How do I check if an integer is even or odd?

You guys are waaaaaaaay too efficient. What you really want is:

public boolean isOdd(int num) {
  int i = 0;
  boolean odd = false;

  while (i != num) {
    odd = !odd;
    i = i + 1;
  }

  return odd;
}

Repeat for isEven.

Of course, that doesn't work for negative numbers. But with brilliance comes sacrifice...

Use the modulo (%) operator to check if there's a remainder when dividing by 2:

if (x % 2) { /* x is odd */ }

A few people have criticized my answer above stating that using x & 1 is "faster" or "more efficient". I do not believe this to be the case.

Out of curiosity, I created two trivial test case programs:

/* modulo.c */
#include <stdio.h>

int main(void)
{
    int x;
    for (x = 0; x < 10; x++)
        if (x % 2)
            printf("%d is odd\n", x);
    return 0;
}

/* and.c */
#include <stdio.h>

int main(void)
{
    int x;
    for (x = 0; x < 10; x++)
        if (x & 1)
            printf("%d is odd\n", x);
    return 0;
}

I then compiled these with gcc 4.1.3 on one of my machines 5 different times:

• With no optimization flags.
• With -O
• With -Os
• With -O2
• With -O3

I examined the assembly output of each compile (using gcc -S) and found that in each case, the output for and.c and modulo.c were identical (they both used the andl $1, %eax instruction). I doubt this is a "new" feature, and I suspect it dates back to ancient versions. I also doubt any modern (made in the past 20 years) non-arcane compiler, commercial or open source, lacks such optimization. I would test on other compilers, but I don't have any available at the moment.

If anyone else would care to test other compilers and/or platform targets, and gets a different result, I'd be very interested to know.

Finally, the modulo version is guaranteed by the standard to work whether the integer is positive, negative or zero, regardless of the implementation's representation of signed integers. The bitwise-and version is not. Yes, I realise two's complement is somewhat ubiquitous, so this is not really an issue.