如果有的话,下面两个循环之间的性能差异是什么?

for (Object o: objectArrayList) {
    o.DoSomething();
}

and

for (int i=0; i<objectArrayList.size(); i++) {
    objectArrayList.get(i).DoSomething();
}

当前回答

所有这些循环都是一样的,我只是想在发表我的观点之前展示一下。

首先,循环List的经典方法:

for (int i=0; i < strings.size(); i++) { /* do something using strings.get(i) */ }

其次,这是首选的方法,因为它更不容易出错(你有多少次做过“哎呀,在这些循环中循环中混合变量i和j”的事情?)

for (String s : strings) { /* do something using s */ }

第三,微优化for循环:

int size = strings.size();
for (int i = -1; ++i < size;) { /* do something using strings.get(i) */ }

现在真正的两美分:至少当我测试这些时,第三个是最快的,当计算每种类型的循环所花费的毫秒数时,其中一个简单的操作重复了数百万次——这是在Windows上使用Java 5和jre1.6u10,如果有人感兴趣的话。

While it at least seems to be so that the third one is the fastest, you really should ask yourself if you want to take the risk of implementing this peephole optimization everywhere in your looping code since from what I've seen, actual looping isn't usually the most time consuming part of any real program (or maybe I'm just working on the wrong field, who knows). And also like I mentioned in the pretext for the Java for-each loop (some refer to it as Iterator loop and others as for-in loop) you are less likely to hit that one particular stupid bug when using it. And before debating how this even can even be faster than the other ones, remember that javac doesn't optimize bytecode at all (well, nearly at all anyway), it just compiles it.

如果你喜欢微观优化,或者你的软件使用了很多递归循环,那么你可能会对第三种循环感兴趣。只需要记住,在更改for循环之前和之后,都要对软件进行良好的基准测试。

其他回答

for-each循环通常是首选的。如果您使用的List实现不支持随机访问,那么“get”方法可能会慢一些。例如,如果使用LinkedList,则会产生遍历代价,而For -each方法使用迭代器跟踪其在列表中的位置。关于for-each循环的细微差别的更多信息。

我想文章现在在这里:新的位置

这里显示的链接已经失效。

所有这些循环都是一样的,我只是想在发表我的观点之前展示一下。

首先,循环List的经典方法:

for (int i=0; i < strings.size(); i++) { /* do something using strings.get(i) */ }

其次,这是首选的方法,因为它更不容易出错(你有多少次做过“哎呀,在这些循环中循环中混合变量i和j”的事情?)

for (String s : strings) { /* do something using s */ }

第三,微优化for循环:

int size = strings.size();
for (int i = -1; ++i < size;) { /* do something using strings.get(i) */ }

现在真正的两美分:至少当我测试这些时,第三个是最快的,当计算每种类型的循环所花费的毫秒数时,其中一个简单的操作重复了数百万次——这是在Windows上使用Java 5和jre1.6u10,如果有人感兴趣的话。

While it at least seems to be so that the third one is the fastest, you really should ask yourself if you want to take the risk of implementing this peephole optimization everywhere in your looping code since from what I've seen, actual looping isn't usually the most time consuming part of any real program (or maybe I'm just working on the wrong field, who knows). And also like I mentioned in the pretext for the Java for-each loop (some refer to it as Iterator loop and others as for-in loop) you are less likely to hit that one particular stupid bug when using it. And before debating how this even can even be faster than the other ones, remember that javac doesn't optimize bytecode at all (well, nearly at all anyway), it just compiles it.

如果你喜欢微观优化,或者你的软件使用了很多递归循环,那么你可能会对第三种循环感兴趣。只需要记住,在更改for循环之前和之后,都要对软件进行良好的基准测试。

1. for(Object o: objectArrayList){
    o.DoSomthing();
}
and

2. for(int i=0; i<objectArrayList.size(); i++){
    objectArrayList.get(i).DoSomthing();
}

两者都做同样的事情,但为了方便和安全的编程使用,在第二种使用方式中有容易出错的可能性。

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Ubuntu 12.10 alpha。

一般来说,HotSpot优化了大量的间接操作和简单的冗余操作,所以一般情况下,您不必担心它们,除非它们有很多顺序或嵌套严重。

另一方面,LinkedList上的索引get比LinkedList上的next On迭代器要慢得多,所以当你使用迭代器(显式或隐式地在for-each循环中)时,你可以避免性能损失,同时保持可读性。

Accepted answer回答了这个问题,除了ArrayList…

因为大多数开发人员都依赖于ArrayList(至少我是这么认为的)

所以我有义务在这里加上正确答案。

直接从开发人员文档:-

增强的for循环(有时也称为“for-each”循环)可用于实现Iterable接口的集合和数组。对于集合,会分配一个迭代器来对hasNext()和next()进行接口调用。使用ArrayList,手工编写的计数循环大约快3倍(有或没有JIT),但对于其他集合,增强的for循环语法将完全等同于显式迭代器的使用。

有几种迭代数组的方法:

static class Foo {
    int mSplat;
}

Foo[] mArray = ...

public void zero() {
    int sum = 0;
    for (int i = 0; i < mArray.length; ++i) {
        sum += mArray[i].mSplat;
    }
}

public void one() {
    int sum = 0;
    Foo[] localArray = mArray;
    int len = localArray.length;

    for (int i = 0; i < len; ++i) {
        sum += localArray[i].mSplat;
    }
}

public void two() {
    int sum = 0;
    for (Foo a : mArray) {
        sum += a.mSplat;
    }
}

zero()是最慢的,因为JIT还不能优化掉每次循环迭代获取数组长度的成本。

一个()比较快。它将所有内容提取到局部变量中,避免了查找。只有数组长度能带来性能上的好处。

two()对于没有JIT的设备来说是最快的,对于有JIT的设备来说与one()是没有区别的。它使用Java编程语言1.5版中引入的增强for循环语法。

因此,默认情况下您应该使用增强的for循环,但是可以考虑使用手写的计数循环来进行性能关键的ArrayList迭代。