Logical gates with integer values

I have created this code for both boolean and integer values to display a truth table for an "AND","OR","XOR", "NOT" gate. However I think that my code needs reviewing as it could be simplified.

```
public class LogicalOpTable {
public static void main(String[] args){

boolean p,q;

System.out.println("P\tQ\tAND\tOR\tXOR\tNOT");

p = false;
q = false;
System.out.print(p + "\t" + q + "\t" + (p&&q) + "\t");
System.out.println((p||q)+"\t"+(p^q)+"\t"+(!p));

p = false;
q = true;
System.out.print(p + "\t" + q + "\t" + (p&&q) + "\t");
System.out.println((p||q)+"\t"+(p^q)+"\t"+(!p));

p = true;
q = false;
System.out.print(p + "\t" + q + "\t" + (p&&q) + "\t");
System.out.println((p||q)+"\t"+(p^q)+"\t"+(!p));

p = true;
q = true;
System.out.print(p + "\t" + q + "\t" + (p&&q) + "\t");
System.out.println((p||q)+"\t"+(p^q)+"\t"+(!p));

System.out.println();

withBinary();

}

public static void withBinary(){

System.out.println("A\tB\tAND\tOR\tXOR\tNOT");
int a = 0;
int b = 0;
int and = a&b;
int or = a|b;
int xor = a^b;
int not = a;

if(a==0 && b == 0 )
not = 1;
System.out.println(a + "\t" + b + "\t" + and + "\t" + or + "\t" + xor + "\t" + (not));

b=1;
and = a&b;
or = a|b;
xor = a^b;
not = a;
if(a==0 && b == 1)
System.out.println(a + "\t" + b + "\t" + and + "\t" + or + "\t" + xor + "\t" + (not));

a=1;
b=0;
not = b;
and = a&b;
or = a|b;
xor = a^b;
not = a;
if(a==1 && b == 0)
System.out.println(a + "\t" + b + "\t" + and + "\t" + or + "\t" + xor + "\t" + (not));

a=1;
b=1;
not = 0;
and = a&b;
or = a|b;
xor = a^b;
if(a==1 &&

First part

You want to iterate through all possible combinations of for a pair of booleans, so you can make the code reflect that explicitly and make it simpler:

System.out.println("P\tQ\tAND\tOR\tXOR\tNOT");

    for (boolean p : new boolean[] {true, false}) {
        for (boolean q : new boolean[] {true, false}) {
            System.out.print(p + "\t" + q + "\t" + (p&&q) + "\t");
            System.out.println((p||q)+"\t"+(p^q)+"\t"+(!p));
        }
    }
    System.out.println();

Second part

You have something suspicious in your if statements. As you don't use braces, this code smells of a copy&paste bug:

if(a==0 && b == 0 )
        not = 1;
        System.out.println(a + "\t" + b + "\t" + and + "\t" + or + "\t" + xor + "\t" + (not));

If the condition is met, it will execute the line not = 1;. The second line will be executed independently of the condition. I strongly suggest you use braces even for 1-line blocks, like this:

if (a == 0 && b == 0) {
        not = 1;
        System.out.println(a + "\t" + b + "\t" + and + "\t" + or + "\t" + xor + "\t" + (not));
    }

This way you can avoid this kind of bugs.

Now, applying the same reasoning as with the previous method, you can rewrite it as:

System.out.println("A\tB\tAND\tOR\tXOR\tNOT");
    for (int a : new int[] {0, 1}) {
        for (int b : new int[] {0, 1} ) {
            System.out.println(a + "\t" + b + "\t" + (a & b) + "\t" + (a | b) + "\t" + (a ^ b) + "\t" + ~a);
        }
    }

BUT What do you want to accomplish with the NOT operation? Do you mean the Bitwise Complement? I used that, but maybe you want a function that returns 0 when it's 1, and 1 when it's 0. In that case, you need to replace the ~a with something like (a == 0) ? 1 : 0.

So the whole code could be reduced to just:

public static void printTable() {
    System.out.println("P\tQ\tAND\tOR\tXOR\tNOT");
    for (boolean p : new boolean[]{true, false}) {
        for (boolean q : new boolean[]{true, false}) {
            System.out.print(p + "\t" + q + "\t" + (p&&q) + "\t");
            System.out.println((p||q)+"\t"+(p^q)+"\t"+(!p));
          }
    }
    System.out.println();
    System.out.println("A\tB\tAND\tOR\tXOR\tNOT");
    for (int a : new int[]{ 0, 1}) {
        for (int b : new int[]{0, 1} ) {
            System.out.println(a + "\t" + b + "\t" + (a & b) + "\t" + (a | b) + "\t" + (a ^ b) + "\t" + ~a);
        }
    }
}

StackExchange Code Review Q#99185, answer score: 8