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

Question

How can I check if a given number is even or odd in C?

Answer 1

i % 2 == 0

Answer 2

This is a follow up to the discussion with @RocketRoy regarding his answer, but it might be useful to anyone who wants to compare these results.

tl;dr From what I've seen, Roy's approach ((0xFFFFFFFF == (x | 0xFFFFFFFE)) is not completely optimized to x & 1 as the mod approach, but in practice running times should turn out equal in all cases.

So, first I compared the compiled output using Compiler Explorer:

Functions tested:

int isOdd_mod(unsigned x) {
    return (x % 2);
}

int isOdd_and(unsigned x) {
    return (x & 1);
}

int isOdd_or(unsigned x) {
    return (0xFFFFFFFF == (x | 0xFFFFFFFE));
}   

CLang 3.9.0 with -O3:

isOdd_mod(unsigned int):                          # @isOdd_mod(unsigned int)
        and     edi, 1
        mov     eax, edi
        ret

isOdd_and(unsigned int):                          # @isOdd_and(unsigned int)
        and     edi, 1
        mov     eax, edi
        ret

isOdd_or(unsigned int):                           # @isOdd_or(unsigned int)
        and     edi, 1
        mov     eax, edi
        ret

GCC 6.2 with -O3:

isOdd_mod(unsigned int):
        mov     eax, edi
        and     eax, 1
        ret

isOdd_and(unsigned int):
        mov     eax, edi
        and     eax, 1
        ret

isOdd_or(unsigned int):
        or      edi, -2
        xor     eax, eax
        cmp     edi, -1
        sete    al
        ret

Hats down to CLang, it realized that all three cases are functionally equal. However, Roy's approach isn't optimized in GCC, so YMMV.

It's similar with Visual Studio; inspecting the disassembly Release x64 (VS2015) for these three functions, I could see that the comparison part is equal for "mod" and "and" cases, and slightly larger for the Roy's "or" case:

// x % 2
test bl,1  
je (some address) 

// x & 1
test bl,1  
je (some address) 

// Roy's bitwise or
mov eax,ebx  
or eax,0FFFFFFFEh  
cmp eax,0FFFFFFFFh  
jne (some address)

However, after running an actual benchmark for comparing these three options (plain mod, bitwise or, bitwise and), results were completely equal (again, Visual Studio 2005 x86/x64, Release build, no debugger attached).

Release assembly uses the test instruction for and and mod cases, while Roy's case uses the cmp eax,0FFFFFFFFh approach, but it's heavily unrolled and optimized so there is no difference in practice.

My results after 20 runs (i7 3610QM, Windows 10 power plan set to High Performance):

[Test: Plain mod 2 ] AVERAGE TIME: 689.29 ms (Relative diff.: +0.000%)
[Test: Bitwise or  ] AVERAGE TIME: 689.63 ms (Relative diff.: +0.048%)
[Test: Bitwise and ] AVERAGE TIME: 687.80 ms (Relative diff.: -0.217%)

The difference between these options is less than 0.3%, so it's rather obvious the assembly is equal in all cases.

Here is the code if anyone wants to try, with a caveat that I only tested it on Windows (check the #if LINUX conditional for the get_time definition and implement it if needed, taken from this answer).

#include <stdio.h>

#if LINUX
#include <sys/time.h>
#include <sys/resource.h>
double get_time()
{
    struct timeval t;
    struct timezone tzp;
    gettimeofday(&t, &tzp);
    return t.tv_sec + t.tv_usec*1e-6;
}
#else
#include <windows.h>
double get_time()
{
    LARGE_INTEGER t, f;
    QueryPerformanceCounter(&t);
    QueryPerformanceFrequency(&f);
    return (double)t.QuadPart / (double)f.QuadPart * 1000.0;
}
#endif

#define NUM_ITERATIONS (1000 * 1000 * 1000)

// using a macro to avoid function call overhead
#define Benchmark(accumulator, name, operation) { \
    double startTime = get_time(); \
    double dummySum = 0.0, elapsed; \
    int x; \
    for (x = 0; x < NUM_ITERATIONS; x++) { \
        if (operation) dummySum += x; \
    } \
    elapsed = get_time() - startTime; \
    accumulator += elapsed; \
    if (dummySum > 2000) \
        printf("[Test: %-12s] %0.2f ms\r\n", name, elapsed); \
}

void DumpAverage(char *test, double totalTime, double reference)
{
    printf("[Test: %-12s] AVERAGE TIME: %0.2f ms (Relative diff.: %+6.3f%%)\r\n",
        test, totalTime, (totalTime - reference) / reference * 100.0);
}

int main(void)
{
    int repeats = 20;
    double runningTimes[3] = { 0 };
    int k;

    for (k = 0; k < repeats; k++) {
        printf("Run %d of %d...\r\n", k + 1, repeats);
        Benchmark(runningTimes[0], "Plain mod 2", (x % 2));
        Benchmark(runningTimes[1], "Bitwise or", (0xFFFFFFFF == (x | 0xFFFFFFFE)));
        Benchmark(runningTimes[2], "Bitwise and", (x & 1));
    }

    {
        double reference = runningTimes[0] / repeats;
        printf("\r\n");
        DumpAverage("Plain mod 2", runningTimes[0] / repeats, reference);
        DumpAverage("Bitwise or", runningTimes[1] / repeats, reference);
        DumpAverage("Bitwise and", runningTimes[2] / repeats, reference);
    }

    getchar();

    return 0;
}

Answer 3

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.

Answer 4

Checking even or odd is a simple task.

We know that any number exactly divisible by 2 is even number else odd.

We just need to check divisibility of any number and for checking divisibility we use % operator

Checking even odd using if else

if(num%2 ==0)  
{
    printf("Even");
}
else
{
    printf("Odd");
}

C program to check even or odd using if else

Using Conditional/Ternary operator

(num%2 ==0) printf("Even") : printf("Odd");

C program to check even or odd using conditional operator.

Using Bitwise operator

if(num & 1)  
{
    printf("Odd");
}
else 
{
    printf("Even");
}

Answer 5

A number is even if, when divided by two, the remainder is 0. A number is odd if, when divided by 2, the remainder is 1.

// Java
public static boolean isOdd(int num){
    return num % 2 != 0;
}

/* C */
int isOdd(int num){
    return num % 2;
}

Methods are great!

Answer 6

One more solution to the problem
(children are welcome to vote)

bool isEven(unsigned int x)
{
  unsigned int half1 = 0, half2 = 0;
  while (x)
  {
     if (x) { half1++; x--; }
     if (x) { half2++; x--; }

  }
  return half1 == half2;
}