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

我实现了一个简单的定时器,我认为它真的很有用:

public class Timer{
    private static long start_time;

    public static double tic(){
        return start_time = System.nanoTime();
    }

    public static double toc(){
        return (System.nanoTime()-start_time)/1000000000.0;
    }

}

这样你就可以计算一个或多个动作的时间:

Timer.tic();
// Code 1
System.out.println("Code 1 runtime: "+Timer.toc()+" seconds.");
// Code 2
System.out.println("(Code 1 + Code 2) runtime: "+Timer.toc()+"seconds");
Timer.tic();
// Code 3
System.out.println("Code 3 runtime: "+Timer.toc()+" seconds.");

其他回答

如果你想要时间

long start_time = System.currentTimeMillis();
object.method();
long end_time = System.currentTimeMillis();
long execution_time = end_time - start_time;

使用Java 8的新API中的即时和持续时间,

Instant start = Instant.now();
Thread.sleep(5000);
Instant end = Instant.now();
System.out.println(Duration.between(start, end));

输出,

PT5S

也可以实现定时器接口,并在类的任何方法上执行

import java.util.function.*;

public interface Timer {

    default void timeIt(Runnable r) {
        timeIt(() -> { r.run(); return 0;});
    }

    default <S,T> T timeIt(Function<S,T> fun, S arg) {
        long start = System.nanoTime();
        T result = fun.apply(arg);
        long stop = System.nanoTime();
        System.out.println("Time: " + (stop-start)/1000000.0 + " msec");
        return result;
    }

    default <T> T timeIt(Supplier<T> s) {
        return timeIt(obj -> s.get(), null);
    }
}

用法:

class MyClass implements Timer ..

timeIt(this::myFunction); 

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。

这里有很多有效的答案,它们都是在方法中实现的。为了制作一个通用的计时方法,我通常有一个timing类,它由以下内容组成。

public record TimedResult<T>(T result, Duration duration) {}

public static Duration time(Runnable r) {
    var s = Instant.now();
    r.run();
    var dur = Duration.between(s, Instant.now());
    return dur;
}

public static <T> TimedResult<T> time(Callable<T> r) throws Exception {
    var s = Instant.now();
    T res = r.call();
    var dur = Duration.between(s, Instant.now());
    return new TimedResult<>(res, dur);
}

这足够通用,可以传递Runnable或Callable对象。

Duration result = Timing.time(() -> {
    // do some work.
});

TimedResult<String> result = Timing.time(() -> {
    // do some work.
    return "answer";
});

Duration timeTaken = result.duration();
String answer = result.result();