Microsoft KB Archive/172338: Difference between revisions
m (Text replacement - "<" to "<") |
m (Text replacement - ">" to ">") |
||
Line 123: | Line 123: | ||
Count2 = GetTickCount() | Count2 = GetTickCount() | ||
Loops = Loops + 1 | Loops = Loops + 1 | ||
Loop Until Count1 < | Loop Until Count1 <> Count2 | ||
Debug.Print "GetTickCount minimum resolution: "; _ | Debug.Print "GetTickCount minimum resolution: "; _ | ||
(Count2 - Count1); "ms" | (Count2 - Count1); "ms" | ||
Line 136: | Line 136: | ||
Count2 = timeGetTime() | Count2 = timeGetTime() | ||
Loops = Loops + 1 | Loops = Loops + 1 | ||
Loop Until Count1 < | Loop Until Count1 <> Count2 | ||
Debug.Print "timeGetTime minimum resolution: "; _ | Debug.Print "timeGetTime minimum resolution: "; _ | ||
(Count2 - Count1); "ms" | (Count2 - Count1); "ms" |
Revision as of 20:47, 20 July 2020
Article ID: 172338
Article Last Modified on 1/20/2007
APPLIES TO
- Microsoft Excel 2000 Standard Edition
- Microsoft Visual Basic 5.0 Learning Edition
- Microsoft Visual Basic 6.0 Learning Edition
- Microsoft Visual Basic 5.0 Professional Edition
- Microsoft Visual Basic 6.0 Professional Edition
- Microsoft Visual Basic 5.0 Enterprise Edition
- Microsoft Visual Basic 6.0 Enterprise Edition
- Microsoft Visual Basic 5.0 Control Creation Edition
- Microsoft Visual Basic 4.0 Standard Edition
- Microsoft Visual Basic 4.0 Professional Edition
- Microsoft Visual Basic 4.0 32-Bit Enterprise Edition
- Microsoft Access 2002 Standard Edition
- Microsoft Access 2000 Standard Edition
- Microsoft Access 97 Standard Edition
- Microsoft Access 95 Standard Edition
- Microsoft Excel 2002 Standard Edition
- Microsoft Excel 97 Standard Edition
- Microsoft Excel 95 Standard Edition
- Microsoft Word 2002 Standard Edition
- Microsoft Word 2000 Standard Edition
- Microsoft Word 97 Standard Edition
This article was previously published under Q172338
SUMMARY
When timing code to identify performance bottlenecks, you want to use the highest resolution timer the system has to offer. This article describes how to use the QueryPerformanceCounter function to time application code.
MORE INFORMATION
Several timers of differing accuracy are offered by the operating system:
Function Units Resolution --------------------------------------------------------------------------- Now, Time, Timer seconds 1 second GetTickCount milliseconds approx. 10 ms TimeGetTime milliseconds approx. 10 ms QueryPerformanceCounter QueryPerformanceFrequency same
If your system supports a high-resolution counter, you can use QueryPerformanceCounter and QueryPerformanceFrequency to do high-resolution timings.
The following sample code compares the various counters:
WARNING: ANY USE BY YOU OF THE CODE PROVIDED IN THIS ARTICLE IS AT YOUR OWN RISK. Microsoft provides this code "as is" without warranty of any kind, either express or implied, including but not limited to the implied warranties of merchantability and/or fitness for a particular purpose.
Step-by-Step Procedures
Enter the following code into a Module. If you enter it into a class, form, or report module, make the declarations Private.
Option Explicit Declare Function QueryPerformanceCounter Lib "Kernel32" _ (X As Currency) As Boolean Declare Function QueryPerformanceFrequency Lib "Kernel32" _ (X As Currency) As Boolean Declare Function GetTickCount Lib "Kernel32" () As Long Declare Function timeGetTime Lib "winmm.dll" () As Long Sub Test_Timers() Dim Ctr1 As Currency, Ctr2 As Currency, Freq As Currency Dim Count1 As Long, Count2 As Long, Loops As Long ' ' Time QueryPerformanceCounter ' If QueryPerformanceCounter(Ctr1) Then QueryPerformanceCounter Ctr2 Debug.Print "Start Value: "; Format$(Ctr1, "0.0000") Debug.Print "End Value: "; Format$(Ctr2, "0.0000") QueryPerformanceFrequency Freq Debug.Print "QueryPerformanceCounter minimum resolution: 1/" & _ Freq * 10000; " sec" Debug.Print "API Overhead: "; (Ctr2 - Ctr1) / Freq; "seconds" Else Debug.Print "High-resolution counter not supported." End If ' ' Time GetTickCount ' Debug.Print Loops = 0 Count1 = GetTickCount() Do Count2 = GetTickCount() Loops = Loops + 1 Loop Until Count1 <> Count2 Debug.Print "GetTickCount minimum resolution: "; _ (Count2 - Count1); "ms" Debug.Print "Took"; Loops; "loops" ' ' Time timeGetTime ' Debug.Print Loops = 0 Count1 = timeGetTime() Do Count2 = timeGetTime() Loops = Loops + 1 Loop Until Count1 <> Count2 Debug.Print "timeGetTime minimum resolution: "; _ (Count2 - Count1); "ms" Debug.Print "Took"; Loops; "loops" End Sub
- Run the function from the Debug/Immediate window. Your output should appear similar to the following:
Start Value: 3516284.3498
End Value: 3516284.3521
QueryPerformanceCounter minimum resolution: 1/1193182 sec
API Overhead: 1.92761875388667E-05 seconds
GetTickCount minimum resolution: 10 ms
Took 650 loops
timeGetTime minimum resolution: 10 ms
Took 1565 loops
Multiple statements execute before either GetTickCount or timeGetTime record a change. The actual number of loops will vary depending on the background tasks the operating system is executing.
On the other hand, QueryPerformanceCounter changes value between successive API calls, indicating its usefulness in high-resolution timing. The resolution in this case is on the order of a microsecond. Because the resolution is system-dependent, there are no standard units that it measures. You have to divide the difference by the QueryPerformanceFrequency to determine the number of seconds elapsed. In the case above, the overhead for just calling the API is about 19 microseconds. This would have to be subtracted when timing other code as follows:
Private Sub Time_Addition() Dim Ctr1 As Currency, Ctr2 As Currency, Freq As Currency Dim Overhead As Currency, A As Long, I As Long QueryPerformanceFrequency Freq QueryPerformanceCounter Ctr1 QueryPerformanceCounter Ctr2 Overhead = Ctr2 - Ctr1 ' determine API overhead QueryPerformanceCounter Ctr1 ' time loop For I = 1 To 100 A = A + I Next I QueryPerformanceCounter Ctr2 Debug.Print "("; Ctr1; "-"; Ctr2; "-"; Overhead; ") /"; Freq Debug.Print "100 additions took"; Debug.Print (Ctr2 - Ctr1 - Overhead) / Freq; "seconds" End Sub
Sample output:
( 3630876.6256 - 3630876.6388 - 0.0013 ) / 119.3182
100 additions took 9.97333181358753E-05 seconds
NOTE: Because currency variables are used, the values returned are 10000 times smaller than the actual counters. Because the calculation of seconds involves a division operation, this factor is cancelled out.
REFERENCES
Microsoft Developer Network; topics: timeGetTime GetTickCount QueryPerformanceCounter QueryPerformanceFrequency
Keywords: kbhowto kbprogramming KB172338