如何从组合布尔值中获取十六进制值？

Question

I want to get a hexadecimal value from four boolean variables something like this example:

boolean[] m = {true, false, true, true};

I want to get a String or char that hold B which means 1011 in binary.

PS: I am working on an Android app.

Answer 1

You can use below code to get your binary string, integer value and hexDecimal value.

    boolean[] m = {true,false,true,true};
    String binaryStr = "";
    for (boolean bit : m) {
        binaryStr = binaryStr + ((bit) ? "1" : "0" );
    }
    int decimal = Integer.parseInt(binaryStr , 2);
    String hexStr = Integer.toString(decimal , 16);

In above code, binaryStr is your Binary String ie 1011 and its equivalent decimal and hexa decimal value are decimal and hexStr

Answer 2

boolean[] booleanArray = new boolean[] { true, false, false, true };
String binaryString = Arrays.toString(booleanArray).replace("true", "1").replace("false", "0");

and just convert that binaryString to hexadecimalvalue

Answer 3

While the above answers do work, they have their limitations. The first limit being a cap at 32 booleans because of the 32 bit integer limit. But the above code also uses built-in methods, they don't allow a beginner to grasp what is really going on.

If you just want to get the job done, use something from above. It'll probably be more efficient. I'm just posting this solution because I feel it will allow someone who wants to grasp the actual concept do so better.

(Hopefully the comments make a little sense)

public static String booleanArrayToHex1(boolean[] arr){

    if(!(arr.length%4==0)){arr=makeMultOf4(arr);} //If you are using arrays not in multiples of four, you can put a method here to add the correct number of zeros to he front.
    //Remove the above line if your array will always have a length that is a multiple of 4

    byte[] hexValues=new byte[arr.length/4]; 
    //The array above is where the decimal hex values are stored, it is byte because the range we need is 0-16, bytes are the closest to that with the range of 0-255

    int hexCounter=arr.length/4-1; //counts what hex number you are calculating

    /* The below loop calculates chunks of 4 bianary numbers to it's hex value
     * this works because 16 is a power of 2
     * so if we simpily squish those numbers together it will be the same as finding the accual integer value of the whole thing
     * This can go to higher numbers because we do not have the 32 bit integer limit
     * it runs in reverse because the lowest value is on the right of the array (the ones place is on the right)
     */
    for(int i=arr.length-1;i>=0;i--){
        if(arr[i]){
            int place=1; //will count the value of a bianary number in terms of its place

            for(int j=3;j>i%4;j--) 
            //loop multiplies the bianary number by 2 j times, j being what place it's in (two's, four's, eight's). This gives the correct value for the number within the chunk.
            //This is why the array needs to be a multiple of 4
                place*=2;

            hexValues[hexCounter]+=place; //this will add that place value to the current chunk, only if this place in the boolean array is true (because of the above if - statement)
        }

        if(i%4==0)//moves the chunk over one every 4 binary values
            hexCounter--;
    }

    //Everything below simpily takes the array of decimal hex values, and converts to a alpha-numeric string (array to normal hex numbers)
    String ret="";
    for(byte b:hexValues)
        switch(b){
            case 10:
            ret=ret+"A";
            break;
            case 11:
            ret=ret+"B";
            break;
            case 12:
            ret=ret+"C";
            break;
            case 13:
            ret=ret+"D";
            break;
            case 14:
            ret=ret+"E";
            break;
            case 15:
            ret=ret+"F";
            break;
            default:
            ret=ret+b;
            break;
        }
    return ret;
}

If your array length is not a multiple of 4, to make this work you will need to make the array a multiple of 4. I have some code below that will do that. It's not commented, and probably not that efficient, but it works.

public static boolean[] makeMultOf4(boolean[] arr){
    if(arr.length%4==0){return arr;}
    boolean[] array=new boolean[arr.length+(arr.length%4==1?3:arr.length%4==2?2:1)];
    for(int i=0;i<array.length;i++){
        try{
            array[i]=arr[i-(arr.length%4==1?3:arr.length%4==2?2:1)];
        }catch(Exception e){
            array[i]=false;
        }
    }
    return array;
}

Answer 4

You can use this logic :

String x = ""; 
for(int i  = 0 ; i < m.length ; i++){
    if(m[i])
      x += "1";
    else
       x += "0";
 }

Log.i("values = ",x);

Answer 5

For each boolean value append to binary string 1 if true 0 otherwise, then using Integer.parseInt() convert binary string to Integer instance, finally convet integer to hex string using Integer.toHexString() method

@org.junit.Test
public void test() throws Exception{

    boolean[] m = {true, false, true, true};
    String binary = "";

    for(boolean b : m) binary += b ? 1 : 0;

    String hex = Integer.toHexString(Integer.parseInt(binary, 2));

    System.out.println(hex);
}