This is in c++ How would I see if a certain number is odd or even? Would it be something like this (btw this doesnt work for me)?

#include <iostream.h>

int main()
{
int number;
double newnumber;
cout<<"Enter a number";
cin>>number;
newnumber=(double((number*(1/2))%2)); /*a even number should have a remainder of 0 after division*/
if(newnumber==0)
{
cout<<"The number is even";
}
else
{
cout<<"The number is odd";
}
return 0;
}
 

Now the problem is that even if I put in a odd number like 55, the program says ''even'', not odd like it should.

I think that works

edit:
yup, it works.

Now to explain it... every number is stored in binary, of course. Lets look at a table of all the numbers 0 through 10, in binary.



Notice something about the last bit in each of those numbers? The last bit on all even numbers (0, 2, 4, 6, 8, 10) is a 0. The last bit on all odd numbers (1, 3, 5, 7, 9) is 1.

So it can be theorized that if the least-significant bit in all integers is a 0, the number is even. If it is 1, it is odd. Therefor, to see if a number is even, we just need to look at the last bit.

How do we do that? We use the C/C++ bitwise and operator (&). This operator compares all the bits of two given values, and returns a third. For each bit in the original two numbers, the corresponding bit in the third number is a 1 if and ONLY IF the two original bits are both 1. A few examples:

5 & 7:

0000 0101 &
0000 0110
---------
0000 0100 = 4

36 & 2:
0010 0100 &
0000 0010
---------
0000 0000 = 0


Now, take the number 1, in binary.

0000 0001

The ONLY binary digit that has a 1 in the binary number 1 is the last one, or the least significant. So, if we & a number with 1, and that last digit in the answer comes out to be 1, then the number is odd!

example:

13 & 1:

0000 1101 &
0000 0001
---------
0000 0001

We got a result with the last digit being 1! So the number 13 is odd. Lets try an even number:

12 & 1:
0000 1100 &
0000 0001
---------
0000 0000

We got 0, or false! So the number 12 is Even!

[edited by - Ronin Magus on March 8, 2003 7:26:36 PM]

[edited by - Ronin Magus on March 8, 2003 7:36:31 PM]

[edited by - Ronin Magus on March 8, 2003 7:58:26 PM]

this way seems to work for me.
As far as i know, every even number is divisible by two.
if(number%2==0)
... number is even
else
... odd

[edited by - ph4ntasyc0d3 on March 8, 2003 7:13:34 PM]

As you said, an even number has a remainder of 0 when divided by 2. Hence,

if(n%2 == 0) /* it''s even */else /* it''s odd */ 

I tried the following and it worked as well - pick your solution

#include <stdio.h>

int main(int argc, char *argv[])
{
int a = 2;
int b = 55;

if (a%2 == 0)
printf("even\n");
else
printf("odd\n");

if (b%2 == 0)
printf("even\n");
else
printf("odd\n");

return 0;
}

For a I got even and for b I got odd.

I think you over complicated the problem a bit. After a minute, this is what I came up with:

#include <iostream>int main(){	long number;	std::cout << "Enter a number:\t";	std::cin >> number;	if(number%2 == 0)	{		std::cout << "It''s even\n";	}	else	{		std::cout << "It''s odd\n";	}	return 0;} 

It just takes the number you inputted and finds the remainder from dividing by 2. Since even numbers don''t have a remainder if you divide by 2, it says it''s even. Otherwise, if there is a remainder, it''s odd.

Oh I got it to work, thanks. I think I did overcomplicate the problem. Thanks guys

Oh I got it to work, thanks. I think I did overcomplicate the problem. Thanks guys

All those solutions work.. but remember that a bitwise & doesn''t require a divide, which can be much faster than %.

Wow, I like that little binary "trick". That was well written and easy to understand, even for me. I was going to suggest using the modulus operator as others had said, but the binary comparison is very interesting idea. I''ll have to try that too.

Peon