Explain this snippet which finds the maximum of two integers without using if-else or any other comparison operator?

int getMax(int a, int b) {
    int c = a - b;
    int k = (c >> 31) & 0x1;
    int max = a - k * c;
    return max;
}

Let's dissect this. This first line appears to be straightforward - it stores the difference of a and b. This value is negative if a < b and is nonnegative otherwise. But there's actually a bug here - if the difference of the numbers a and b is so big that it can't fit into an integer, this will lead to undefined behavior - oops! So let's assume that doesn't happen here.

In the next line, which is

int k = (c >> 31) & 0x1;

the idea is to check if the value of c is negative. In virtually all modern computers, numbers are stored in a format called two's complement in which the highest bit of the number is 0 if the number is positive and 1 if the number is negative. Moreover, most ints are 32 bits. (c >> 31) shifts the number down 31 bits, leaving the highest bit of the number in the spot for the lowest bit. The next step of taking this number and ANDing it with 1 (whose binary representation is 0 everywhere except the last bit) erases all the higher bits and just gives you the lowest bit. Since the lowest bit of c >> 31 is the highest bit of c, this reads the highest bit of c as either 0 or 1. Since the highest bit is 1 iff c is 1, this is a way of checking whether c is negative (1) or positive (0). Combining this reasoning with the above, k is 1 if a < b and is 0 otherwise.

The final step is to do this:

int max = a - k * c;

If a < b, then k == 1 and k * c = c = a - b, and so

a - k * c = a - (a - b) = a - a + b = b

Which is the correct max, since a < b. Otherwise, if a >= b, then k == 0 and

a - k * c = a - 0 = a

Which is also the correct max.

Here we go: (a + b) / 2 + |a - b| / 2

Use bitwise hacks

r = x ^ ((x ^ y) & -(x < y)); // max(x, y)

If you know that INT_MIN <= x - y <= INT_MAX, then you can use the following, which is faster because (x - y) only needs to be evaluated once.

r = x - ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1))); // max(x, y)

Source : Bit Twiddling Hacks by Sean Eron Anderson

(sqrt( a*a + b*b - 2*a*b ) + a + b) / 2

This is based on the same technique as mike.dld's solution, but it is less "obvious" here what I am doing. An "abs" operation looks like you are comparing the sign of something but I here am taking advantage of the fact that sqrt() will always return you the positive square root so I am squaring (a-b) writing it out in full then square-rooting it again, adding a+b and dividing by 2.

You will see it always works: eg the user's example of 10 and 5 you get sqrt(100 + 25 - 100) = 5 then add 10 and 5 gives you 20 and divide by 2 gives you 10.

If we use 9 and 11 as our numbers we would get (sqrt(121 + 81 - 198) + 11 + 9)/2 = (sqrt(4) + 20) / 2 = 22/2 = 11

The simplest answer is below.

#include <math.h>

int Max(int x, int y)
{
    return (float)(x + y) / 2.0 + abs((float)(x - y) / 2);
}

int Min(int x, int y)
{
    return (float)(x + y) / 2.0 - abs((float)(x - y) / 2);
}