Question

How to get the minimum (min) of two integers with bitwise operators in C

Answer 1

#include <stdio.h>
  
int main() {
    int x = 5, y = 9;   
    printf("(x ^ y) & -(x < y) = (%i) & -(%i) = %i\n", x ^ y, x < y, (x ^ y) & -(x < y));    
    int min = y ^ ((x ^ y) & -(x < y)); 
    printf("min = %i\n", min);
    
    x = 7, y = 2;   
    printf("(x ^ y) & -(x < y) = (%i) & -(%i) = %i\n", x ^ y, x < y, (x ^ y) & -(x < y));    
    min = y ^ ((x ^ y) & -(x < y)); 
    printf("min = %i\n", min);
    
    x = -5, y = 3;   
    printf("(x ^ y) & -(x < y) = (%i) & -(%i) = %i\n", x ^ y, x < y, (x ^ y) & -(x < y));    
    min = y ^ ((x ^ y) & -(x < y)); 
    printf("min = %i\n", min);
    
    x = 3, y = -1;   
    printf("(x ^ y) & -(x < y) = (%i) & -(%i) = %i\n", x ^ y, x < y, (x ^ y) & -(x < y));    
    min = y ^ ((x ^ y) & -(x < y)); 
    printf("min = %i\n", min);

}
  
  
  
/*
run:
  
(x ^ y) & -(x < y) = (12) & -(1) = 12
min = 5
(x ^ y) & -(x < y) = (5) & -(0) = 0
min = 2
(x ^ y) & -(x < y) = (-8) & -(1) = -8
min = -5
(x ^ y) & -(x < y) = (-4) & -(0) = 0
min = -1
 
*/

 

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