Most efficient way to find the greatest of three ints

问题

Below is my pseudo code.

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

I think that works, but is that the most efficient way in C++?

回答1:

To find the greatest you need to look at exactly 3 ints, no more no less. You're looking at 6 with 3 compares. You should be able to do it in 3 and 2 compares.

int ret = max(i,j);
ret = max(ret, k);
return ret;

回答2:

Pseudocode:

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

回答3:

How about

return i > j? (i > k? i: k): (j > k? j: k);

two comparisons, no use of transient temporary stack variables...

回答4:

Your current method: http://ideone.com/JZEqZTlj (0.40s)

Chris's solution:

int ret = max(i,j);
ret = max(ret, k);
return ret;

http://ideone.com/hlnl7QZX (0.39s)

Solution by Ignacio Vazquez-Abrams:

result = i;
if (j > result)
  result = j;
if (k > result)
  result = k;
return result;

http://ideone.com/JKbtkgXi (0.40s)

And Charles Bretana's:

return i > j? (i > k? i: k): (j > k? j: k);

http://ideone.com/kyl0SpUZ (0.40s)

Of those tests, all the solutions take within 3% the amount of time to execute as the others. The code you are trying to optimize is extremely short as it is. Even if you're able to squeeze 1 instruction out of it, it's not likely to make a huge difference across the entirety of your program (modern compilers might catch that small optimization). Spend your time elsewhere.

EDIT: Updated the tests, turns out it was still optimizing parts of it out before. Hopefully it's not anymore.

回答5:

For a question like this, there is no substitute for knowing just what your optimizing compiler is doing and just what's available on the hardware. If the fundamental tool you have is binary comparison or binary max, two comparisons or max's are both necessary and sufficient.

I prefer Ignacio's solution:

result = i;
if (j > result)
  result = j;
if (k > result)
  result = k;
return result;

because on the common modern Intel hardware, the compiler will find it extremely easy to emit just two comparisons and two cmov instructions, which place a smaller load on the I-cache and less stress on the branch predictor than conditional branches. (Also, the code is clear and easy to read.) If you are using x86-64, the compiler will even keep everything in registers.

Note you are going to be hard pressed to embed this code into a program where your choice makes a difference...

回答6:

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.

回答7:

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);

回答8:

There is a proposal to include this into the C++ library under N2485. The proposal is simple, so I've included the meaningful code below. Obviously, this assumes variadic templates

template < typename T >
const T & max ( const T & a )
{ return a ; }

template < typename T , typename ... Args >
const T & max( const T & a , const T & b , const Args &... args )
{ return  max ( b > a ? b : a , args ...); }

回答9:

public int maximum(int a,int b,int c){
    int max = a;
    if(b>max)
        max = b;
    if(c>max)
        max = c;
    return max;
}

回答10:

I think by "most efficient" you are talking about performance, trying not to waste computing resources. But you could be referring to writing fewer lines of code or maybe about the readability of your source code. I am providing an example below, and you can evaluate if you find something useful or if you prefer another version from the answers you received.

/* Java version, whose syntax is very similar to C++. Call this program "LargestOfThreeNumbers.java" */
class LargestOfThreeNumbers{
    public static void main(String args[]){
        int x, y, z, largest;
        x = 1;
        y = 2;
        z = 3;
        largest = x;
        if(y > x){
            largest = y;
            if(z > y){
                largest = z;
            }
        }else if(z > x){
            largest = z;
        }
        System.out.println("The largest number is: " + largest);
    }
}

回答11:

#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;
}

回答12:

Greatest of 3 numbers

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

来源：https://stackoverflow.com/questions/2233166/most-efficient-way-to-find-the-greatest-of-three-ints

标签

c++

performance

max

min