How do I check if an int is even

Question

This is likely an easy solution that is simply eluding me. Specifically, I am creating locations on a canvas dynamically using a sin() function for equidistant points on a circle. Once these points are created, I am animating a shape moving from one point to the next by calculating the slope between points and redrawing the shape at each slope step.

Problem is, depending on the coordinate values, the slope step may just be one step from point a to point b. I need the shape to move along the path, not just jump point to point.

What I want to do is force the location coordinates (x, y) to be even numbers allowing for slope values to always be reducible. So, the simple part of question is...

How do I check if an int value is even? If it is not, I will simply add 1 to the coordinate value.

Answer 1

int newNumber = someInt % 2 == 0 ? someInt : someInt + 1;

Answer 2

Do a mod 2 on it. If the remainder is 0, it's an even number.

VB:

Dim even = (3 mod 2 = 0 )

Answer 3

It's hard to give you specifics since you haven't given many, but you should look into the Modulo operation.

Answer 4

To see if an integer is even:

1. Check if its value is congruent to 0 modulo 2. That is value MOD 2 == 0. In C-style languages this is usually expressed as value % 2 == 0.
2. Alternatively, check the value of bit 0. That is value BITWISE-AND 0x01 == 0. In C-style languages this is usually expressed as (value & 0x01) == 0.

If you do not care which direction you round, you can even-ize an integer in a single operation by taking its value bitwise-and a mask of 0xFFFE (of course padded to the width of your integer), which will force set the 0 bit to zero. That is value := value BITWISE-AND 0xFFFE, or in C-style languages value &= 0xFFFE.

Answer 5

For integers, while using modulo will give you the correct answer, it requires division. Division isn't as fast as bitwise operations. For what you need, a bitwise AND is sufficient.

if(x & 0x1)
{
   sdt::cout << "x is odd" << std::endl;
}
else
{
   std::cout << "x is even" << std::endl;
}

The key is that all positive powers of two are even. Therefore the only way a binary representation of an integer can be odd is if the first bit is set.