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

Question

How best can I check if an integer is even or odd in C? I considered how I'd do this in Java, but I couldn't come up with an answer either. Thanks.

Answer 1

I'd say just divide it by 2 and if there is a 0 remainder, it's even, otherwise it's odd.

Using the modulus (%) makes this easy.

eg. 4 % 2 = 0 therefore 4 is even 5 % 2 = 1 therefore 5 is odd

Answer 2

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 3

i % 2 == 0

Answer 4

Use bit arithmetic:

if((x & 1) == 0)
    printf("EVEN!\n");
else
    printf("ODD!\n");

This is faster than using division or modulus.

Answer 5

// C#
bool isEven = ((i % 2) == 0);

Answer 6

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 7

The bitwise method depends on the inner representation of the integer. Modulo will work anywhere there is a modulo operator. For example, some systems actually use the low level bits for tagging (like dynamic languages), so the raw x & 1 won't actually work in that case.

Answer 8

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...

Answer 9

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;
}

Answer 10

In response to ffpf - I had exactly the same argument with a colleague years ago, and the answer is no it doesn't work with negative numbers.

The C standard stipulates that negative numbers can be represented in 3 ways: - 2's compliment - 1's compliment - "sign and magnitude".

Checking like this:

  isEven = (x & 1);

will work for 2's compliment and sign and magnitude formats, but not for 1's compliment.

However, I believe that the following will work for all cases:

  isEven = (x & 1) ^ ((-1 & 1) | ((x < 0) ? 0 : 1)));

edit: thanks to ffpf for pointing out that the text box was eating everything after my lessthan character!

Answer 11

A nice one is:

bool isEven(unsigned int n)
{
  if (n == 0) 
    return true ;  // I know 0 is even
  else if (n == 1)
    return isOdd(n-1) ; // n is even if n-1 is odd
}

bool isOdd(unsigned int n)
{
  if (n == 0)
    return false ;
  else
    return isEven(n-1) ; // n is odd if n-1 is even
}

Note that this method use tail recursion involving two functions. It can be implemented efficiently (turned into a while/until kind of loop) if your compiler supports tail recursion like a Scheme compiler. In this case the stack should not overflow !

Answer 12

I know this is just syntactic sugar and only applicable in .net but what about extension method...

public static class RudiGroblerExtensions
{
    public static bool IsOdd(this int i)
    {
        return ((i % 2) != 0);
    }
}

Now you can do the following

int i = 5;
if (i.IsOdd())
{
    // Do something...
}

Answer 13

[Joke mode="on"]

public enum Evenness
{
  Unknown = 0,
  Even = 1,
  Odd = 2
}

public static Evenness AnalyzeEvenness(object o)
{

  if (o == null)
    return Evenness.Unknown;

  string foo = o.ToString();

  if (String.IsNullOrEmpty(foo))
    return Evenness.Unknown;

  char bar = foo[foo.Length - 1];

  switch (bar)
  {
     case '0':
     case '2':
     case '4':
     case '6':
     case '8':
       return Evenness.Even;
     case '1':
     case '3':
     case '5':
     case '7':
     case '9':
       return Evenness.Odd;
     default:
       return Evenness.Unknown;
  }
}

[Joke mode="off"]

EDIT: Added confusing values to the enum.

Answer 14

IsOdd(int x) { return true; }

Proof of correctness - consider the set of all positive integers and suppose there is a non-empty set of integers that are not odd. Because positive integers are well-ordered, there will be a smallest not odd number, which in itself is pretty odd, so clearly that number can't be in the set. Therefore this set cannot be non-empty. Repeat for negative integers except look for the greatest not odd number.

Answer 15

Portable:

i % 2 ? odd : even;

Unportable:

i & 1 ? odd : even;

i << (BITS_PER_INT - 1) ? odd : even;

Answer 16

    MOV EAX, VALUE_TO_TEST
    AND EAX, 01
    JNZ VALUE_ODD
VALUE_EVEN:
    // Even code goes here
    JMP DONE:
VALUE_ODD:
    // Odd code goes here
DONE:

Answer 17

For the sake of discussion...

You only need to look at the last digit in any given number to see if it is even or odd. Signed, unsigned, positive, negative - they are all the same with regards to this. So this should work all round: -

void tellMeIfItIsAnOddNumberPlease(int iToTest){
  int iLastDigit;
  iLastDigit = iToTest - (iToTest / 10 * 10);
  if (iLastDigit % 2 == 0){
    printf("The number %d is even!\n", iToTest);
  } else {
    printf("The number %d is odd!\n", iToTest);
  }
}

The key here is in the third line of code, the division operator performs an integer division, so that result are missing the fraction part of the result. So for example 222 / 10 will give 22 as a result. Then multiply it again with 10 and you have 220. Subtract that from the original 222 and you end up with 2, which by magic is the same number as the last digit in the original number. ;-) The parenthesis are there to remind us of the order the calculation is done in. First do the division and the multiplication, then subtract the result from the original number. We could leave them out, since the priority is higher for division and multiplication than of subtraction, but this gives us "more readable" code.

We could make it all completely unreadable if we wanted to. It would make no difference whatsoever for a modern compiler: -

printf("%d%s\n",iToTest,0==(iToTest-iToTest/10*10)%2?" is even":" is odd");

But it would make the code way harder to maintain in the future. Just imagine that you would like to change the text for odd numbers to "is not even". Then someone else later on want to find out what changes you made and perform a svn diff or similar...

If you are not worried about portability but more about speed, you could have a look at the least significant bit. If that bit is set to 1 it is an odd number, if it is 0 it's an even number. On a little endian system, like Intel's x86 architecture it would be something like this: -

if (iToTest & 1) {
  // Even
} else {
  // Odd
}

Answer 18

If you want to be efficient, use bitwise operators (x & 1), but if you want to be readable use modulo 2 (x % 2)

Answer 19

In the "creative but confusing category" I offer:

int isOdd(int n) { return n ^ n * n ? isOdd(n * n) : n; }

A variant on this theme that is specific to Microsoft C++:

__declspec(naked) bool __fastcall isOdd(const int x)
{
 __asm
 {
  mov eax,ecx
  mul eax
  mul eax
  mul eax
  mul eax
  mul eax
  mul eax
  ret
 }
}

Answer 20

int isOdd(int i){
  return(i % 2);
}

done.

Answer 21

in c language

main()

{

int n;

clrscr();

printf("enter integer value: ");

scanf("%d",&n);

if(n%2= =0)

printf("given value is even");

if(n%2!=0)

printf("given value is odd");

}

Answer 22

I would build a table of the parities (0 if even 1 if odd) of the integers (so one could do a lookup :D), but gcc won't let me make arrays of such sizes:

typedef unsigned int uint;

char parity_uint [UINT_MAX];
char parity_sint_shifted [((uint) INT_MAX) + ((uint) abs (INT_MIN))];
char* parity_sint = parity_sint_shifted - INT_MIN;

void build_parity_tables () {
    char parity = 0;
    unsigned int ui;
    for (ui = 1; ui <= UINT_MAX; ++ui) {
        parity_uint [ui - 1] = parity;
        parity = !parity;
    }
    parity = 0;
    int si;
    for (si = 1; si <= INT_MAX; ++si) {
        parity_sint [si - 1] = parity;
        parity = !parity;
    }
    parity = 1;
    for (si = -1; si >= INT_MIN; --si) {
        parity_sint [si] = parity;
        parity = !parity;
    }
}

char uparity (unsigned int n) {
    if (n == 0) {
        return 0;
    }
    return parity_uint [n - 1];
}

char sparity (int n) {
    if (n == 0) {
        return 0;
    }
    if (n < 0) {
        ++n;
    }
    return parity_sint [n - 1];
}

So let's instead resort to the mathematical definition of even and odd instead.

An integer n is even if there exists an integer k such that n = 2k.

An integer n is odd if there exists an integer k such that n = 2k + 1.

Here's the code for it:

char even (int n) {
    int k;
    for (k = INT_MIN; k <= INT_MAX; ++k) {
        if (n == 2 * k) {
            return 1;
        }
    }
    return 0;
}

char odd (int n) {
    int k;
    for (k = INT_MIN; k <= INT_MAX; ++k) {
        if (n == 2 * k + 1) {
            return 1;
        }
    }
    return 0;
}

Let C-integers denote the possible values of int in a given C compilation. (Note that C-integers is a subset of the integers.)

Now one might worry that for a given n in C-integers that the corresponding integer k might not exist within C-integers. But with a little proof it is can be shown that for all integers n, |n| <= |2n| (*), where |n| is "n if n is positive and -n otherwise". In other words, for all n in integers at least one of the following holds (exactly either cases (1 and 2) or cases (3 and 4) in fact but I won't prove it here):

Case 1: n <= 2n.

Case 2: -n <= -2n.

Case 3: -n <= 2n.

Case 4: n <= -2n.

Now take 2k = n. (Such a k does exist if n is even, but I won't prove it here. If n is not even then the loop in even fails to return early anyway, so it doesn't matter.) But this implies k < n if n not 0 by (*) and the fact (again not proven here) that for all m, z in integers 2m = z implies z not equal to m given m is not 0. In the case n is 0, 2*0 = 0 so 0 is even we are done (if n = 0 then 0 is in C-integers because n is in C-integer in the function even, hence k = 0 is in C-integers). Thus such a k in C-integers exists for n in C-integers if n is even.

A similar argument shows that if n is odd, there exists a k in C-integers such that n = 2k + 1.

Hence the functions even and odd presented here will work properly for all C-integers.

Answer 23

// to determine the biggest of three numbers entered by the user.

include

using namespace std; int main()

{ int num; cout <<"\t enter the number \t";
cin>>num; if(num%==0); cout<<" the number is even ";

else
cout<< "\t the number is odd " ;
return 0;

}