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up vote 3 down vote favorite

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 && b == 1)
            System.out.println(a + "\t" + b + "\t" + and + "\t" + or + "\t" + xor + "\t" + (not));  
        }
}
share|improve this question
    
I don't really get the point of this program. Has it any purpose other than print some debug-values? –  tkausl Aug 6 at 11:35
    
Its just a simple program to print out a truth table for the logic gates, once in boolean and then in integers. It was a question in a book. They already completed the boolean part, and then it says attempt to create a new logic table with the binary equivalent values. –  mp252 Aug 6 at 11:42

1 Answer 1

up vote 8 down vote accepted

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);
        }
    }
}
share|improve this answer
    
You could even wrap your formatting "print" code in named method for enhanced readability. –  Diego Martinoia Aug 6 at 14:05

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