如果您不使用工具，并且希望对执行时间较短的方法进行计时，那么只需执行多次，每次将执行次数增加一倍，直到达到1秒左右。因此，系统调用的时间。纳米时间等，也没有系统的准确性。nanoTime确实对结果有很大影响。

    int runs = 0, runsPerRound = 10;
    long begin = System.nanoTime(), end;
    do {
        for (int i=0; i<runsPerRound; ++i) timedMethod();
        end = System.nanoTime();
        runs += runsPerRound;
        runsPerRound *= 2;
    } while (runs < Integer.MAX_VALUE / 2 && 1000000000L > end - begin);
    System.out.println("Time for timedMethod() is " + 
        0.000000001 * (end-begin) / runs + " seconds");

当然，使用挂钟也有一些注意事项:jit编译、多线程/进程等的影响。因此，您需要首先执行该方法很多次，以便JIT编译器完成它的工作，然后重复此测试多次，并使用最短的执行时间。

总有一些过时的方法:

long startTime = System.nanoTime();
methodToTime();
long endTime = System.nanoTime();

long duration = (endTime - startTime);  //divide by 1000000 to get milliseconds.

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java ee中对我有效的策略是:

Create a class with a method annotated with @AroundInvoke; @Singleton public class TimedInterceptor implements Serializable { @AroundInvoke public Object logMethod(InvocationContext ic) throws Exception { Date start = new Date(); Object result = ic.proceed(); Date end = new Date(); System.out.println("time: " + (end.getTime - start.getTime())); return result; } } Annotate the method that you want to monitoring: @Interceptors(TimedInterceptor.class) public void onMessage(final Message message) { ...

我希望这能有所帮助。

非常好的代码。

http://www.rgagnon.com/javadetails/java-0585.html

import java.util.concurrent.TimeUnit;

long startTime = System.currentTimeMillis();
........
........
........
long finishTime = System.currentTimeMillis();

String diff = millisToShortDHMS(finishTime - startTime);


  /**
   * converts time (in milliseconds) to human-readable format
   *  "<dd:>hh:mm:ss"
   */
  public static String millisToShortDHMS(long duration) {
    String res = "";
    long days  = TimeUnit.MILLISECONDS.toDays(duration);
    long hours = TimeUnit.MILLISECONDS.toHours(duration)
                   - TimeUnit.DAYS.toHours(TimeUnit.MILLISECONDS.toDays(duration));
    long minutes = TimeUnit.MILLISECONDS.toMinutes(duration)
                     - TimeUnit.HOURS.toMinutes(TimeUnit.MILLISECONDS.toHours(duration));
    long seconds = TimeUnit.MILLISECONDS.toSeconds(duration)
                   - TimeUnit.MINUTES.toSeconds(TimeUnit.MILLISECONDS.toMinutes(duration));
    if (days == 0) {
      res = String.format("%02d:%02d:%02d", hours, minutes, seconds);
    }
    else {
      res = String.format("%dd%02d:%02d:%02d", days, hours, minutes, seconds);
    }
    return res;
  }

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);