我有一個案例，其中switch語句包含大約40個每個返回基於輸入的不同配置對象的情況。這種方法在度量標準中顯示爲具有太高的圈複雜度，通常我會將其更改爲處理程序對象的映射。但是這個開關位於一塊代碼中，性能是非常重要的，所以我提出了HashMap查找和處理程序調用與開關塊執行性能比較的問題。任何人都比較過呢？值得考慮嗎？還是有更快的查找對象有限數量的int鍵？java switch cyclomatic複雜性與性能
乾杯， 凱
似乎搜索的值比開關方法稍快。 因此，如果您的代碼更改經常關注散列方法，那麼如果它經過了打磨並且不會經常更改，則可以使用不帶偏轉的切換方法。如果你的應用程序正在由幾個程序員開發，你應該更喜歡prevent future bugs的散列方法。
切換版本：27.32191395 ns。哈希版本：26.98444367 ns。
用散列獲得的時間：1.23％。
我用下面的代碼測試了它：
進口java.util.Random中;
/**
* @author ruslan.lopez
*/
public class CyclomaticVsHash {
private static final Random RNG = new Random();
static int[] myHash = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40 };
/**
* @param args
* the command line arguments
*/
public static void main(String[] args) {
long iterations = 100000000;
warmUp(iterations);
System.out.println("Cycle1");
double individualTime = getAverageTimePerIterationc1(iterations);
iterations = 10000;
double totalTime = getTotalTimec1(iterations);
System.out.println("ns/iteration: " + individualTime);
System.out.println("Total time for " + iterations + " runs: "
+ totalTime);
System.out.println("Cycle2");
iterations = 100000000;
warmUp(iterations);
double individualTime1 = getAverageTimePerIterationc2(iterations);
iterations = 10000;
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() {
Integer num = RNG.nextInt();
int newnum;
switch (num) {
case 1:
newnum = num * 1;
break;
case 2:
newnum = num * 2;
break;
case 3:
newnum = num * 3;
break;
case 4:
newnum = num * 4;
break;
case 5:
newnum = num * 5;
break;
case 6:
newnum = num * 6;
break;
case 7:
newnum = num * 7;
break;
case 8:
newnum = num * 8;
break;
case 9:
newnum = num * 9;
break;
case 10:
newnum = num * 10;
break;
case 11:
newnum = num * 11;
break;
case 12:
newnum = num * 12;
break;
case 13:
newnum = num * 13;
break;
case 14:
newnum = num * 14;
break;
case 15:
newnum = num * 15;
break;
case 16:
newnum = num * 16;
break;
case 17:
newnum = num * 17;
break;
case 18:
newnum = num * 18;
break;
case 19:
newnum = num * 19;
break;
case 20:
newnum = num * 20;
break;
case 21:
newnum = num * 21;
break;
case 22:
newnum = num * 22;
break;
case 23:
newnum = num * 23;
break;
case 24:
newnum = num * 24;
break;
case 25:
newnum = num * 25;
break;
case 26:
newnum = num * 26;
break;
case 27:
newnum = num * 7;
break;
case 28:
newnum = num * 28;
break;
case 29:
newnum = num * 29;
break;
case 30:
newnum = num * 30;
break;
case 31:
newnum = num * 31;
break;
case 32:
newnum = num * 32;
break;
case 33:
newnum = num * 33;
break;
case 34:
newnum = num * 34;
break;
case 35:
newnum = num * 35;
break;
case 36:
newnum = num * 36;
break;
case 37:
newnum = num * 37;
break;
case 38:
newnum = num * 38;
break;
case 39:
newnum = num * 39;
break;
default:
newnum = num * 40;
break;
}
}
private static void runCycles1() {
Integer num = RNG.nextInt();
int nwenum = num > 0 && num < 39
? myHash[num - 1]
: myHash[39];
}
}
我敢肯定，一個switch的內部表示使用某種類型的查找表，將有類似結構的HashMap，所以我的猜測是，任何差別可以忽略不計的。
我會說在這裏不值得考慮性能，只要用哪個解決方案就可以給出最乾淨的代碼。
試試看，並測量其性能。 – 2013-05-16 20:12:11