Smallest of 2/3 integers with & without using comparison operators

Write a C program to find the smallest of 2 integers with & without using any of the comparison operators.Also use this to calculate maximum of 3 integers.

On some rare machines where branching is expensive, the below obvious approach to find minimum can be slow as it uses branching.
/* The obvious approach to find minimum (involves branching) */
int min(int x, int y)
{
  return (x < y) ? x : y
}

Below are the methods to get minimum(or maximum) without using branching. Typically, the obvious approach is best, though.

Method 1(Use XOR and comparison operator)

Minimum of x and y will be

y ^ ((x ^ y) & -(x < y))

It works because if x < y, then -(x < y) will be all ones, so r = y ^ (x ^ y) & ~0 = y ^ x ^ y = x. Otherwise, if x >= y, then -(x < y) will be all zeros, so r = y ^ ((x ^ y) & 0) = y. On some machines, evaluating (x < y) as 0 or 1 requires a branch instruction, so there may be no advantage.

To find the maximum, use

x ^ ((x ^ y) & -(x < y));

#include<stdio.h>

int min(int x, int y)
{
  return y ^ ((x ^ y) & -(x < y));
}

int max(int x, int y)
{
  return x ^ ((x ^ y) & -(x < y));
}

int main()
{
  int x = 15;
  int y = 6;
  printf("Minimum of %d and %d is ", x, y);
  printf("%d", min(x, y));
  printf("\nMaximum of %d and %d is ", x, y);
  printf("%d", max(x, y));
  getchar();
}

Method 2(Use subtraction and shift)
If we know that

INT_MIN <= (x - y) <= INT_MAX

, then we can use the following, which are faster because (x - y) only needs to be evaluated once.

Minimum of x and y will be

y + ((x - y) & ((x - y) >>(sizeof(int) * CHAR_BIT - 1)))

This method shifts the subtraction of x and y by 31 (if size of integer is 32). If (x-y) is smaller than 0, then (x -y)>>31 will be 1. If (x-y) is greater than or equal to 0, then (x -y)>>31 will be 0.
So if x >= y, we get minimum as y + (x-y)&0 which is y.
If x < y, we get minimum as y + (x-y)&1 which is x.

Similarly, to find the maximum use

x - ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1)))

#include<stdio.h>
#define CHAR_BIT 8

/*Function to find minimum of x and y*/
int min(int x, int y)
{
  return  y + ((x - y) & ((x - y) >>
            (sizeof(int) * CHAR_BIT - 1)));
}

/*Function to find maximum of x and y*/
int max(int x, int y)
{
  return x - ((x - y) & ((x - y) >>
            (sizeof(int) * CHAR_BIT - 1)));
}

int main()
{
  int x = 15;
  int y = 6;
  printf("Minimum of %d and %d is ", x, y);
  printf("%d", min(x, y));
  printf("\nMaximum of %d and %d is ", x, y);
  printf("%d", max(x, y));
  getchar();
}

Note that the 1989 ANSI C specification doesn't specify the result of signed right-shift, so above method is not portable. If exceptions are thrown on overflows, then the values of x and y should be unsigned or cast to unsigned for the subtractions to avoid unnecessarily throwing an exception, however the right-shift needs a signed operand to produce all one bits when negative, so cast to signed there. 

Maximum of 3 integers:

Let 3 input numbers be x, y and z.

Method 1 (Repeated Subtraction)
Take a counter variable c and initialize it with 0. In a loop, repeatedly subtract x, y and z by 1 and increment c. The number which becomes 0 first is the smallest. After the loop terminates, c will hold the minimum of 3.
#include<stdio.h>

int smallest(int x, int y, int z)
{
  int c = 0;
  while ( x && y && z )
  {
      x--;  y--; z--; c++;
  }
  return c;
}

int main()
{
   int x = 12, y = 15, z = 5;
   printf("Minimum of 3 numbers is %d", smallest(x, y, z));
   return 0;
}

This methid doesn’t work for negative numbers. Method 2 works for negative nnumbers also.

Method 2 (Use Bit Operations)
Once we have functionality to find minimum of 2 numbers, we can use this to find minimum of 3 numbers.

#include<stdio.h>
#define CHAR_BIT 8

int min(int x, int y)
{
  return  y + ((x - y) & ((x - y) >>
            (sizeof(int) * CHAR_BIT - 1)));
}

int smallest(int x, int y, int z)
{
    return min(x, min(y, z));
}

int main()
{
   int x = 12, y = 15, z = 5;
   printf("Minimum of 3 numbers is %d", smallest(x, y, z));
   return 0;
}