如何获得方法的执行时间? 是否有Timer实用程序类来计时任务所需的时间等?

在谷歌上的大多数搜索都返回调度线程和任务的计时器的结果,这不是我想要的。


当前回答

使用AOP/AspectJ和来自jcabi-aspects的@Loggable注释,你可以简单而紧凑地完成:

@Loggable(Loggable.DEBUG)
public String getSomeResult() {
  // return some value
}

对该方法的每个调用都将发送到SLF4J日志记录工具,具有DEBUG日志记录级别。每个日志消息都将包括执行时间。

其他回答

在Java 8中引入了一个名为Instant的新类。根据文件:

Instant represents the start of a nanosecond on the time line. This class is useful for generating a time stamp to represent machine time. The range of an instant requires the storage of a number larger than a long. To achieve this, the class stores a long representing epoch-seconds and an int representing nanosecond-of-second, which will always be between 0 and 999,999,999. The epoch-seconds are measured from the standard Java epoch of 1970-01-01T00:00:00Z where instants after the epoch have positive values, and earlier instants have negative values. For both the epoch-second and nanosecond parts, a larger value is always later on the time-line than a smaller value.

这可以用于:

Instant start = Instant.now();
try {
    Thread.sleep(7000);
} catch (InterruptedException e) {
    e.printStackTrace();
}
Instant end = Instant.now();
System.out.println(Duration.between(start, end));

打印pt7.001。

加油,伙计们!没有人提到用番石榴来做这件事(可以说是很棒):

import com.google.common.base.Stopwatch;

Stopwatch timer = Stopwatch.createStarted();
//method invocation
LOG.info("Method took: " + timer.stop());

Stopwatch.toString()很好地为测量选择了时间单位。也就是说,如果值很小,它将输出38ns,如果值很长,它将显示5m 3s

甚至更好的:

Stopwatch timer = Stopwatch.createUnstarted();
for (...) {
   timer.start();
   methodToTrackTimeFor();
   timer.stop();
   methodNotToTrackTimeFor();
}
LOG.info("Method took: " + timer);

注意:谷歌Guava需要Java 1.6+

new Timer(""){{
    // code to time 
}}.timeMe();



public class Timer {

    private final String timerName;
    private long started;

    public Timer(String timerName) {
        this.timerName = timerName;
        this.started = System.currentTimeMillis();
    }

    public void timeMe() {
        System.out.println(
        String.format("Execution of '%s' takes %dms.", 
                timerName, 
                started-System.currentTimeMillis()));
    }

}

在我的机器上进行性能测量

System.nanoTime(): 750ns System.currentTimeMillis(): 18ns

如前所述,System.nanoTime()被认为是度量经过的时间。只要注意在循环中使用的代价就可以了。

我修改了代码,从正确答案得到的结果在几秒钟内:

long startTime = System.nanoTime();

methodCode ...

long endTime = System.nanoTime();
double duration = (double)(endTime - startTime) / (Math.pow(10, 9));
Log.v(TAG, "MethodName time (s) = " + duration);