:0) + 
                (bool[6]?1<<1:0) + (bool[7]?1:0));

For the input :

boolean[] bool = new boolean[] {false,false,true,false,true,false,true,false};

you get the byte 42.

edited Aug 17, 2015 at 10:37

answered Aug 17, 2015 at 10:35

Eran

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2 Comments

grizeldi Over a year ago

Is the last bool[1] there for a reason or should it be bool[7]?

Eran Over a year ago

@grizeldi Thanks, that was a copy and paste error :)

André R. · Accepted Answer · 2015-08-17 10:47:08Z

3

if i understood your question correct, then you want to convert a byte that is represented by it's bit pattern as an array of booleans ... if that is the case then this is one solution :

 public static void main(String[] args) {
        // The integer where the result will be build up
        int result = 0;
        // The bit-pattern as an array of booleans
        // 2^0 Position is at bools[0] !!
        // so this is the byte '0 1 0 1 0 1 0 1' --> decimal : 85
        boolean[] bools = new boolean[]{ true, false, true, false, true, false, true, false };
        // iterate through the 'bits'
        // and set the corresponding position in the result to '1' if the 'bit' is set
        for (int i = 0; i < bools.length; i++) {
            if (bools[i]) {
                result |= 1 << i;
            }
        }
        System.out.println(result);  // Prints '85'
    }

answered Aug 17, 2015 at 10:47

André R.

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1 Comment

grizeldi Over a year ago

Yes I wanted something like that, but @Eran's oneliner is what I'm using now.

jlxip · Accepted Answer · 2018-06-30 20:42:41Z

0

I'd do it like this:

byte b = 0;
for(int i=0;i<8;i++) if(binaryValues[i]) b |= (128 >> i);
return b;

edited Jun 30, 2018 at 20:42

answered Jul 3, 2017 at 11:33

jlxip

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