May 14th, 1999, 09:00 PM
Can someone help me figure out how to use TimeGetTime(); ?? Please.
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Click to See Complete Forum and Search --> : timegettime May 14th, 1999, 09:00 PM Can someone help me figure out how to use TimeGetTime(); ?? Please. Jaeyeon Lee May 14th, 1999, 10:05 PM You can use timeGetTime to get the duration between execution points as follows. DWORD start = timeGetTime(); ...... Do Some Job ...... DWORD end = timeGetTime(); DWORD ellapsed = end - start; // in milisecond unit arf May 17th, 1999, 02:16 PM // CStopWatch.h // #ifndef CSTOPWATCH_H #define CSTOPWATCH_H #include <windows.h> #include <mmsystem.h> ////////////////////////////////////////////////////////////////////////////// // CStopWatch class CStopWatch { public: // // construction/destruction // CStopWatch(); CStopWatch(int resolution); // // operators // void Start(); // start stopwatch float Stop(); // stop stopwatch, return = number of seconds private: int Resolution; // set timer resolution in milliseconds (only effective on WinNT) int StartTime; // start time }; #endif // !#define CSTOPWATCH_H // CStopWatch.cpp // #include "stdafx.h" #include "CStopWatch.h" #if _MSC_VER >= 1000 // // include the multimedia library for linking // #pragma comment( lib, "winmm" ) #pragma once #endif // _MSC_VER >= 1000 CStopWatch :: CStopWatch() { Resolution = 1; StartTime = 0; } CStopWatch :: CStopWatch(int res) { Resolution = res; StartTime = 0; } void CStopWatch :: Start() { #ifdef _WIN32 StartTime = timeBeginPeriod(Resolution); if (StartTime != 0) throw(0,"Unable to set requested resolution","CStopWatch"); StartTime = timeGetTime(); #endif } float CStopWatch :: Stop() { float RT = 0.0; #ifdef _WIN32 RT = (timeGetTime() - StartTime)/1000.0F; StartTime = timeEndPeriod(Resolution); #endif return RT; } Michel Wassink May 20th, 1999, 03:45 AM Add winmm.lib to linker settings if necessary ,,,^..^,,, Michel May 21st, 1999, 09:06 PM Note I would like this to work under MFC if at all possible. I would like it to function as a timer for a simple 2D animation. (I need more help understanding) codeguru.com
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