alternative way to compare if three ints are equal

Question

int a = 10;
int b = 10;
int c = 10;

I am trying to find out if there is alternate way to compare if three ints or anything are equal.

The way i am currently doing is

if( (a == b) && (a == c) && (b == c)){

}

I was wondering if there is an alternate and more concise way to do this.

Forwhy

For many such ints you might use some set implementation and after adding them check number of elements – Deviant

Answer 1

Equality is transitive; you don't need the last comparison b == c. If a == b and a == c, then b == c.

Try

if ((a == b) && (a == c)){

Dric

Answer 2

If all you need is to know whether three are equal then you can use:

if ((a==b) && (b==c)) {

}

Macdermot

Answer 3

alternative variant:

int t1 = a-b;
int t2 = c-b;
if ( (t1|t2) == 0 ) // both are equal

Vicious

This isn't really more compact than the other answers here, and is certainly less clear ;) – Amparoampelopsis

@OliCharlesworth it is not clear, but it is "alternative" :) also, can be a little bit faster (sometimes, at least for c/c++) – Imperturbation

Are you saying that 0/0 equals 0 ? – Macdermot

@PM77-1: No, but 0|0 certainly does. – Amparoampelopsis

@PM77-1 I'm not using division, I'm using binary OR, 0|0 = 0 in deed – Imperturbation

if (0 == ((a - b) | (b - c))) is equivalent, more compact, but even less clear. (Edit because didn't see Java tag) – Nyala

@OliCharlesworth - I guess it's time to put my reading glasses on. – Macdermot

@Nyala for java you need to add == 0 or != 0 as it cannot cast int to bool automatically, for c/c++ it is of course shorter notation, but of course less clear :) – Imperturbation

Answer 4

public static int equal(int a, int b, int c) {
    int count = 0;

    if (a==b && b==c) { 
      count = 3; 
    } else if ((a==b && b!=c) || (a==c && a!=b)) { 
      count = 2; 
    }

    return count;
}

Kal

Please add some explanation to your answer such that others can learn from it – Anklet

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value. Consider reading How to Answer and edit your answer to improve it. – Welter

Answer 5

You can add them into a list, then remove duplicates and compare the size of the actual list minus the unique elements

List elements = Arrays.asList(a,b,c);
List distinct = elements.stream().distinct().toList();

return elements.size() == distinct.size() ? 0 : elements.size() - distinct.size() + 1;

distinct list contains one entry of the duplicate so we have to increment with 1 because we already decreased it.

Or use a HashSet that holds only unique elements

List elements = Arrays.asList(a,b,c);
HashSet<ArrayList> set = new HashSet<>();
set.addAll(elements);

return elements.size() == set.size() ? 0 : elements.size() - set.size() + 1;

Lavern