hashcode implementation on boolean fields

Question

How do I implement a good hashcode if there are two boolean fields? Usually people just add the integer values to their hashcode values. But if I simply add 1 or 0 to my hashcode, I do not think it is good. Because if I have two objects of class A:

obj1.b = true, obj1.c = false.

obj2.b = false, obj2.c = true.

Everyting else is the same. Then the hashcode for these two unequal objects are the same. I know this situation is okay. But imagine if there are 100 boolean fields, then there would be too much collision right? I do not want so many different objects to fall in the same bucket.

What I did below is to assign different numbers to different truth values for each field so objects hashcodes can be very different.

public class A {

    private final String a;
    private final boolean b;
    private final boolean c;

...

@Override public int hashCode(){
            int i,j;
            if(b) {
                    i = 10;
            }
            else {
                    i = 0;
            }
            if(c) {
                    j = 88;
            }
            else {
                    j = 3;
            }
            int result = 0;
            result = 31*result + i + j;
            result = 31*result + (a != null ? a.hashCode() : 0);
            return result;
    }
}

Answer 1

When you have two or even more booleans, the hashcode algorithm is already taken care of that.

Look a little bit closer:

// Very simple example
public int hashCode() {
    int result = 31;

    for(boolean val : booleanList) {
        // This is the important part:
        result = 31 * result + Boolean.hashCode(val);
    }

    return result;
}

Notice the main part of the for loop, in this case, we treat each boolean differently, as we always multiply the result by 31 (or any good prime number) before adding it into the result.

If we visualize the whole hashcode as a number of base 31, so we can understand that the position and the value of each boolean are all taken into account in the final result. Each boolean can be treated as a digit in the hashcode, so for case (true, false) and case (false, true), they will have two different hashcodes.