下面两个循环之间的性能差异是什么?
for (Object o: objectArrayList) {
o.DoSomething();
}
和
for (int i=0; i<objectArrayList.size(); i++) {
objectArrayList.get(i).DoSomething();
}
发布于 2008-11-02 13:07:58
通常应该首选for-each循环。如果您使用的列表实现不支持随机访问,则"get“方法可能会较慢。例如,如果使用LinkedList,则会产生遍历开销,而for-each方法使用迭代器来跟踪它在列表中的位置。有关nuances of the for-each loop的更多信息。
我想文章现在已经在这里了:new location
这里显示的链接是死的。
发布于 2014-09-27 04:57:58
不幸的是,这似乎是有区别的。
如果您查看为这两种循环生成的字节代码,您会发现它们是不同的。
下面是Log4j源代码中的一个示例。
在/log4j-api/src/main/java/org/apache/logging/log4j/MarkerManager.java中,我们有一个名为Log4jMarker的静态内部类,它定义:
/*
* Called from add while synchronized.
*/
private static boolean contains(final Marker parent, final Marker... localParents) {
//noinspection ForLoopReplaceableByForEach
for (final Marker marker : localParents) {
if (marker == parent) {
return true;
}
}
return false;
}
使用标准循环:
private static boolean contains(org.apache.logging.log4j.Marker, org.apache.logging.log4j.Marker...);
Code:
0: iconst_0
1: istore_2
2: aload_1
3: arraylength
4: istore_3
5: iload_2
6: iload_3
7: if_icmpge 29
10: aload_1
11: iload_2
12: aaload
13: astore 4
15: aload 4
17: aload_0
18: if_acmpne 23
21: iconst_1
22: ireturn
23: iinc 2, 1
26: goto 5
29: iconst_0
30: ireturn
使用for-each:
private static boolean contains(org.apache.logging.log4j.Marker, org.apache.logging.log4j.Marker...);
Code:
0: aload_1
1: astore_2
2: aload_2
3: arraylength
4: istore_3
5: iconst_0
6: istore 4
8: iload 4
10: iload_3
11: if_icmpge 34
14: aload_2
15: iload 4
17: aaload
18: astore 5
20: aload 5
22: aload_0
23: if_acmpne 28
26: iconst_1
27: ireturn
28: iinc 4, 1
31: goto 8
34: iconst_0
35: ireturn
那个先知是怎么回事?
我已经在Windows7上的Java7和8上尝试过了。
发布于 2012-08-29 05:53:42
以下代码:
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.List;
interface Function<T> {
long perform(T parameter, long x);
}
class MyArray<T> {
T[] array;
long x;
public MyArray(int size, Class<T> type, long x) {
array = (T[]) Array.newInstance(type, size);
this.x = x;
}
public void forEach(Function<T> function) {
for (T element : array) {
x = function.perform(element, x);
}
}
}
class Compute {
int factor;
final long constant;
public Compute(int factor, long constant) {
this.factor = factor;
this.constant = constant;
}
public long compute(long parameter, long x) {
return x * factor + parameter + constant;
}
}
public class Main {
public static void main(String[] args) {
List<Long> numbers = new ArrayList<Long>(50000000);
for (int i = 0; i < 50000000; i++) {
numbers.add(i * i + 5L);
}
long x = 234553523525L;
long time = System.currentTimeMillis();
for (int i = 0; i < numbers.size(); i++) {
x += x * 7 + numbers.get(i) + 3;
}
System.out.println(System.currentTimeMillis() - time);
System.out.println(x);
x = 0;
time = System.currentTimeMillis();
for (long i : numbers) {
x += x * 7 + i + 3;
}
System.out.println(System.currentTimeMillis() - time);
System.out.println(x);
x = 0;
numbers = null;
MyArray<Long> myArray = new MyArray<Long>(50000000, Long.class, 234553523525L);
for (int i = 0; i < 50000000; i++) {
myArray.array[i] = i * i + 3L;
}
time = System.currentTimeMillis();
myArray.forEach(new Function<Long>() {
public long perform(Long parameter, long x) {
return x * 8 + parameter + 5L;
}
});
System.out.println(System.currentTimeMillis() - time);
System.out.println(myArray.x);
myArray = null;
myArray = new MyArray<Long>(50000000, Long.class, 234553523525L);
for (int i = 0; i < 50000000; i++) {
myArray.array[i] = i * i + 3L;
}
time = System.currentTimeMillis();
myArray.forEach(new Function<Long>() {
public long perform(Long parameter, long x) {
return new Compute(8, 5).compute(parameter, x);
}
});
System.out.println(System.currentTimeMillis() - time);
System.out.println(myArray.x);
}
}
在我的系统上给出以下输出:
224
-699150247503735895
221
-699150247503735895
220
-699150247503735895
219
-699150247503735895
我运行的是带有OracleJDK 1.7更新6的Ubuntu12.10alpha。
一般来说,HotSpot优化了很多间接和简单的冗余操作,所以通常你不应该担心它们,除非它们在序列中有很多或者它们是大量嵌套的。
另一方面,在LinkedList上索引的get比在LinkedList的迭代器上调用next慢得多,因此在使用迭代器(显式或隐式地在for-each循环中)时,可以避免这种性能影响,同时保持可读性。
https://stackoverflow.com/questions/256859
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