Most efficient way to find the greatest of three ints

Question

Below is my pseudo code.

function highest(i, j, k)
{
  if(i > j && i > k)
  {
    return i;
  }
  else         


        

Answer 1

Pseudocode:

result = i
if j > result:
  result = j
if k > result:
  result = k
return result

Answer 2

Greatest of 3 numbers

int greater = a>b ? (a>c? a:c) :(b>c ? b:c); 
System.out.println(greater);

Answer 3

I like to eliminate conditional jumps as an intellectual exercise. Whether this has any measurable effect on performance I have no idea though :)

#include <iostream>
#include <limits>

inline int max(int a, int b)
{
    int difference = a - b;
    int b_greater = difference >> std::numeric_limits<int>::digits;
    return a - (difference & b_greater);
}

int max(int a, int b, int c)
{
    return max(max(a, b), c);
}

int main()
{
    std::cout << max(1, 2, 3) << std::endl;
    std::cout << max(1, 3, 2) << std::endl;
    std::cout << max(2, 1, 3) << std::endl;
    std::cout << max(2, 3, 1) << std::endl;
    std::cout << max(3, 1, 2) << std::endl;
    std::cout << max(3, 2, 1) << std::endl;
}

This bit twiddling is just for fun, the cmov solution is probably a lot faster.

Answer 4

The easiest way to find a maximum or minimum of 2 or more numbers in c++ is:-

int a = 3, b = 4, c = 5;
int maximum = max({a, b, c});

int a = 3, b = 4, c = 5;
int minimum = min({a, b, c});

You can give as many variables as you want.

Interestingly enough it is also incredibly efficient, at least as efficient as Ignacio Vazquez-Abrams'solution (https://godbolt.org/z/j1KM97):

        mov     eax, dword ptr [rsp + 8]
        mov     ecx, dword ptr [rsp + 4]
        cmp     eax, ecx
        cmovl   eax, ecx
        mov     ecx, dword ptr [rsp]
        cmp     eax, ecx
        cmovl   eax, ecx

Similar with GCC, while MSVC makes a mess with a loop.

Answer 5

Not sure if this is the most efficient or not, but it might be, and it's definitely shorter:

int maximum = max( max(i, j), k);

Answer 6

#include<stdio.h>
int main()
{
    int a,b,c,d,e;
    scanf("%d %d %d",&a,&b,&c);
    d=(a+b+abs(a-b))/2;
    e=(d+c+abs(c-d))/2;
    printf("%d is Max\n",e);
    return 0;
}