Comparing two integers without any comparison

Question

Is it possible to find the greatest of two integers without any comparison? I found some solutions:

if(!(a/b)) // if a is less than b then division result          


        

Answer 1

char c;
c=0x3D + (!(b/a) && (a-b)) - (!(a/b) && (a-b));
printf("a %c b",c);

Answer 2

(!(a/b) ?  cout << " b is greater than a" : (!(b-a) ? cout << "a and b are equal" :  cout << "a is greater than b") :  cout << "a is greater than b");

That gets a bit messy though

Edit: Is this homework?

Answer 3

Try this, tested it, works well.

public static int compare(int a, int b)
{
    int c = a - b;
    return (c >> 31) & 1 ^ 1;
}

Answer 4

To get the greatest number without using comparison/relational operator

void PrintGreatestNumber(int a, int b)
{
   int [] x = new int[] { -1, 0, 1 };
   int greatestNumber =  ((a+b)+ x[ 1 + ((a-b) >> 31) - (-(a-b) >> 31)] * (a-b)) /2;  
   Console.WriteLine(greatestNumber);
}

Answer 5

Subtract them and check the sign using nasty bit twiddling hacks
http://graphics.stanford.edu/~seander/bithacks.html

Don't do this in production code if the other programmers know where you live.

Answer 6

I think this method is better than others, you can use this logic c and java both programming languages but int should be of 4 byte if int is of 2 byte then make 15 byte right shift instead of 31 byte.

enter code here

#include<stdio.h>

main()
{
   int a, b;
   printf("Enter three numbers\n");
   scanf("%d %d", &a, &b);
   printf("Largest number is %d \n",findMax( a,b ));
}
int findMax( int x, int y)
{
  int z = x - y;
  int i  = (z  >>  31)  &  0x1;
  printf("i = %d shift = %d \n", i, (z>>31));
  int  max  =  x - i  *  z;
  return max;
}