此代碼繪製從上面兩個垂直條紋到窗口的底部在兩個線程(對應函數爲每個線程是thread_func
)。第一個線程繪製左側條紋的一部分,然後第二個繪製右側的一部分,然後再繪製第一個線條,依此類推。一個信號量和一個關鍵部分用於確保這個順序。不同程序行爲
#include <windows.h>
#include <cstdint>
HDC hDC;
HDC hDCMem;
HBITMAP hbitmap;
HWND hwnd;
int ScreenMaxX;
int ScreenMaxY;
short pattern[8]={~0xFF, ~0xFF, ~0xFF, ~0xFF, ~0xFF, ~0xFF, ~0xFF, ~0xFF};
HBRUSH brush=::CreatePatternBrush(::CreateBitmap(8, 8, 1, 1, pattern));
void bar(int nLeft, int nTop, int nRight, int nBottom)
{
RECT rect;
rect.left = nLeft;
rect.right = nRight;
rect.top = nTop;
rect.bottom = nBottom;
::SetTextColor(hDCMem, 0xFF00FF);
::SetBkColor(hDCMem, 0xFF00FF);
//brush=::CreatePatternBrush(::CreateBitmap(8, 8, 1, 1, pattern));
::FillRect(hDCMem, &rect, brush);
}
void flush(){
::BitBlt(hDC, 0, 0, ScreenMaxX, ScreenMaxY, hDCMem, 0, 0, SRCCOPY);
}
CRITICAL_SECTION graphics_cs;
uint8_t thread_cnt=0;
uint8_t total_threads=2;
HANDLE turnstile1=CreateSemaphoreW(nullptr, 0, 2, nullptr);
void thread_func(int num){
int x,y;
if(num==0){
x=20; y=0;
} else {
x=110; y=0;
}
while(true) {
while(true) {
EnterCriticalSection(&graphics_cs);
if (thread_cnt == num) {
thread_cnt++;
bar(x, y, x+40, y+40);
y+=1;
//flush();
if(thread_cnt==total_threads){
thread_cnt = 0;
flush();
ReleaseSemaphore(turnstile1, total_threads, nullptr);
}
LeaveCriticalSection(&graphics_cs);
break;
} else {
LeaveCriticalSection(&graphics_cs);
}
}
WaitForSingleObject(turnstile1, INFINITE);
Sleep(100);
}
}
void mainx()
{
InitializeCriticalSection(&graphics_cs);
for(int i=0; i<total_threads; i++){
CreateThread (nullptr, 0, (LPTHREAD_START_ROUTINE)thread_func, (LPVOID)i, 0, nullptr);
}
}
DWORD Th(LPVOID param)
{
(void)param;
::SetWindowPos(hwnd, HWND_TOP,
10,
10,
400,
500,
SWP_SHOWWINDOW
);
mainx();
flush();
return 0;
}
DWORD g_nMainThreadID;
//processing main window messages
long FAR PASCAL WindowProc(HWND hWnd,UINT message, WPARAM wParam,LPARAM lParam)
{
switch (message)
{
case WM_PAINT: flush();
break;
case WM_DESTROY: PostQuitMessage(0);
break;
}
return DefWindowProc(hWnd, message, wParam, lParam);
}
int PASCAL WinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPSTR lpCmdLine, int nShowCmd)
{
(void)hPrevInstance, (void)lpCmdLine;
WNDCLASS wc;
MSG msg;
wc.style = CS_HREDRAW | CS_VREDRAW;
wc.lpfnWndProc = WindowProc;
wc.cbClsExtra = 0;
wc.cbWndExtra = 0;
wc.hInstance = hInstance;
wc.hIcon = NULL;
wc.hCursor = LoadCursor(NULL, IDC_ARROW);
wc.hbrBackground = (HBRUSH) GetStockObject(WHITE_BRUSH);
wc.lpszMenuName = "Menu_one";
wc.lpszClassName = "NAME";
if (!RegisterClass(&wc)) {return 0; };
//main window
hwnd = CreateWindow("NAME",
"!",
WS_OVERLAPPEDWINDOW,
CW_USEDEFAULT,
CW_USEDEFAULT,
CW_USEDEFAULT,
CW_USEDEFAULT,
HWND_DESKTOP,
NULL,
hInstance,
NULL
);
ScreenMaxX = ::GetSystemMetrics(SM_CXSCREEN);
ScreenMaxY = ::GetSystemMetrics(SM_CYSCREEN);
hDC = ::GetDC(hwnd);
hDCMem = ::CreateCompatibleDC(hDC);
hbitmap = ::CreateCompatibleBitmap(hDC, ScreenMaxX, ScreenMaxY);
::SelectObject(hDCMem, hbitmap);
auto hbrush = (HBRUSH)::GetStockObject(WHITE_BRUSH);
::SelectObject(hDCMem, hbrush);
::PatBlt(hDCMem, 0,0, ScreenMaxX, ScreenMaxY, PATCOPY);
::DeleteObject(hbrush);
CreateThread(NULL,0,(LPTHREAD_START_ROUTINE)Th, (LPVOID)hwnd, 0,&g_nMainThreadID);
ShowWindow(hwnd, nShowCmd);
UpdateWindow(hwnd);
while (GetMessage(&msg, NULL, 0, 0))
{
TranslateMessage(&msg);
DispatchMessage(&msg);
}
return msg.wParam;
}
在Windows 7中,條紋以與預期相同的速度繪製。但是,在Windows XP中的速度是不同的:
如果我取消或者//brush=::CreatePatternBrush(::CreateBitmap(8, 8, 1, 1, pattern));
或//flush();
線,在Windows XP中繪製的速度將是相同的。爲什麼這可以解決問題,爲什麼不同版本的Windows中初始代碼的行爲有所不同?
更新
當我後thread_func
bar
和flush
通話添加std::cout<<"num = "<<num<<" : bar call\n";
和std::cout<<"num = "<<num<<" : flush call\n";
,輸出
num = 0 : bar call
num = 1 : bar call
num = 1 : flush call
num = 0 : bar call
num = 1 : bar call
num = 1 : flush call
num = 0 : bar call
num = 1 : bar call
num = 1 : flush call
num = 0 : bar call
num = 1 : bar call
num = 1 : flush call
...
的順序似乎是正確的,但左邊的條紋是不是後立即被吸引同花順呼叫。
如果線程沒有得到愛處理器的相同金額,然後一會落後。如果一個人總是運行在一個超線程的核心上,那就是Bummer。測量實際耗用的時間來計算繪製多少,不要只假設40. GetTickCount()可以完成任務。現在既沒有調度,也沒有核心性能,也沒有實際的Sleep()的數量,也沒有什麼刷你創造的問題了。 –
@HansPassant如果一個線程的優先級更高,那麼它將等待'turnstile1'上的其他線程。我嘗試在'flush'調用並協調更改後向控制檯輸出消息,看起來正確,但屏幕上的圖像沒有反映出這一點。我甚至在單核CPU上運行這個程序。 –
是的,你可以刪除旋轉門代碼,少一件事要調試。 –