Would you use num%2 or num&1 to check if a number is even?

Question

Well, there are at least two low-level ways of determining whether a given number is even or not:

 1. if (num%2 == 0) { /* even */ } 
 2. if ((num&1) ==          


        

Answer 1

I code for readability first so my choice here is num % 2 == 0. This is far more clear than num & 1 == 0. I'll let the compiler worry about optimizing for me and only adjust if profiling shows this to be a bottleneck. Anything else is premature.

I consider the second option to be far more elegant and meaningful

I strongly disagree with this. A number is even because its congruency modulo two is zero, not because its binary representation ends with a certain bit. Binary representations are an implementation detail. Relying on implementation details is generally a code smell. As others have pointed out, testing the LSB fails on machines that use ones' complement representations.

Another point is that the second one might be a little harder to comprehend for less experienced programmers. On that I'd answer that it will probably only benefit everybody if these programmers take that short time to understand statements of this kind.

I disagree. We should all be coding to make our intent clearer. If we are testing for evenness the code should express that (and a comment should be unnecessary). Again, testing congruency modulo two more clearly expresses the intent of the code than checking the LSB.

And, more importantly, the details should be hidden away in an isEven method. So we should see if(isEven(someNumber)) { // details } and only see num % 2 == 0 once in the definition of isEven.