有关C＃中DateTime.Now的问题

Question

DateTime startTime = System.DateTime.Now;

//do the search
Graph_SearchBFS BFS = new Graph_SearchBFS(Graph, sourceCell, targetCell);

DateTime endTime = System.DateTime.Now;

How come startTime equals endTime? What does that mean? By the way my goal is to measure how much time it takes for the graph search.

Thanks

Answer 1

It means that the system clock granularity is too large to measure this sort of thing accurately - in other words, calling the constructor happens pretty quickly.

However, you can use a class which is more appropriate for precision timing - or at least, it can take advantage of a high-precision clock if one is available: Stopwatch .

Stopwatch sw = Stopwatch.StartNew();
Graph_SearchBFS BFS = new Graph_SearchBFS(Graph, sourceCell, targetCell);
sw.Stop();

Now you can look at the sw.Elapsed property to see how long it took.

Of course, it's entirely possible that it will still be too fast to measure. Typically for benchmarking you do something enough that you can time over the course of many seconds, rather than a single operation.

As an aside, is your Graph_SearchBFS constructor actually doing the work? If so, that's a bit of a design smell as well. Typically I'd expect code more like this:

Graph_SearchBFS BFS = new Graph_SearchBFS(Graph);
Stopwatch sw = Stopwatch.StartNew();
BFS.FindCell(sourceCell, targetCell);
sw.Stop();

or something similar. Then you can loop lots of times over the "finding" part:

Graph_SearchBFS BFS = new Graph_SearchBFS(Graph);
Stopwatch sw = Stopwatch.StartNew();
for (int i = 0; i < Iterations; i++)
{
    BFS.FindCell(sourceCell, targetCell);
}
sw.Stop();

Answer 2

Probably because your method ran faster than the resolution of the clock.

Use the StopWatch class instead; it's more precise.

Answer 3

Use Stopwatch to measure execution speed

  Stopwatch sw = new Stopwatch();
  sw.Start();

  Graph_SearchBFS BFS = new Graph_SearchBFS(Graph, sourceCell, targetCell);

  sw.Stop();
  Debug.WriteLine("Time to query: " + sw.ElapsedMilliseconds + " ms");

Answer 4

Could be that the method is very fast. Have you looked at the milliseconds or ticks?

http://msdn.microsoft.com/en-us/library/system.datetime.millisecond.aspx

http://msdn.microsoft.com/en-us/library/system.datetime.ticks.aspx