How do I programmatically return the max of two integers without using any comparison operators and without using if, else, etc?

Question

How do I programmatically return the maximum of two integers without using any comparison operators and without using if, else, etc?

Answer 1

max: // Will put MAX(a,b) into a

a -= b;
a &= (~a) >> 31;
a += b;

And:

int a, b;

min: // Will put MIN(a,b) into a

a -= b;
a &= a >> 31;
a += b;

from here.

Answer 2

From z0mbie's (famous virii writer) article "Polymorphic Games", maybe you'll find it useful:

#define H0(x)       (((signed)(x)) >> (sizeof((signed)(x))*8-1))
#define H1(a,b)     H0((a)-(b))

#define MIN1(a,b)   ((a)+(H1(b,a) & ((b)-(a))))
#define MIN2(a,b)   ((a)-(H1(b,a) & ((a)-(b))))
#define MIN3(a,b)   ((b)-(H1(a,b) & ((b)-(a))))
#define MIN4(a,b)   ((b)+(H1(a,b) & ((a)-(b))))
//#define MIN5(a,b)   ((a)<(b)?(a):(b))
//#define MIN6(a,b)   ((a)>(b)?(b):(a))
//#define MIN7(a,b)   ((b)>(a)?(a):(b))
//#define MIN8(a,b)   ((b)<(a)?(b):(a))

#define MAX1(a,b)   ((a)+(H1(a,b) & ((b)-(a))))
#define MAX2(a,b)   ((a)-(H1(a,b) & ((a)-(b))))
#define MAX3(a,b)   ((b)-(H1(b,a) & ((b)-(a))))
#define MAX4(a,b)   ((b)+(H1(b,a) & ((a)-(b))))
//#define MAX5(a,b)   ((a)<(b)?(b):(a))
//#define MAX6(a,b)   ((a)>(b)?(a):(b))
//#define MAX7(a,b)   ((b)>(a)?(b):(a))
//#define MAX8(a,b)   ((b)<(a)?(a):(b))

#define ABS1(a)     (((a)^H0(a))-H0(a))
//#define ABS2(a)     ((a)>0?(a):-(a))
//#define ABS3(a)     ((a)>=0?(a):-(a))
//#define ABS4(a)     ((a)<0?-(a):(a))
//#define ABS5(a)     ((a)<=0?-(a):(a))

cheers

Answer 3

int max(int a, int b)
{
   return ((a - b) >> 31) ? b : a;
}

Answer 4

This is kind of cheating, using assembly language, but it's interesting nonetheless:

// GCC inline assembly
int max(int a, int b)
{
  __asm__("movl %0, %%eax\n\t"   // %eax = a
          "cmpl %%eax, %1\n\t"   // compare a to b
          "cmovg %1, %%eax"      // %eax = b if b>a
         :: "r"(a), "r"(b));
}

If you want to be strict about the rules and say that the cmpl instruction is illegal for this, then the following (less efficient) sequence will work:

int max(int a, int b)
{
  __asm__("movl %0, %%eax\n\t"
      "subl %1, %%eax\n\t"
          "cmovge %0, %%eax\n\t"
          "cmovl %1, %%eax"
         :: "r"(a), "r"(b)
         :"%eax");
}

Answer 5

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

r = x - ((x - y) & -(x < y)); // max(x, y)

You can have fun with arithmetically shifting (x - y) to saturate the sign bit, but this is usually enough. Or you can test the high bit, always fun.

Answer 6

I think I've got it.

int data[2] = {a,b};
int c = a - b;
return data[(int)((c & 0x80000000) >> 31)];

Would this not work? Basically, you take the difference of the two, and then return one or the other based on the sign bit. (This is how the processor does greater than or less than anyway.) So if the sign bit is 0, return a, since a is greater than or equal to b. If the sign bit is 1, return b, because subtracting b from a caused the result to go negative, indicating that b was greater than a. Just make sure that your ints are 32bits signed.