如果你想测试这样的东西，你真的必须优化你的微基准来衡量你关心的东西.

If you want to test something like this, you really must optimize your microbenchmark to measure what you care about.

首先，使循环廉价但不可能跳过.计算总和通常可以解决问题.

First, make the loop inexpensive but impossible to skip. Computing a sum usually does the trick.

其次，比较两个时序.

这里有一些代码可以做到这两点:

Here's some code that does both:

import java.util.*;

public class Test {

public static long run1() {
  final List<String> myList = Arrays.asList("A", "B", "C", "D");
  final long start = System.nanoTime();
  int sum = 0;
  for (int i = 1000000000; i > myList.size(); i--) sum += i;
  final long stop = System.nanoTime();
  System.out.println("Finish: " + (stop - start)*1e-9 + " ns/op; sum = " + sum);
  return stop-start;
}

public static long run2() {
  final List<String> myList = Arrays.asList("A", "B", "C", "D");
  final long start = System.nanoTime();
  int sum = 0;
  int limit = myList.size();
  for (int i = 1000000000; i > limit; i--) sum += i;
  final long stop = System.nanoTime();
  System.out.println("Finish: " + (stop - start)*1e-9 + " ns/op; sum = " + sum);
  return stop-start;
}

public static void main(String[] args) {
  for (int i=0 ; i<5 ; i++) {
    long t1 = run1();
    long t2 = run2();
    System.out.println("  Speedup = " + (t1-t2)*1e-9 + " ns/op
");
  }
}

}

如果我们运行它，在我的系统上我们会得到:

And if we run it, on my system we get:

Finish: 0.481741256 ns/op; sum = -243309322
Finish: 0.40228402 ns/op; sum = -243309322
  Speedup = 0.079457236 ns/op

Finish: 0.450627151 ns/op; sum = -243309322
Finish: 0.43534661700000005 ns/op; sum = -243309322
  Speedup = 0.015280534 ns/op

Finish: 0.47738474700000005 ns/op; sum = -243309322
Finish: 0.403698331 ns/op; sum = -243309322
  Speedup = 0.073686416 ns/op

Finish: 0.47729349600000004 ns/op; sum = -243309322
Finish: 0.405540508 ns/op; sum = -243309322
  Speedup = 0.071752988 ns/op

Finish: 0.478979617 ns/op; sum = -243309322
Finish: 0.36067492700000003 ns/op; sum = -243309322
  Speedup = 0.11830469 ns/op

这意味着方法调用的开销大约为 0.1 ns.如果您的循环执行的时间不超过 1-2 ns，那么您应该关心这一点.否则，不要.

which means that the overhead of the method call is approximately 0.1 ns. If your loop does things that take no more than 1-2 ns, then you should care about this. Otherwise, don't.