Which is the correct way to determine if two numbers have the same parity?

Question

I found two solutions to find out if two numbers have the same parity (both are odd numbers or both are even). In C++, they look like this:

if ((a+b)%2 == 0)

and

if (a%2 == b%2)

The problem is that the first one works on 100% of cases and the second one only on 80% of the cases (the tests of a problem that I have submitted on a website) and I do not understand why. For me, both lines of code should work fine on all cases. Can somebody explain me why does the first line of code work in all cases and the second one does not? And which method (those showed by me or another) would you recommend?

I am not allowed to see the tests on this problem and I could not find any cases on paper on which the second statement does not work. — Tudor Gălățan, Nov 22 '17 at 21:28
I think the problem must be in code you haven't posted. Please provide a [mcve]. — Fred Larson, Nov 22 '17 at 21:35
@Yunnosch, I have tried and I can not see anything wrong in both cases. — Tudor Gălățan, Nov 22 '17 at 21:43
@FredLarson, this is the problem that I am talking about: [link](https://www.pbinfo.ro/?pagina=probleme&id=1366) — Tudor Gălățan, Nov 22 '17 at 21:43
Right. And what result do you get when you take the remainder of a negative number? — 1201ProgramAlarm, Nov 22 '17 at 21:44
@FredLarson, if I use your example I do not see anything wrong. `-49943 % 2 = 1` and `8833 % 2 = 1`, so both numbers have the same parity and if I sum up them, I get `-49943 + 8833 = -41110`. `-41110 % 2 = 0`, so, again, they have the same parity. — Tudor Gălățan, Nov 22 '17 at 21:51
Are you computing that, or is C++ computing that? I get `-49943%2 == -1`. — Fred Larson, Nov 22 '17 at 21:53
Nope. It is unspecified in the language if it is negative or positive. On many hardware platforms, including x86, it will be a negative number. — 1201ProgramAlarm, Nov 22 '17 at 21:53
Thanks for the information! It is very useful. But if you sum the numbers in your example, you also get a negative number, so even my first line of code (which worked on all tests) should not work in this case. — Tudor Gălățan, Nov 22 '17 at 22:03
Only if it's odd, and then the sign doesn't matter in the first case. -1 is still not 0. — Fred Larson, Nov 22 '17 at 22:08

Michaël Roy · Accepted Answer · 2017-11-22T22:26:35.450

I would not recommend either of the methods in your post, you should use one of these instead:

if ((a & 1) == (b & 1)) {} // this is clearer

if (((a ^ b) & 1) == 0) {} // this is faster

if (!((a ^ b) & 1)) {}     // this is as fast as faster

These depend on the fact that bit 0 will be set for odd values, even if negative. Avoid integer division (and modulo) whenever possible, it's one of the slowest instructions on any CPU.

Which is the correct way to determine if two numbers have the same parity?

1 Answers1