怎样才能最准确地看到某个东西(例如一个方法调用)在代码中花费了多长时间?
我猜最简单和最快的方法是:
DateTime start = DateTime.Now;
{
// Do some work
}
TimeSpan timeItTook = DateTime.Now - start;
但这有多精确呢?有没有更好的办法?
怎样才能最准确地看到某个东西(例如一个方法调用)在代码中花费了多长时间?
我猜最简单和最快的方法是:
DateTime start = DateTime.Now;
{
// Do some work
}
TimeSpan timeItTook = DateTime.Now - start;
但这有多精确呢?有没有更好的办法?
当前回答
更好的方法是使用Stopwatch类:
using System.Diagnostics;
// ...
Stopwatch sw = new Stopwatch();
sw.Start();
// ...
sw.Stop();
Console.WriteLine("Elapsed={0}",sw.Elapsed);
其他回答
正如其他人所说,秒表是一个很好的课程使用这里。你可以用一个有用的方法来包装它:
public static TimeSpan Time(Action action)
{
Stopwatch stopwatch = Stopwatch.StartNew();
action();
stopwatch.Stop();
return stopwatch.Elapsed;
}
(注意Stopwatch.StartNew()的使用。我更喜欢这样创建一个秒表,然后调用Start()在简单性方面。)显然,这会引起调用委托的冲击,但在绝大多数情况下,这是无关紧要的。你可以这样写:
TimeSpan time = StopwatchUtil.Time(() =>
{
// Do some work
});
你甚至可以为此制作一个ITimer接口,在可用的地方使用StopwatchTimer, CpuTimer等实现。
更好的方法是使用Stopwatch类:
using System.Diagnostics;
// ...
Stopwatch sw = new Stopwatch();
sw.Start();
// ...
sw.Stop();
Console.WriteLine("Elapsed={0}",sw.Elapsed);
正如其他人所说,秒表应该是正确的工具。虽然它可以有一些改进,具体看这个线程:在c#中测试小代码样本,这个实现可以改进吗?
我在这里看到了Thomas Maierhofer的一些有用的建议
他的代码大致如下:
//prevent the JIT Compiler from optimizing Fkt calls away
long seed = Environment.TickCount;
//use the second Core/Processor for the test
Process.GetCurrentProcess().ProcessorAffinity = new IntPtr(2);
//prevent "Normal" Processes from interrupting Threads
Process.GetCurrentProcess().PriorityClass = ProcessPriorityClass.High;
//prevent "Normal" Threads from interrupting this thread
Thread.CurrentThread.Priority = ThreadPriority.Highest;
//warm up
method();
var stopwatch = new Stopwatch()
for (int i = 0; i < repetitions; i++)
{
stopwatch.Reset();
stopwatch.Start();
for (int j = 0; j < iterations; j++)
method();
stopwatch.Stop();
print stopwatch.Elapsed.TotalMilliseconds;
}
另一种方法是依赖Process。TotalProcessTime用于度量CPU一直忙于运行特定代码/进程的时间,如下所示。这可以反映更真实的场景,因为没有其他进程影响度量。它是这样做的:
var start = Process.GetCurrentProcess().TotalProcessorTime;
method();
var stop = Process.GetCurrentProcess().TotalProcessorTime;
print (end - begin).TotalMilliseconds;
在这里可以找到相同内容的详细实现。
我写了一个助手类,以一种易于使用的方式来执行这两项:
public class Clock
{
interface IStopwatch
{
bool IsRunning { get; }
TimeSpan Elapsed { get; }
void Start();
void Stop();
void Reset();
}
class TimeWatch : IStopwatch
{
Stopwatch stopwatch = new Stopwatch();
public TimeSpan Elapsed
{
get { return stopwatch.Elapsed; }
}
public bool IsRunning
{
get { return stopwatch.IsRunning; }
}
public TimeWatch()
{
if (!Stopwatch.IsHighResolution)
throw new NotSupportedException("Your hardware doesn't support high resolution counter");
//prevent the JIT Compiler from optimizing Fkt calls away
long seed = Environment.TickCount;
//use the second Core/Processor for the test
Process.GetCurrentProcess().ProcessorAffinity = new IntPtr(2);
//prevent "Normal" Processes from interrupting Threads
Process.GetCurrentProcess().PriorityClass = ProcessPriorityClass.High;
//prevent "Normal" Threads from interrupting this thread
Thread.CurrentThread.Priority = ThreadPriority.Highest;
}
public void Start()
{
stopwatch.Start();
}
public void Stop()
{
stopwatch.Stop();
}
public void Reset()
{
stopwatch.Reset();
}
}
class CpuWatch : IStopwatch
{
TimeSpan startTime;
TimeSpan endTime;
bool isRunning;
public TimeSpan Elapsed
{
get
{
if (IsRunning)
throw new NotImplementedException("Getting elapsed span while watch is running is not implemented");
return endTime - startTime;
}
}
public bool IsRunning
{
get { return isRunning; }
}
public void Start()
{
startTime = Process.GetCurrentProcess().TotalProcessorTime;
isRunning = true;
}
public void Stop()
{
endTime = Process.GetCurrentProcess().TotalProcessorTime;
isRunning = false;
}
public void Reset()
{
startTime = TimeSpan.Zero;
endTime = TimeSpan.Zero;
}
}
public static void BenchmarkTime(Action action, int iterations = 10000)
{
Benchmark<TimeWatch>(action, iterations);
}
static void Benchmark<T>(Action action, int iterations) where T : IStopwatch, new()
{
//clean Garbage
GC.Collect();
//wait for the finalizer queue to empty
GC.WaitForPendingFinalizers();
//clean Garbage
GC.Collect();
//warm up
action();
var stopwatch = new T();
var timings = new double[5];
for (int i = 0; i < timings.Length; i++)
{
stopwatch.Reset();
stopwatch.Start();
for (int j = 0; j < iterations; j++)
action();
stopwatch.Stop();
timings[i] = stopwatch.Elapsed.TotalMilliseconds;
print timings[i];
}
print "normalized mean: " + timings.NormalizedMean().ToString();
}
public static void BenchmarkCpu(Action action, int iterations = 10000)
{
Benchmark<CpuWatch>(action, iterations);
}
}
就叫
Clock.BenchmarkTime(() =>
{
//code
}, 10000000);
or
Clock.BenchmarkCpu(() =>
{
//code
}, 10000000);
The last part of the Clock is the tricky part. If you want to display the final timing, its up to you to choose what sort of timing you want. I wrote an extension method NormalizedMean which gives you the mean of the read timings discarding the noise. I mean I calculate the the deviation of each timing from the actual mean, and then I discard the values which was farer (only the slower ones) from the mean of deviation (called absolute deviation; note that its not the often heard standard deviation), and finally return the mean of remaining values. This means, for instance, if timed values are { 1, 2, 3, 2, 100 } (in ms or whatever), it discards 100, and returns the mean of { 1, 2, 3, 2 } which is 2. Or if timings are { 240, 220, 200, 220, 220, 270 }, it discards 270, and returns the mean of { 240, 220, 200, 220, 220 } which is 220.
public static double NormalizedMean(this ICollection<double> values)
{
if (values.Count == 0)
return double.NaN;
var deviations = values.Deviations().ToArray();
var meanDeviation = deviations.Sum(t => Math.Abs(t.Item2)) / values.Count;
return deviations.Where(t => t.Item2 > 0 || Math.Abs(t.Item2) <= meanDeviation).Average(t => t.Item1);
}
public static IEnumerable<Tuple<double, double>> Deviations(this ICollection<double> values)
{
if (values.Count == 0)
yield break;
var avg = values.Average();
foreach (var d in values)
yield return Tuple.Create(d, avg - d);
}
stopwatch就是为这个任务设计的。
秒表是好的,但循环工作10^6次,然后除以10^6。 你会得到更精确的结果。