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。
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
对于java 8+,另一种可能的解决方案(更通用,函数风格,没有方面)可能是创建一些实用程序方法,将代码作为参数接受
public static <T> T timed (String description, Consumer<String> out, Supplier<T> code) {
final LocalDateTime start = LocalDateTime.now ();
T res = code.get ();
final long execTime = Duration.between (start, LocalDateTime.now ()).toMillis ();
out.accept (String.format ("%s: %d ms", description, execTime));
return res;
}
调用代码可以是这样的smth:
public static void main (String[] args) throws InterruptedException {
timed ("Simple example", System.out::println, Timing::myCode);
}
public static Object myCode () {
try {
Thread.sleep (1500);
} catch (InterruptedException e) {
e.printStackTrace ();
}
return null;
}
其他回答
使用分析器(JProfiler, Netbeans profiler, Visual VM, Eclipse profiler等)。您将得到最准确的结果,并且是最少的干扰。它们使用内置的JVM机制进行概要分析,还可以为您提供额外的信息,如堆栈跟踪、执行路径以及必要时更全面的结果。
当使用完全集成的分析器时,对方法进行分析是非常简单的。右击,分析器->添加到根方法。然后像运行测试运行或调试器一样运行剖析器。
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。
对于java 8+,另一种可能的解决方案(更通用,函数风格,没有方面)可能是创建一些实用程序方法,将代码作为参数接受
public static <T> T timed (String description, Consumer<String> out, Supplier<T> code) {
final LocalDateTime start = LocalDateTime.now ();
T res = code.get ();
final long execTime = Duration.between (start, LocalDateTime.now ()).toMillis ();
out.accept (String.format ("%s: %d ms", description, execTime));
return res;
}
调用代码可以是这样的smth:
public static void main (String[] args) throws InterruptedException {
timed ("Simple example", System.out::println, Timing::myCode);
}
public static Object myCode () {
try {
Thread.sleep (1500);
} catch (InterruptedException e) {
e.printStackTrace ();
}
return null;
}
我的答案很简单。对我有用。
long startTime = System.currentTimeMillis();
doReallyLongThing();
long endTime = System.currentTimeMillis();
System.out.println("That took " + (endTime - startTime) + " milliseconds");
它运行得很好。分辨率显然只有毫秒级,使用System.nanoTime()可以做得更好。这两种方法都有一些限制(操作系统调度切片等),但效果很好。
平均几次运行(越多越好),你就会得到一个不错的想法。
我已经编写了一个方法,以易于阅读的形式打印方法执行时间。 例如,要计算100万的阶乘,大约需要9分钟。因此,执行时间打印为:
Execution Time: 9 Minutes, 36 Seconds, 237 MicroSeconds, 806193 NanoSeconds
代码在这里:
public class series
{
public static void main(String[] args)
{
long startTime = System.nanoTime();
long n = 10_00_000;
printFactorial(n);
long endTime = System.nanoTime();
printExecutionTime(startTime, endTime);
}
public static void printExecutionTime(long startTime, long endTime)
{
long time_ns = endTime - startTime;
long time_ms = TimeUnit.NANOSECONDS.toMillis(time_ns);
long time_sec = TimeUnit.NANOSECONDS.toSeconds(time_ns);
long time_min = TimeUnit.NANOSECONDS.toMinutes(time_ns);
long time_hour = TimeUnit.NANOSECONDS.toHours(time_ns);
System.out.print("\nExecution Time: ");
if(time_hour > 0)
System.out.print(time_hour + " Hours, ");
if(time_min > 0)
System.out.print(time_min % 60 + " Minutes, ");
if(time_sec > 0)
System.out.print(time_sec % 60 + " Seconds, ");
if(time_ms > 0)
System.out.print(time_ms % 1E+3 + " MicroSeconds, ");
if(time_ns > 0)
System.out.print(time_ns % 1E+6 + " NanoSeconds");
}
}