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

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 2

I execute this code for ODD & EVEN:

#include <stdio.h>
int main()
{
    int number;
    printf("Enter an integer: ");
    scanf("%d", &number);

    if(number % 2 == 0)
        printf("%d is even.", number);
    else
        printf("%d is odd.", number);
}

Answer 3

Use bit arithmetic:

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

This is faster than using division or modulus.

Answer 4

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 5

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

done.

Answer 6

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