為什么在Java中進行循環內聲明似乎比在Java中進行循環外聲明更快？

Question

當我檢查此問題時，我不敢相信，因此我進行了測試，似乎是對的。 循環內聲明似乎比循環外聲明快。 有人可以解釋為什么嗎？

這是我的測試代碼：

public class CycleTest {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        long iterations = 1000000;
        warmUp(iterations);
        System.out.println("Cycle1");
        double individualTime = getAverageTimePerIterationc1(iterations);
        iterations = 1000;
        double totalTime = getTotalTimec1(iterations);

        System.out.println("ns/iteration: " + individualTime);
        System.out.println("Total time for " + iterations + " runs: " + totalTime);

        System.out.println("Cycle2");
        iterations = 1000000;
        double individualTime1 = getAverageTimePerIterationc2(iterations);
        iterations = 1000;
        double totalTime1 = getTotalTimec2(iterations);

        System.out.println("ns/iteration: " + individualTime1);
        System.out.println("Total time for " + iterations + " runs: " + totalTime1);

    }

    public static void warmUp(long iterations) {
        System.out.println("Starting warmup");
        for (int i = 0; i < iterations; i++) {
            runCycles();
            runCycles1();
        }
    }

    public static double getAverageTimePerIterationc1(long iterations) {
        // test
        System.out.println("Starting individual time test");
        long timeTaken = 0;
        for (int i = 0; i < iterations; i++) {
            long startTime = System.nanoTime();
            runCycles();
            timeTaken += System.nanoTime() - startTime;
        }
        return (double) timeTaken / iterations;
    }

    public static long getTotalTimec1(long iterations) {
        // test
        System.out.println("Starting total time test");
        long timeTaken = 0;
        for (int i = 0; i < iterations; i++) {
            long startTime = System.nanoTime();
            runCycles();
            timeTaken += System.nanoTime() - startTime;
        }
        return timeTaken;
    }

    public static double getAverageTimePerIterationc2(long iterations) {
        // test
        System.out.println("Starting individual time test");
        long timeTaken = 0;
        for (int i = 0; i < iterations; i++) {
            long startTime = System.nanoTime();
            runCycles1();
            timeTaken += System.nanoTime() - startTime;
        }
        return (double) timeTaken / iterations;
    }

    public static long getTotalTimec2(long iterations) {
        // test
        System.out.println("Starting total time test");
        long timeTaken = 0;
        for (int i = 0; i < iterations; i++) {
            long startTime = System.nanoTime();
            runCycles1();
            timeTaken += System.nanoTime() - startTime;
        }
        return timeTaken;
    }

    private static void runCycles() {
        double intermediateResult;
        for (int i = 0; i < 1000; i++) {
            intermediateResult = i;
            intermediateResult += 1;
        }
    }

    private static void runCycles1() {
        for (int i = 0; i < 1000; i++) {
            double intermediateResult = i;
            intermediateResult += 1;
        }
    }
}

額外的新信息：我已經在Windows計算機上的JDK 1.6.0_27上運行了它。

Answer 1

它們生成相同的代碼。 這兩種方法：

private static void runCycles1() {
    double intermediateResult;
    for (int i = 0; i < 1000; i++) {
        intermediateResult = i;
    }
}
private static void runCycles2() {
    for (int i = 0; i < 1000; i++) {
        double intermediateResult = i;
    }
}

在Java 8（jdk1.8.0_51）中生成以下字節碼：

runCycles1()                runCycles2()
Code:                       Code:
   0: iconst_0                 0: iconst_0                i = 0
   1: istore_2                 1: istore_0
   2: goto          11         2: goto          11        goto 11

   5: iload_2                  5: iload_0                 intermediateResult = (double)i
   6: i2d                      6: i2d
   7: dstore_0                 7: dstore_1
   8: iinc          2, 1       8: iinc          0, 1      i++

  11: iload_2                 11: iload_0                 if (i < 1000) goto 5
  12: sipush        1000      12: sipush        1000
  15: if_icmplt     5         15: if_icmplt     5
  18: return                  18: return                  return

實際的聲明不會生成任何代碼，因此如果您發現性能有所不同，我會感到驚訝。

如此測試：

for (int j = 0; j < 10; j++) {
    long t1 = System.nanoTime();
    for (int i = 0; i < 1000_000_000; i++)
        runCycles1();
    long t2 = System.nanoTime();
    for (int i = 0; i < 1000_000_000; i++)
        runCycles2();
    long t3 = System.nanoTime();
    System.out.printf("%d  %d%n", t2 - t1, t3 - t2);
}

並得到：

4250095  2020120
4067898  0
3904236  0
0  0
0  0
0  0
0  0
0  0
0  0
0  0

他們最終都編譯成空！！！