Review implementation of boolean triple

Question

Let a,b,c be a boolean value. I need to print all values of expression a && b && c. I'm writing the class

public class BooleanTriple{
    private boolean a,b,c;
    public BooleanTriple(boolean a, boolean b, boolean c){ this.a =a; this.b=b; this.c=c;}

    /**
    *Increments the value of triple with lexicographical ordering
    */
    public void incr(){
        if (!a) a=true;
        else if(!b){
            a= false;
            b= true;
        else if (!c){
            a= false;        
            b= false;
            c=true;  
        }
    }
    public boolean logProduct(){ return a && b && c;}
}

And Main class:

public class Main{
    BooleanTriple bTriple = new BooleanTriple(false,false,false);
    public static void main(String[] args){
        for (int i=0; i<8; i++){
            System.out.println(bTriple.logProduct());
            bTriple.incr();
        }
    }
}

But i think, that it's bad implementation. Can you correct me?

Answer 1

if I quite understand your problem, to do a proper incr() method I would do so :

public class BooleanTriple {
    private boolean a,b,c;
    private int i;
    public BooleanTriple(boolean a, boolean b, boolean c){
        this.a =a; this.b=b; this.c=c;
        this.i = (a ? 1 : 0)*4+(b ? 1 : 0)*2+(c ? 1 : 0)*1;  // not sure if this compile, but the idea is here
    }

    public void incr(){
        i = (i + 1) % 8;
        a = (i & 0x4) != 0;
        b = (i & 0x2) != 0;
        c = (i & 0x1) != 0;
    }
    public boolean logProduct(){ return a && b && c;}
}

With i going from 0 (000 binary) to 7 (111 binary) you have all the possible values for a, b and c

Answer 2

Try this:

public class TribleBool {
    public static void main(String[] args) {
        boolean a, b, c;
        for (int i = 0; i <= 7; i++) {
            int x = i;
            a = (x / 4) > 0;
            x = x % 4;
            b = (x / 2) > 0;
            x = x % 2;
            c = x > 0;

            System.out.println(logProduct(a, b, c));
        }
    }

    public static boolean logProduct(boolean a, boolean b, boolean c) {
        return a && b && c;
    }
}

0 = 0 + 0 + 0

1 = 0 + 0 + 1

2 = 0 + 2 + 0

3 = 0 + 2 + 1

4 = 4 + 0 + 0

5 = 4 + 0 + 1

6 = 4 + 2 + 0

7 = 4 + 2 + 1

Translate ever column into 0 or 1 and you will find required binary combinations