java - 为什么单独的方法比内联相同的代码工作得更快

Question

例如在大数组中搜索最大值。单独方法中带有循环的代码比 main 中相同的内联代码运行得更快。这里运行 10 次以获得平均消耗时间:

  Max generated: 999999  Max is 999999  Time consumed main: 1.8210982999999998 ms  Max is 999999 Time consumed method: 0.6809502000000001 ms

Can someone please explain why such a difference? Thanks.

import java.util.Random;

public class Main {
    private static int[] data = new int[1000000];
    private final static int COUNT = 10;


    public static void main(String[] args) {
        initData(data);
        long consumedTime;
        long startTime = System.nanoTime();
        int max = 0;
        for (int x = 0; x < COUNT; x++) {
            max  = data[0];
            for (int i = 0; i < data.length; i++) {
                if (data[i] > max) {
                    max = data[i];
                }
            }
        }
        consumedTime =(System.nanoTime() - startTime);
        System.out.println("Max is "+max+ "  Time consumed main: " + (double) (consumedTime)/1000000/COUNT + " ms");


        startTime = System.nanoTime();
        for (int x = 0; x < COUNT; x++) {
            max = getMaxOnPart();
        }
        consumedTime=(System.nanoTime() - startTime);
        System.out.println("Max is "+max+" Time consumed method: " + (double) (consumedTime)/1000000/COUNT + " ms");
    }

    private static void initData(int[] data) {
        final Random random = new Random();
        int max = 0;
        for (int i = 0; i < data.length; i++) {
            data[i] = random.nextInt(1000000);
            if (max < data[i]) {
                max = data[i];
            }
        }
        System.out.println("  Max generated: " + max);
    }

    private static int getMaxOnPart() {
        int max = data[0];
        for (int i = 0; i < data.length; i++) {
            if (data[i] > max) {
                max = data[i];
            }
        }
        return max;
    }
}

Answer 1

您的测量不正确/不一致。在第一个循环中

long startTime = System.nanoTime();

在 10 次重复循环之外，因此

consumedTime+=(System.nanoTime() - startTime);

测量 1x + 2x + 3x + ... 会将结果夸大 55 倍(如果我的数学是正确的)。

而第二个调用的函数测量重置了 repeat x 10 循环内的 startTime:

 startTime = System.nanoTime();
 max = getMaxOnPart();
 consumedTime+=(System.nanoTime() - startTime);

这将仅对方法中花费的时间求和。