To find a>b or a<b without using of Logical & relational Operators!!!

Discussion in 'C' started by asadullah.ansari, Jan 12, 2008.

  1. asadullah.ansari

    asadullah.ansari TechCake

    Joined:
    Jan 9, 2008
    Messages:
    356
    Likes Received:
    14
    Trophy Points:
    0
    Occupation:
    Developer
    Location:
    NOIDA
    Program to find max(x,y) or Min(x,y) without using any relational and logical operator.

    Code:
    int maximum(int x, int y)
     {   
       int fNum[2],temp,num[2]; 
       num[0] = x;  
       num[1] = y;    
       temp=fNum[ Fun4Sign(x - y)];   
       return( temp); 
    }
    
    int Fun4Sign(int sgn) 
    {   
         int temp;   
         temp= unsigned(sgn) >>  (sizeof(int)*8-1);    return(temp); 
    }
    EXTRA DOSE: One More thing to find min and max...
    Code:
    int  MinimumOf( int a, int b)  
    {
        return(   b + ((a - b) & ((a - b) >> (sizeof(int) * CHAR_BIT - 1)))  );
    }
    Code:
    int  MaximumOf( int a, int b)  
    {
        return(   a - ((a - b) & ((a - b) >> (sizeof(int) * CHAR_BIT - 1)))  );
    }
     
  2. shabbir

    shabbir Administrator Staff Member

    Joined:
    Jul 12, 2004
    Messages:
    15,375
    Likes Received:
    388
    Trophy Points:
    83
    Looks like there is some problem in your code where you have not defined num
     
  3. oogabooga

    oogabooga New Member

    Joined:
    Jan 9, 2008
    Messages:
    115
    Likes Received:
    11
    Trophy Points:
    0
    That's pretty neat, the perhaps not universally portable.
    Your code does better than just avoiding relational and logical operators,
    it also avoids conditional (and loop) statements (including ?: ).
    Here's a re-write:
    Code:
    #include <limits.h> /* for CHAR_BIT: number of bits in a char */
    
    #define INT_BITS (sizeof(int) * CHAR_BIT)
    
    int maximum( int x, int y ) {
        int xy[] = { x, y };
        return xy[ getSign( x - y )]; /* Index xy by sign of (x-y) */
    }
    
    int getSign( int n ) { /* Returns sign bit of n */
        /* Shift sign into bit position 0 */
        return (unsigned)n >> INT_BITS - 1;
    }
    
    One could replace int xy[] = { x, y } by int* xy = &x, but
    that assumes (minimally) that x has a lower address than y.
    The method is already making an assumption about manipulating
    the sign bit that may not work on all computers.

    I also found your MinimumOf and MaximumOf interesting:
    Code:
        /* Fill n with its sign bit, becoming all 1's or 0's */
    #define spread_sign(n) ((n) >> (INT_BITS - 1))
    
        /* Return zero if n is positive, otherwise return n */
    #define zero_if_positive(n) ((n) & spread_sign(n))
    
    int minimumOf( int a, int b ) {
        return b + zero_if_positive( a - b );
    }
    
    int maximumOf( int a, int b ) {
        return a - zero_if_positive( a - b );
    }
    
     
  4. asadullah.ansari

    asadullah.ansari TechCake

    Joined:
    Jan 9, 2008
    Messages:
    356
    Likes Received:
    14
    Trophy Points:
    0
    Occupation:
    Developer
    Location:
    NOIDA

    Very Very Sorry for this mistake.

    correct code will be



    Code: C
    Code:
    int maximum(int x, int y)
     {   
       int fNum[2],temp,num[2]; 
       num[0] = x;  
       num[1] = y;    
      temp=fNum[ Fun4Sign(x - y)];   
      return( temp); 
    }
    
     int Fun4Sign(int sgn) 
    {   
         int temp;   
         temp= unsigned(sgn) >>  (sizeof(int)*8-1);    return(temp); 
    }
     
    Last edited by a moderator: Jan 14, 2008
  5. shabbir

    shabbir Administrator Staff Member

    Joined:
    Jul 12, 2004
    Messages:
    15,375
    Likes Received:
    388
    Trophy Points:
    83
    I have updated the above article but learn to use the code block in the posts - [thread=168]Before you make a query[/thread]
     

Share This Page

  1. This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
    By continuing to use this site, you are consenting to our use of cookies.
    Dismiss Notice