How do I programmatically return the max of two integers without using any comparison operators and without using if, else, etc?

Question

How do I programmatically return the maximum of two integers without using any comparison operators and without using if, else, etc?

Answer 1

return (a > b ? a : b);

or

int max(int a, int b)
{
        int x = (a - b) >> 31;
        int y = ~x;
        return (y & a) | (x & b); 
}

Answer 2

not as snazzy as the above... but...

int getMax(int a, int b)
{
    for(int i=0; (i<a) || (i<b); i++) { }
    return i;
}

Answer 3

In the math world:

max(a+b) = ( (a+b) + |(a-b)| ) / 2
min(a-b) = ( (a+b) - |(a-b)| ) / 2

Apart from being mathematically correct it is not making assumptions about the bit size as shifting operations need to do.

|x| stands for the absolute value of x.

Comment:

You are right, the absolute value was forgotten. This should be valid for all a, b positive or negative

Answer 4

Since this is a puzzle, solution will be slightly convoluted:

let greater x y = signum (1+signum (x-y))

let max a b = (greater a b)*a + (greater b a)*b

This is Haskell, but it will be the same in any other language. C/C# folks should use "sgn" (or "sign"?) instead of signum.

Note that this will work on ints of arbitrary size and on reals as well.