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.
/*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(); }
printf("%d",(14&16)|14); for minimum
ReplyDeleteprintf("%d",(2&3)|3); for maximum
ReplyDeleteYashesh.. Bhai u wrote both r same.. but make function 4 both min(a,b) max(a,b).. and write just 1 line code in it..
ReplyDeleteMethod 1(Use XOR and comparison operator)
ReplyDeleteMinimum 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();
}