为什么这段代码在java中比在C++和C#中更快？

Question

我正在做一个简单的家庭作业,我必须在C中开发一个软件,找到100个之间的两个最近的点.

当我完成时,我很想知道需要花多少时间才能运行它并获得更多积分并启用完整的VC++优化.我尝试了10000,花了大约8~9秒.然后我很想知道C#和Java花多少时间做同样的事情.正如预期的那样,C#需要更长的时间,9~10秒; 然而,Java只花了大约400毫秒!为什么会这样？!

这是我在C,C#和Java中的代码:

C:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <Windows.h>

long perfFrequency = 0;

typedef struct
{
    double X;
    double Y;
} Point;

double distance(Point p1, Point p2)
{
    return sqrt(pow(p1.X - p2.X, 2) + pow(p1.Y - p2.Y, 2));
}

double smallerDistance(Point *points, int size, Point *smallerA, Point  *smallerB)
{
    int i, j;
    double smaller = distance(points[0], points[1]);

    for (i = 0; i < size; i++)
    {
        for (j = i + 1; j < size; j++)
        {
            double dist = distance(points[i], points[j]);
            if (dist < smaller)
            {
                smaller= dist;
                *smallerA = points[i];
                *smallerB = points[j];
            }
        }
    }
    return smaller;
}

void main()
{
    // read size and points from file.
    int size;
    Point *points= (Point *)malloc(size * sizeof(Point));

    // just to make sure everything is ready before the benchmark begins    
    system("pause");

    Point smallerA, smallerB;
    if (!QueryPerformanceFrequency((LARGE_INTEGER *)&perfFrequency))
        printf("Couldn't query performance frequency.");

    long long start, end;   
    double smaller;
    QueryPerformanceCounter((LARGE_INTEGER *)&start);

    smaller= smallerDistance(points, size, &smallerA, &smallerB);

    QueryPerformanceCounter((LARGE_INTEGER *)&end);

    printf("The smaller distance is: %lf. The coordinates of the most close points are: (%lf, %lf) and (%lf, %lf). Time taken: %lfms\n",
        smaller, smallerA.X, smallerA.Y, smallerB.X, smallerB.Y, (end - start) * 1000.0 / perfFrequency);

}

C#:

using System;
using System.IO;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Diagnostics;

namespace StructuredTest
{
    struct Point
    {
        public double X;
        public double Y;
    }

    class Program
    {
        static double Distance(Point p1, Point p2)
        {
            return Math.Sqrt(Math.Pow(p1.X - p2.X, 2) + Math.Pow(p1.Y - p2.Y, 2));
        }

        static double SmallerDistance(Point[] points, int size, out Point smallerA, out Point smallerB)
        {
            int i, j;
            double smaller = Distance(points[0], points[1]);
            smallerA = default(Point);
            smallerB = default(Point);

            for (i = 0; i < size; i++)
            {
                for (j = i + 1; j < size; j++)
                {
                    double dist = Distance(points[i], points[j]);
                    if (dist < smaller)
                    {
                        smaller = dist;
                        smallerA = points[i];
                        smallerB = points[j];
                    }
                }
            }

            return smaller;
        }

        static void Main(string[] args)
        {
            // read size and points from file 
            int size = int.Parse(file[0]);
            Point[] points= new Point[size];                   

            // make sure everything is ready
            Console.WriteLine("Press any key to continue...");
            Console.ReadKey(true);

            Point smallerA, smallerB;
            double smaller;

            Stopwatch sw = new Stopwatch();
            sw.Restart();

            smaller = SmallerDistance(points, size, out smallerA, out smallerB);

            sw.Stop();

            Console.WriteLine($"The smaller distance is: {smaller}. The coordinates of the most close points are: ({smallerA.X}, {smallerA.Y}) and " +
                $"({smallerB.X}, {smallerB.Y}). Time taken: {sw.ElapsedMilliseconds}ms.");

        }
    }
}

Java的:

package structuredtest;

import java.io.IOException;
import java.nio.charset.Charset;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.List;

class Point {

    public Point(double X, double Y) {
        this.X = X;
        this.Y = Y;
    }

    double X;
    double Y;
}

class Result {

    double distance;
    Point p1;
    Point p2;
}

public class StructuredTest {

    static double distance(Point p1, Point p2) {
        return Math.sqrt(Math.pow(p1.X - p2.X, 2) + Math.pow(p1.Y - p2.Y, 2));
    }

    static Result smallerDistance(Point[] points, int size) {
        int i, j;
        double smaller = distance(points[0], points[1]);
        Result r = new Result();

        for (i = 0; i < size; i++) {
            for (j = i + 1; j < size; j++) {
                double dist = distance(points[i], points[j]);
                if (dist < smaller) {
                    smaller = dist;
                    r.p1 = points[i];
                    r.p2 = points[j];
                }
            }
        }

        r.distance = smaller;
        return r;
    }

    public static void main(String[] args) throws IOException {
        // read size and points from file
        int size = Integer.parseInt(file[0]);
        Point[] points = new Point[size];

        // make sure everything is ready    
        System.out.println("Press any key to continue...");
        System.in.read();

        double start = System.nanoTime(), end;

        Result r = smallerDistance(points, size);

        end = System.nanoTime();

        System.out.println("The smaller distance is: " + r.distance + ". The most close points are: ("
                + r.p1.X + "," + r.p1.Y + ") and " + r.p2.X + "," + r.p2.Y + "). Time taken: " + (end - start) / 1000000 + "ms.");

    }

}

如果java以小幅度击败C和C#我不会感到惊讶,但速度快20倍？!

该文件采用以下格式:

3 // number of points in the file. Note that there no comments in the actual file
(3.7098722472288, 4.49056397953787) // point (X,Y)
(8.90232811621332, 9.67982769279173)
(5.68254334818822, 1.71918922506136)
(6.22585901842366, 9.51660500242835)

有趣的是:首先,我之前提到的10000点的文件,我用来进行基准测试,实际上只是100个随机点的另一个文件的100倍复制粘贴.像这样:

(Point 1)
(Point 2)
(Point 3)
(Point 1)
(Point 2)
(Point 3)

我认为不需要生成10000个随机点,因为因为代码必须遍历所有数字,所以它几乎没有区别(只有更多的赋值).但后来我决定生成10000个随机点来看看它们会如何反应:C和C#在大约相同的时间内仍然运行(增加约50ms); 另一方面,Java增加了~500ms.

此外,我认为值得注意的是,在NetBeans中运行时,java大约需要11秒(即使在"运行"模式下,而不是"调试").

我也尝试编译为C++而不是C,但它没有任何区别.

我正在使用VS 2015进行C和C#.

这些是每种语言的设置:

C:

x64
Optimization: Maximize Speed (/O2)
Intrinsic Functions: Yes (/Oi)
Favor Size or Speed: Favor fast code (/Ot)
Enable Fiber-Safe Optimizations: Yes (/GT)
Security Check: Disable Security Check (/GS-)
Floating point model: Fast (/fp:fast)
Everything else: default

C#:

x64
Release Mode
Optimize Code: Enabled
Check for arithmetic overflow: Disabled
.NET 4.5.2 

Java的:

JRE/JDK 1.8
Default settings (if there are any)

编辑:

好的,我按照建议重新做了测试:

首先,我Result在C和C#中使用了class/struct.我在java中使用它而不是在C/C#中使用它的原因是因为java无法通过引用传递.其次,我现在在main()函数中重复测试.并感谢@Tony D捕获该错误!:)

我不会发布代码,因为更改很小:只需在其他测试中完全实现java版本,这就是我所做的.

这次我只测试了7000点(不是10000点),只进行了30次迭代测试,因为它需要很长时间才能测试,而且它在这里很晚.

结果没有太大变化:C#平均需要5228ms,C 4424ms和Java 223ms.Java仍然以20倍或更快的速度获胜.

然后我尝试删除对Math.Pow的调用(只需更改((p1.X - p2.X) * (p1.X - p2.X)) + ((p1.Y - p2.Y) * (p1.Y - p2.Y))),然后一切都改变了.新结果:

Java:平均220ms

C#:平均195ms

C:平均195ms

如果我之前只检查过:p

正如我评论的那样,我考虑过这样做,但后来决定测试每个编译器内联函数和优化这种简单调用的能力会更好.然而,当我获得那些奇怪的结果时,我应该回去做这件事,但我很紧张,以至于我忘记了这样做.

无论如何,说实话,我很惊讶Java编译器能够完全优化那行代码,而C#和C++却没有.虽然我知道角落案例检查和C#上的内部调用,但我发现Java编译器能够注意到在该代码中不需要任何角落案例检查真的很有趣.

Answer 1

正如解释这里并检查在这里,在纯C有没有超载与整数次幂,像这样的:

double pow(double base, int exponent );

这意味着当您pow在C中调用时,它的处理方式类似于:

double pow(double base, double exponent) {
    return exp(log(base) * exponent);
}

还应该检查负基数和整数幂的情况,这是以特殊方式处理的.这是不加条件是一个好主意if (exponent == 1.0),并if (exponent == 2.0)在这里,因为它会减慢,真正使用数学代码pow为正当的用途.结果,平方在十二次或类似的事情中变慢.

原则上,优化了唯一合理的途径pow(x, 2)来x * x就是让编译器识别这样的反模式,并为他们产生特殊的代码.它恰好发生了,因此您的C和C#编译器无法执行此操作,而Java编译器可以执行此操作.请注意,它与内联功能无关:只是内联exp(log(a) * b)不会使此代码更快.

作为结论,我想指出,没有编写好数学代码的程序员会pow(x, 2)在他的代码中写入.