[RESOLVED] wait function

[killo]

if i use this wait function in my program:

Private Declare Function GetTickCount Lib "kernel32" () As Long

Public Sub Wait(Time As Long)
  Dim StartTime As Long
  
  StartTime = GetTickCount()
  
  While (StartTime + Time) > GetTickCount()
    DoEvents
  Wend
End Sub

i find that there is no difference between
wait(10)
and
wait(1)

i need my program to do this because i have a loop which i want to run at the same speed even on slow comps so i have programmed it so that it times how long it takes for the comp to complete the instructions in one instance of the loop and then it waits for 10 milliseconds minus the time it took to complete the instructions, for example if the instructions took 4 milliseconds, it would wait 6 milliseconds before looping round again, so the loop always takes 10 milliseconds. (except on rediculously slow comps where the instructions take more than 10 milliseconds, which isnt likely) but this isnt working because wait(10) and wait(1), wait for the same amount of time!
how can i get a more accurate wait function?

[Milk]

Both the native vb Timer function and the GetTickcount API function have a limited resolution of 0.015625 (1/64) seconds i.e. they can only measure time in multiples of that resolution. QueryPerformanceCounter combined with QueryPerformanceFrequency can be used to measure very small increments of time.

[killo]

sorry I dont know what QueryPerformanceCounter and QueryPerformanceFrequency are, what are they and how would i use them to measure a small amount of time?

[Milk]

Option Explicit
Private Declare Function QueryPerformanceFrequency Lib "kernel32" (lpFrequency As Currency) As Long
Private Declare Function QueryPerformanceCounter Lib "kernel32" (lpPerformanceCount As Currency) As Long
Dim cTFreq As Currency
Dim cTStart As Currency
Dim cTEnd As Currency
 
 
Private Sub Form_Load()
  Dim T As Single
  QueryPerformanceFrequency cTFreq 'Generally the frequency of your processor
  
  QueryPerformanceCounter cTStart 'Generally the number of cycles of your processor
  Wait 0.001
  QueryPerformanceCounter cTEnd
  T = (cTEnd - cTStart) / cTFreq  'cycles / frequency = time 
  Debug.Print "And that took " & Round(T, 4) & " seconds"
End Sub
 
Public Sub Wait(ByVal sSeconds As Single)
  Dim T1 As Currency, T2 As Currency
  QueryPerformanceCounter T1
  T1 = T1 + sSeconds * cTFreq
  Do
    DoEvents 'Comment out for more accuracy, leave in if waiting for long periods
    QueryPerformanceCounter T2
  Loop Until T2 => T1
End Sub

[killo]

thanks alot, thats just what i needed