How do I find if a number is exactly divisable by 2?

**Once-ler** · 01-02-2012

I am a little frustrated by this I am sure that there is an easy way to do this?

**whiteflags** · 01-02-2012

Code:

double number = whatever / 2.0;
double wholepart;
double fracpart;
wholepart = modf(number, &fracpart);
if (fracpart == 0.0)
   divisible
else
   not

int number = whatever % 2;
if (number == 0)
   divisible
else
   not

**~~CommonTater~~** · 01-02-2012

Code:

int number;

number = 42;

if ((number & 1) == 0)
  puts("Divisible by two");

All integers are stored as binary values... bit 0, is always a 1 for odd numbers, only even numbers are divisible by 2.
So... if bit0 = 0 the number is divisible by 2.

**quzah** · 01-02-2012

Nice. No effort at all on the OP's part, and he has two options already. Guess I may as well round this out:

Code:

r = num / 2;
if( r * 2 == num )
    Yay!
else
    Nay!

Quzah.

**~~CommonTater~~** · 01-02-2012

Ok... how about...

Code:

if ( ! (num - ((num >> 1) << 1)))
  puts("Yup");
else
  puts("Naaaa");

**DeadPlanet** · 01-02-2012

Code:

#include <limits.h>
#include <math.h>

...

int power;

power = log( (double) UINT_MAX + 1 ) / log( 2.0f ) - 1;

if( ( num << (power - 1) ) >> (power - 1)  )
        //Odd
else
        //Even

I don't work with floating point a lot, but I think this would work?

**dmh2000** · 01-02-2012

or

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(int argc,char *argv[])
{
        double x;
        int a;
        int b;

        x = 5.0;
        x = fmod(x,2);
        printf("%d\n",x == 0.0);
        x = 4.0;
        x = fmod(x,2);
        printf("%d\n",x == 0.0);
        x = 4.0000001;
        x = fmod(x,2);
        printf("%d\n",x == 0.0);

        a = 5;
        b = a % 2;
        printf("%d\n",b==0);
        a = 4;
        b = a % 2;
        printf("%d\n",b==0);

        return 0;
}

**Babkockdood** · 01-02-2012

Whoever writes the fanciest program gets a cookie.

**~~CommonTater~~** · 01-02-2012

I like cookies...

Code:

puts(num - ((num >> 1)<<1) ? "No" : "Yes");

**Babkockdood** · 01-02-2012

CommonTater is in the lead.

**iMalc** · 01-02-2012

Code:

const char* answer[] = {"Yes", "No");
puts(answer[num&1]);

Oh wait wait...

Code:

puts("Yes\0No"+((num&1)<<2));

That's better.

Edit: Aaargh, got it this time.

**~~CommonTater~~** · 01-02-2012

Originally Posted by iMalc

Code:

puts("Yes\0No"+((num&1)<<2));

I'll give you the lead after that one...

(Darn, I really wanted that cookie!)

**gardhr** · 01-02-2012

Originally Posted by CommonTater

All integers are stored as binary values...

Except when they aren't, of course.

Code:

#include<stdio.h>
int divisible_by_two(long int data)
{
	return !(data % 2L);
}
int main(int argc, char** argv)
{
	char* errc;
	while(*(++argv))
		printf("%s is%s divisible by two\n", *argv, "\0 not" + !(divisible_by_two(strtol(*argv, &errc, 0)) - !!*errc));
	return 0;
}

I win.

**Libpgeak** · 01-03-2012

Is there a possibility to swap the integer, shift it left, and watch the carry flag? That's the most fancy way i can imagine...
EDIT: Why don't use the parity flag..

**~~CommonTater~~** · 01-03-2012

Originally Posted by gardhr

Except when they aren't, of course.

Riiiiiight. there's one of them in everyone's livingroom.

I win.

Not up to you... this was Babcock's idea...
He gets to judge.

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