Compute the minimum or maximum of two integers without 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

/*Function to find minimum of x and y*/
int min(int x, int y)
{
return y ^ ((x ^ y) & -(x < y));
}

/*Function to find maximum of x and y*/
int max(int x, int y)
{
return x ^ ((x ^ y) & -(x < y));
}

/* Driver program to test above functions */
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
#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)));
}

/* Driver program to test above functions */
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();
}

Source http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax