2014-02-09 33 views
1

Groovy是否可能在我的应用程序中每隔一段时间都没有自动复选框?Groovy在原语上非常慢

下面的代码...

public class SumTest { 
    public static void main(String[] args) { 
     long[] longs = new long[100000000]; 

     long nanoStart = System.nanoTime(); 
     long counter = 1; 
     for (int i = 0; i < longs.length; i++) { 
      longs[i] = counter++; 
     } 

     double msPop = (System.nanoTime() - nanoStart)/1000000d; 
     System.out.println("Time taken, population: " + msPop + "ms."); 

     nanoStart = System.nanoTime(); 
     long sum = 0; 
     for (int i = 0; i < longs.length; i++) { 
      sum += longs[i]; 
     } 
     double msSum = (System.nanoTime() - nanoStart)/1000000d; 
     System.out.println("Time taken, sum: " + msSum + "ms, total: " + (msPop + msSum) + "ms"); 
     System.out.println(" (sum: " + sum + ")"); 
    } 
} 

...展品截然不同的运行时间从时 '的.java' 更名为 '.groovy作为':

的Java:

Time taken, population: 94.793746ms. 
Time taken, sum: 65.172605ms, total: 159.966351ms 
(sum: 5000000050000000) 

Groovy:

Time taken, population: 2233.995965ms. 
Time taken, sum: 2203.64302ms, total: 4437.638985ms 
(sum: 5000000050000000) 

..这是一个约30倍的差异。

的局势加剧当我坚持长远目标内(如在我真正的代码的情况下):

public class SumTest { 
    static class Holder { 
     long l; 
     Holder(long l) { this.l = l; } 
     long getL() { return l; } 
    } 

    public static void main(String[] args) { 
     Holder[] longs = new Holder[100000000]; 

     long nanoStart = System.nanoTime(); 
     long counter = 1; 
     for (int i = 0; i < longs.length; i++) { 
      longs[i] = new Holder(counter++); 
     } 

     double msPop = (System.nanoTime() - nanoStart)/1000000d; 
     System.out.println("Time taken, population: " + msPop + "ms."); 

     nanoStart = System.nanoTime(); 
     long sum = 0; 
     for (int i = 0; i < longs.length; i++) { 
      sum += longs[i].getL(); 
     } 
     double msSum = (System.nanoTime() - nanoStart)/1000000d; 
     System.out.println("Time taken, sum: " + msSum + "ms, total: " + (msPop + msSum) + "ms"); 
     System.out.println(" (sum: " + sum + ")"); 
    } 
} 

运行时间(请注意,我用-Xms16384M -Xmx16384M运行这里) :

的Java

Time taken, population: 1083.784927ms. 
Time taken, sum: 180.518991ms, total: 1264.3039179999998ms 
(sum: 5000000050000000) 

的Groovy:

Time taken, population: 9816.007447ms. 
Time taken, sum: 8685.506864ms, total: 18501.514311ms 
(sum: 5000000050000000) 

..其总数是~15倍,,但最重要的区别来自实际使用这些对象(由总和表示):〜50x。

这可以以某种方式固定吗?当涉及操作只有涉及原语和原始操作时,我可以哄Groovy不自动包含涉及原语的每个操作吗?

回答

2

(哇!真奇怪,彻底地制定一个问题几乎是立刻就导致一个答案?!)

修复:在类的顶部写@CompileStatic

Time taken, population: 1562.978726ms. 
Time taken, sum: 183.388353ms, total: 1746.367079ms 
(sum: 5000000050000000) 
+1

'@ CompileStatic'会做你想做的事,但它也做得更多。这意味着你完全绕过了groovy的所有动态特性。这在很多情况下可以带来很大的性能提升。虽然依赖于动态方法和属性的语言有很多功能,并且不能用于@CompileStatic。只是FYI。 – allTwentyQuestions