Finding whether a signed and an unsigned integer are both even or both odd

Question

I have an int m and an unsigned int j and want to determine whether they are both even or both odd.

In the past I've been using

if((int(j)+m)%2) 

to catch the case that only one is odd. But I'm concerned about the casting to int incorrectly changing the odd-even-ness of j.

Do either of these run into problems?

if(!(j%2)!=!(m%2)) if(bool(j%2)!=bool(j%2)) 

I know that

if(j%2!=m%2) 

doesn't work because 'm%2' will produce -1 when m is negative, which will always evaluate to true no matter what the value of j%2 is.

Answer 1

Don't use %. This is a problem that calls for bitmasks:

bool same_parity = (i & 0x1) == (j & 0x1); 

This works regardless of the sign of i, since the result of that expression will always be 0 or 1.

Answer 2

if (1 & (i ^ j)) { // Getting here if i is even and j is odd // or if i is odd and j is even } 

^ is the exclusive-or bitwise operator, which checks each bit in both numbers if they have the same value. For example, if the binary representation of i is 0101 and j is 1100, then i ^ j will evaluate to 1001, since their first and last bits are different, while the middle bits are the same.

& is the and bitwise operator, which checks each bit in both numbers if they are both 1.

Since just the last bit of each number determines if it's even or odd, i ^ j will evaluate to ...xxx0 if they are both even or odd, and ...xxx1 otherwise (the xs don't matter, we aren't looking at them anyway). Since 1 is actually ...0001, 1 & (i ^ j) evaluates to 0 if i and j are both even or odd, and 1 otherwise.

This works on any combination of unsigned numbers, 2s-complement, and sign-and-magnitude, but not the rare 1s-complement if exactly one is negative.